### 3 Algorithm implementation based on original Kovacic 1985 paper

The following describes the algorithm for each case separately. The easiest one is for case 1, and the hardest is for case 3. Many examples will also be given at the end of the algorithm describing to show how it works.

#### 3.1 Case one algorithm

3.1.1 Step 1
3.1.2 Step 2
3.1.3 Step 3

##### 3.1.1 Step 1

This description is based on KOVACIC 1985 paper and not based on the Saunders paper.

We are given $$y^{\prime \prime }=ry$$. It is assumed that the necessary conditions for case 1 have been met as given in the table above and $$r=\frac {s}{t}$$ where $$\gcd \left ( s,t\right ) =1$$ (in Maple this is done using the normal() command). The ﬁrst step is to ﬁnd the poles of $$r$$ and the order of each pole. If there are no poles, then let the set of poles $$\Gamma$$ will be empty.

If a pole $$x=c$$ is of order 1 which means there is a factor $$\frac {1}{\left ( x-c\right ) }$$ in the partial fractions decomposition of $$r$$, then let \begin {align*} \left [ \sqrt {r}\right ] _{c} & =0\\ \alpha _{c}^{+} & =1\\ \alpha _{c}^{-} & =1 \end {align*}

If the pole $$c$$ is of order 2, which means there is a factor $$\frac {1}{\left ( x-c\right ) ^{2}}$$ in the partial fractions decomposition of $$r$$, then let \begin {align*} \left [ \sqrt {r}\right ] _{c} & =0\\ \alpha _{c}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end {align*}

Where $$b$$ is the coeﬃcient of $$\frac {1}{\left ( x-c\right ) ^{2}}$$ in the partial fraction decomposition of $$r$$. For example, if $$r=\frac {3}{\left ( x-2\right ) ^{2}}$$, then $$x=2$$ is a pole of order $$2$$ and $$b=3$$. The coeﬃcients are found using undetermined coeﬃcients method. (Examples below show how).

If the pole is of order $$4$$ or $$6$$ or $$8$$ and so on, then it is a little bit more complicated. We write $$2v=order$$. For example, if the pole was order 4, then $$v=2$$ and if the pole was order 6, then $$v=3$$ and so on. Notice that for case 1, which we are discussing here, if pole is of order larger than 2, then only poles of order $$4,6,8,\cdots$$ are allowed. This is from the necessary condition. In this case, we add all terms involving $$\frac {1}{\left ( x-c\right ) ^{i}}$$ for $$2\leq i\leq v$$ in the Laurent series expansion of $$\sqrt {r}$$ (not $$r$$) as follows\begin {align} \left [ \sqrt {r}\right ] _{c} & =\sum _{i=2}^{v}\frac {a_{i}}{\left ( x-c\right ) ^{i}}\tag {1}\\ & =\frac {a_{2}}{\left ( x-c\right ) ^{2}}+\frac {a_{3}}{\left ( x-c\right ) ^{3}}+\cdots +\frac {a_{v}}{\left ( x-c\right ) ^{v}}\nonumber \end {align}

For an example if the pole was of order $$6$$, then $$v=3$$. Therefore we need to add all terms in the Laurent series expansion of $$\sqrt {r}$$ from $$v=3$$ down to $$2$$. As follows\begin {align*} \left [ \sqrt {r}\right ] _{c} & =\sum _{i=2}^{3}\frac {a_{i}}{\left ( x-c\right ) ^{i}}\\ & =\frac {a_{2}}{\left ( x-c\right ) ^{2}}+\frac {a_{3}}{\left ( x-c\right ) ^{3}} \end {align*}

Lets look at an example of the above before going to the next step. Assume $r=\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}$ There is only one pole at $$x=0$$ of order $$4$$. Hence $$v=2$$. We need to ﬁnd the Laurent series of $$\sqrt {r}$$ expanding around the speciﬁc pole $$c$$ of order $$2v$$. In Maple this is done using $$\operatorname {series}(\sqrt {r},x=c)$$. \begin {equation} \sqrt {r}\approx \frac {1}{x^{2}}-\frac {5}{2}\frac {1}{x}-\frac {9}{4}-\frac {41}{8}x-\frac {443}{32}x^{2}+\cdots \tag {2} \end {equation} Hence \begin {align} \left [ \sqrt {r}\right ] _{c} & =\sum _{i=2}^{2}\frac {a_{i}}{\left ( x-c\right ) ^{i}}\nonumber \\ & =\frac {a_{2}}{\left ( x-c\right ) ^{2}}\nonumber \\ & =\frac {a_{2}}{\left ( x-0\right ) ^{2}} \tag {3} \end {align}

Comparing the above to Eq. 2 shows that the coeﬃcient is $$a_{2}$$ is (written now as just $$a$$ to make it match the paper and use it in the following equation later on) $a=1$ So the term $$a$$ is the coeﬃcient of $$\frac {a_{v}}{\left ( x-c\right ) ^{v}}$$ in the Laurent series expansion of $$\sqrt {r}$$ around $$x=c$$. In implementation of the algorithm the method of undetermined coeﬃcients is used instead of actually ﬁnding Laurent series for $$\sqrt {r}$$ at $$x=c$$.

Now that we found $$\left [ \sqrt {r}\right ] _{c}$$ for poles $$>2$$, we need to ﬁnd its $$\alpha _{c}^{+},\alpha _{c}^{-}$$ also. In this case $$\alpha _{c}^{+}=\frac {1}{2}\left ( \frac {b}{a}+v\right )$$ and $$\alpha _{c}^{-}=\frac {1}{2}\left ( -\frac {b}{a}+v\right )$$. Where $$a$$ is the one we just found above. But what is $$b$$ here? $$b$$ is the coeﬃcient of the term $$\frac {1}{\left ( x-c\right ) ^{v+1}}$$ in $$r$$ minus the coeﬃcient of $$\frac {1}{\left ( x-c\right ) ^{v+1}}$$ in $$\left [ \sqrt {r}\right ] _{c}$$ which we found above in (2). For an example, using the above $$r$$ its Laurent series expansion around $$x=0$$ is$r\approx x^{2}-2x+3+\frac {1}{x}+\frac {7}{x^{2}}-\frac {5}{x^{3}}+\frac {1}{x^{4}}$ Then since $$a=1$$ from earlier and since $$v=2$$ here (since pole of order 4) then we look above for the coeﬃcient of the term $$\frac {1}{\left ( x-0\right ) ^{v+1}}=\frac {1}{\left ( x-0\right ) ^{3}}$$ in $$r$$ itself. We see this is $$-5$$. Now we need to subtract from this value the coeﬃcient of $$\frac {1}{\left ( x-0\right ) ^{v+1}}=\frac {1}{\left ( x-0\right ) ^{3}}$$ from $$\left [ \sqrt {r}\right ] _{c}$$ series from Eq (3). But since $$\left [ \sqrt {r}\right ] _{c}=\frac {1}{\left ( x-0\right ) ^{2}}$$ then there is no term $$\frac {1}{\left ( x-0\right ) ^{3}}$$. Which means \begin {align*} b & =-5-0\\ & =-5 \end {align*}

Therefore for this example\begin {align*} \alpha _{c}^{+} & =\frac {1}{2}\left ( \frac {b}{a}+v\right ) =\frac {1}{2}\left ( -\frac {5}{1}+2\right ) =-\frac {3}{2}\\ \alpha _{c}^{-} & =\frac {1}{2}\left ( -\frac {b}{a}+v\right ) =\frac {1}{2}\left ( \frac {5}{1}+2\right ) =\frac {7}{2} \end {align*}

We are now done with ﬁnding everything we need related to poles. The above needs to be done for each pole $$c$$ in $$r$$.

We see that for each pole, we need to calculate 3 items. They are $$\left [ \sqrt {r}\right ] _{c},\alpha _{c}^{+},\alpha _{c}^{-}$$.

Now we switch attention to the $$O\left ( \infty \right )$$ order. This is much easier. This is the order of $$r=\frac {s}{t}$$ at inﬁnity which is found from $$\deg \left ( t\right ) -\deg \left ( s\right )$$. There are also three cases to consider.

If $$O\left ( \infty \right ) >2$$ then we write \begin {align*} \left [ \sqrt {r}\right ] _{\infty } & =0\\ \alpha _{\infty }^{+} & =0\\ \alpha _{\infty }^{-} & =1 \end {align*}

If $$O\left ( \infty \right ) =2$$ then $$\left [ \sqrt {r}\right ] _{\infty }=0$$. Now we calculate $$b$$ for this case. This is given by the leading coeﬃcient of $$s$$ divided by the leading coeﬃcient of $$t$$ when $$\gcd \left ( s,t\right ) =1$$. In this case \begin {align*} \left [ \sqrt {r}\right ] _{\infty } & =0\\ \alpha _{\infty }^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{\infty }^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end {align*}

Where here $$b$$ is the coeﬃcient of $$\frac {1}{x^{2}}$$ in the Laurent series expansion of $$r$$ at $$\infty$$. But we do not need to ﬁnd Laurent series expansion of $$r$$ at $$\infty$$ to ﬁnd $$b$$ here. It can be found using $$b=\frac {lcoeff\left ( s\right ) }{lcoeff\left ( t\right ) }$$ where $$r=\frac {s}{t}$$ and $$\gcd \left ( s,t\right ) =1$$. And $$lcoeff$$ is the leading coeﬃcient. For example, if $$r=\frac {1+5x}{2x^{2}}$$ then $$b=\frac {1}{2}$$. If we took the Laurent series of $$r$$ at $$\infty$$ which in Maple can be done using the command $$series(r,x=\infty )$$ then we will get $$\frac {5}{2x}+\frac {1}{2}\frac {1}{x^{2}}$$ which also give $$b=\frac {1}{2}$$.

And ﬁnally, if $$O\left ( \infty \right )$$ $$\leq 0$$, then $$O\left ( \infty \right )$$ has to be negative and even number (conditions for case 1). Let the order of $$r$$ at $$\infty$$ be $$-2v\leq 0$$. Then now $$\left [ \sqrt {r}\right ] _{\infty }$$ is the sum of all terms $$x^{i}$$ for $$0\leq i\leq v$$ in the Laurent series expansion of $$\sqrt {r}$$ at $$\infty$$.$\left [ \sqrt {r}\right ] _{\infty }=ax^{v}+z_{1}x^{v-1}+\cdots +z_{n}$ And $$b$$ is the coeﬃcient of $$x^{v-1}$$ in $$r$$ minus the coeﬃcient of $$x^{v-1}$$ in $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}$$. Then \begin {align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) \\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) \end {align*}

This completes step 1 of the algorithm. We have found $$\left [ \sqrt {r}\right ] _{c}$$ for each pole and associated $$\alpha _{c}^{+},\alpha _{c}^{-}$$ and also found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{+},\alpha _{\infty }^{-}$$. So, what will we do with these? In step 2 these are used to ﬁnd all possible values of what is called $$d$$. For each non negative $$d$$, we will ﬁnd a candidate $$\omega$$. And use this candidate $$\omega$$ to ﬁnd $$P\left ( x\right )$$ by solving $$P^{\prime \prime }+2\omega P^{\prime }+\left ( \omega ^{\prime }+\omega ^{2}-r\right ) P=0$$ (linear algebra problem). If we are able to ﬁnd a $$P\left ( x\right )$$ for any one candidate $$\omega$$ then we stop and we have found the solution $$y=p\left ( x\right ) e^{\int \omega dx}$$ to the $$y^{\prime \prime }=ry$$. Examples below will show how all this works.

##### 3.1.2 Step 2

Recall that from step 1 we have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{+},\alpha _{c}^{-}$$ (this is done for each pole of $$r$$) and we have found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{+},\alpha _{\infty }^{-}$$. From these we now found a possible $$d$$ values and trying each $$d\geq 0$$. The value of $$d$$ is found using the following for each combination of $$s\left ( c\right )$$ where $$s\left ( c\right )$$ is $$+$$ or $$-$$ $d=\alpha _{\infty }^{\pm }-\sum _{c}\alpha _{c}^{\pm }$ We keep only the non negative values of $$d$$. It is important to note that we have to ﬁnd an integer positive value for $$d$$ to continue. If no such value is found from the above, then we stop here as this means no Liouvillian solution exist using case 1. Then we go to case two or case three if it is available.

If we do ﬁnd $$d\geq 0$$, then we now ﬁnd corresponding candidate $$\omega _{d}$$ using\begin {align*} \omega _{d}&=\sum _{c}\left ( (\pm ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}^{\pm }}{x-c}\right ) + (\pm ) \left [\sqrt {r}\right ]_{\infty } \end {align*}

##### 3.1.3 Step 3

In this step we ﬁrst attempt to ﬁnd a polynomial $$p\left ( x\right )$$ of degree $$d$$, for $$\omega$$ found in step 2. This is done by solving $p^{\prime \prime }+2\omega p^{\prime }+\left ( \omega ^{\prime }+\omega ^{2}-r\right ) p=0$ For example, if $$d=2$$, then we let $$p\left ( x\right ) =x^{2}+ax+b$$ and if $$\omega$$ happened to be say $$\frac {1}{x^{2}}-\frac {3}{2x}+x-1$$, then by substituting these in the above, we can solve for $$a,b$$ (if a solution exist). Then the solution to $$y^{\prime \prime }=ry$$ is $$y=p\left ( x\right ) e^{\int \omega dx}$$. If the degree $$d=1$$ then we guess $$p\left ( x\right ) =x+a$$ and try to solve for $$a$$. If the degree $$d=0$$, then we let $$p\left ( x\right ) =1$$, a constant. In the special case of $$p\left ( x\right ) =1$$, there is no coeﬃcients $$a_{i}$$ to solve for. So we would just need to verify that $\omega ^{\prime }+\omega ^{2}-r=0$ In this case.

This completes the full algorithm for case 1. We will now go over many examples for case 1, showing how to implement this algorithm for each example.

The hardest part of the kovacic algorithm is just ﬁnding all the $$\left [ \sqrt {r}\right ] _{c},\alpha _{c}^{\pm },\left [ \sqrt {r}\right ] _{\infty },\alpha _{\infty }^{\pm }$$. Once these are found, the rest of the algorithm is much more direct.

##### 3.1.4 Case one algorithm diagram

The following diagram summarized the above for case one.

#### 3.2 worked examples for case one

3.2.1 Example 1
3.2.2 Example 2
3.2.3 Example 3
3.2.4 Example 4
3.2.5 Example 5
3.2.6 Example 6
3.2.7 Example 7
3.2.8 Example 8
3.2.9 Example 9
3.2.10 Example 10
3.2.11 Example 11
3.2.12 Example 12
3.2.13 Example 13
3.2.14 Example 14
3.2.15 Example 15

##### 3.2.1 Example 1

Let $y^{\prime \prime }=\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}y$ Therefore \begin {align} r & =\frac {s}{t}\nonumber \\ & =\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\nonumber \\ & =x^{2}-2x+3+\frac {1}{x}+\frac {7}{x^{2}}-\frac {5}{x^{3}}+\frac {1}{x^{4}} \tag {1} \end {align}

Step 1 In this we ﬁnd all $$\left [ \sqrt {r}\right ] _{c}$$ and associated $$\alpha _{c}^{\pm }$$ for each pole. There is only one pole at $$x=0$$ of order 4. Hence $$2v=4$$. And $$v=2$$. This is step (C3) now in the paper (1).

We need now to ﬁnd Laurent series of $$\sqrt {r}$$ expanded around $$x=c=0$$. This is given by (using series command on the computer) \begin {equation} \frac {1}{x^{2}}-\frac {5}{2}\frac {1}{x}-\frac {9}{4}-\frac {41}{8}x-\frac {443}{32}x^{2}+\cdots \tag {2} \end {equation} We need to add all terms in the Laurent series expansion of $$\sqrt {r}$$ from $$v=2$$ down to $$2$$. Hence \begin {equation} \left [ \sqrt {r}\right ] _{c}=\frac {1}{\left ( x-0\right ) ^{2}} \tag {3} \end {equation} Is only term from 2. Comparing the above to $$\frac {a}{\left ( x-0\right ) ^{2}}$$ shows that \begin {equation} a=1 \tag {4} \end {equation} Hence\begin {align} \left [ \sqrt {r}\right ] _{c} & =\frac {1}{x^{2}}\tag {5}\\ \alpha _{c}^{+} & =\frac {1}{2}\left ( \frac {b}{a}+v\right ) \nonumber \\ \alpha _{c}^{-} & =\frac {1}{2}\left ( -\frac {b}{a}+v\right ) \tag {6} \end {align}

Where $$v=2$$ and $$a=1$$. We still need to ﬁnd $$b$$. But $$b$$ is the coeﬃcient of the term $$\frac {1}{\left ( x-0\right ) ^{v+1}}$$ in $$r$$ minus the coeﬃcient of $$\frac {1}{\left ( x-0\right ) ^{v+1}}$$ in $$\left [ \sqrt {r}\right ] _{c}$$ which we just found above. Looking at $$r$$ from Eq (1) we see that the term $$\frac {1}{\left ( x-0\right ) ^{v+1}}=\frac {1}{\left ( x-0\right ) ^{3}}$$ has coeﬃcient $$-5$$. And looking at Eq (3) we see that there is no term $$\frac {1}{\left ( x-0\right ) ^{3}}$$ in it. Hence \begin {align} b & =-5-0\nonumber \\ & =-5 \tag {7} \end {align}

Now we found $$a,b$$, then (5,6) becomes (since $$v=2$$)\begin {align} \alpha _{c}^{+} & =\frac {1}{2}\left ( -5+2\right ) =\frac {-3}{2}\tag {8}\\ \alpha _{c}^{-} & =\frac {1}{2}\left ( 5+2\right ) =\frac {7}{2} \tag {9} \end {align}

We are done with all the poles.

Now we consider $$O\left ( \infty \right )$$ which is $$\deg \left ( t\right ) -\deg \left ( s\right ) =4-6=-2$$. Since this is even order and negative then $$-2v=-2$$ or $v=1$ We need the Laurent series of $$\sqrt {r}$$ around $$\infty$$. Using the computer this is $\left [ \sqrt {r}\right ] _{\infty }=x-1+\frac {1}{x}+\frac {3}{2}\frac {1}{x^{2}}+\frac {15}{8x^{3}}+\cdots$ Now we only want the terms $$\,x^{i}$$ where $$0\leq i\leq v$$. This implies the above is reduced to $\left [ \sqrt {r}\right ] _{\infty }=x-1$ The $$a$$ is the coeﬃcient of $$x^{v}=x$$ which is$a=1$ Now we need to ﬁnd $$\alpha _{\infty }^{\pm }$$ associated with $$\left [ \sqrt {r}\right ] _{\infty }$$. For this we need to ﬁrst ﬁnd $$b$$. Recall from above that $$b$$ is the coeﬃcient of $$x^{v-1}$$ or $$x^{0}$$ in $$r$$ minus the coeﬃcient of $$x^{v-1}=x^{0}$$ in $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}$$. Since $$v=1$$ then we want the coeﬃcient of $$x^{0}$$ in $$r$$ and subtract from it the coeﬃcient of $$x^{0}$$ in $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}$$. But\begin {align*} \left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2} & =\left ( x-1\right ) ^{2}\\ & =x^{2}+1-2x \end {align*}

Hence the coeﬃcient of $$x^{0}$$ in $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}$$ is $$1$$.  To ﬁnd the coeﬃcient of $$x^{0}$$ in $$r$$ long division is done\begin {align*} r & =\frac {s}{t}\\ & =\frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\\ & =Q+\frac {R}{4x^{2}} \end {align*}

Where $$Q$$ is the quotient and $$R$$ is the remainder. This gives$r=\left ( x^{2}-2x+3\right ) +\frac {4x^{3}+7x^{2}-20x+4}{4x^{2}}$ For the case of $$v\neq 0$$ then the coeﬃcient is read from $$Q$$ above. Which is $$3$$. Hence\begin {align*} b & =3-1\\ & =2 \end {align*}

For the other case of $$v=0$$ then the coeﬃcient of $$x^{-1}$$ in $$r$$ is found using $$\frac {lcoeff\left ( R\right ) }{lcoeff\left ( t\right ) }$$ which will give $$1$$ in this case. (More examples below).

Now that we found $$a,b$$, then from the above section describing the algorithm, we see in this case that\begin {align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =\frac {1}{2}\left ( \frac {2}{1}-1\right ) =\frac {1}{2}\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( -\frac {2}{1}-1\right ) =-\frac {3}{2} \end {align*}

This completes step 1 of the solution. We have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{\pm }$$ and found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{\pm }$$. Now we go to step 2 which is to ﬁnd the $$d^{\prime }s$$.

step 2 Since we have a pole at zero, and we have one $$O\left ( \infty \right )$$, each with $$\pm$$ signs, then we set up this table to make it easier to work with. This implements$d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{1}\alpha _{c_{i}}^{\pm }$ Therefore we obtain 4 possible $$d$$ values.

 pole $$c$$ $$\alpha _{c}$$ value $$O\left ( \infty \right )$$ value $$d$$ $$d$$ value $$x=0$$ $$\alpha _{c}^{+}=\frac {-3}{2}$$ $$\alpha _{\infty }^{+}=\frac {1}{2}$$ $$\alpha _{\infty }^{+}-\left ( \alpha _{c}^{+}\right ) =\frac {1}{2}-\left ( \frac {-3}{2}\right )$$ $$2$$ $$x=0$$ $$\alpha _{c}^{+}=\frac {-3}{2}$$ $$\alpha _{\infty }^{-}=-\frac {3}{2}$$ $$\alpha _{\infty }^{-}-\left ( \alpha _{c}^{+}\right ) =-\frac {3}{2}-\left ( \frac {-3}{2}\right )$$ $$0$$ $$x=0$$ $$\alpha _{c}^{-}=\frac {7}{2}$$ $$\alpha _{\infty }^{+}=\frac {1}{2}$$ $$\alpha _{\infty }^{+}-\left ( \alpha _{c}^{-}\right ) =\frac {1}{2}-\left ( \frac {7}{2}\right )$$ $$-3$$ $$x=0$$ $$\alpha _{c}^{-}=\frac {7}{2}$$ $$\alpha _{\infty }^{-}=-\frac {3}{2}$$ $$\alpha _{\infty }^{-}-\left ( \alpha _{c}^{-}\right ) =-\frac {3}{2}-\left ( \frac {7}{2}\right )$$ $$-5$$

We see from the above that we took each pole in this problem (there is only one pole here at $$x=0$$) and its associated $$\alpha _{c}^{\pm }$$ with each $$\alpha _{\infty }^{\pm }$$ and generated all possible $$d$$ values from all the combinations. Hence we obtain 4 possible $$d$$ values in this case. If we had 2 poles, then we would have 8 possible $$d$$ values. Hence the maximum possible $$d$$ values we can get is $$2^{p+1}$$ where $$p$$ is number of poles. Now we remove all negative $$d$$ values. Hence the trial $$d$$ values remaining is $d=\left \{ 0,2\right \}$ Now for each $$d$$ value, we generate $$\omega$$ using$\omega =\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$ To apply the above, we update the table above, now using only $$d=0,d=2$$ values, but also add columns for $$\left [ \sqrt {r}\right ] _{c},\left [ \sqrt {r}\right ] _{\infty }$$ to make the computation easier. Here is the new table

pole $$c$$ $$\alpha _{c}$$ value $$s\left ( c\right )$$ $$\left [ \sqrt {r}\right ] _{c}$$ $$O\left ( \infty \right )$$ value $$s\left ( \infty \right )$$ $$\left [ \sqrt {r}\right ] _{\infty }$$ $$d$$ value
 $$\omega$$ value $$\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$$
$$x=0$$ $$\alpha _{c}^{+}=\frac {-3}{2}$$ $$+$$ $$\frac {1}{x^{2}}$$ $$\alpha _{\infty }^{-}=-\frac {3}{2}$$ $$-$$ $$x-1$$ $$0$$ $$\left ( +\left ( \frac {1}{x^{2}}\right ) +\frac {\frac {-3}{2}}{x-0}\right ) +\left ( -\right ) \left ( x-1\right ) =\frac {1}{x^{2}}-\frac {3}{2x}-x+1$$
$$x=0$$ $$\alpha _{c}^{+}=\frac {-3}{2}$$ $$+$$ $$\frac {1}{x^{2}}$$ $$\alpha _{\infty }^{+}=\frac {1}{2}$$ $$+$$ $$x-1$$ $$2$$ $$\left ( +\left ( \frac {1}{x^{2}}\right ) +\frac {\frac {-3}{2}}{x-0}\right ) +\left ( +\right ) \left ( x-1\right ) =x-\frac {3}{2x}+\frac {1}{x^{2}}-1$$

The above are the two candidate $$\omega$$ values. For each $$\omega$$ we need to ﬁnd polynomial $$P$$ by solving \begin {equation} P^{\prime \prime }+2\omega P^{\prime }+\left ( \omega ^{\prime }+\omega ^{2}-r\right ) P=0 \tag {8} \end {equation} If we are able to ﬁnd $$P$$, then we stop and the ode $$y^{\prime \prime }=ry$$ is solved. If we try all candidate $$\omega$$ and can not ﬁnd $$P$$ then this case is not successful and we go to the next case.

step 3 Now for each candidate $$\omega$$ we solve the above Eq (8). Starting with $$\omega =\frac {1}{x^{2}}-\frac {3}{2x}-x+1$$ associated with $$d=0$$ in the table, then (8) becomes

\begin {align*} P^{\prime \prime }+2\left ( \frac {1}{x^{2}}-\frac {3}{2x}-x+1\right ) P^{\prime }+\left ( \left ( \frac {3}{2x^{2}}-\frac {2}{x^{3}}-1\right ) +\left ( \frac {1}{x^{2}}-\frac {3}{2x}-x+1\right ) ^{2}-\left ( \frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\right ) \right ) P & =0\\ P^{\prime \prime }+\left ( \frac {2}{x^{2}}-\frac {3}{2x}-2x+2\right ) P^{\prime }+\left ( \frac {4}{x^{2}}-\frac {6}{x}\right ) P & =0 \end {align*}

Since this the case for $$d=0$$, then $$P$$ has zero degree, Hence $$P$$ is constant. Therefore the above simpliﬁes to$\left ( \frac {4}{x^{2}}-\frac {6}{x}\right ) P=0$ Which means$\frac {4}{x^{2}}-\frac {6}{x}=0$ Which is not possible for all $$x$$. Hence $$d=0$$ do not generate valid $$P$$ polynomial. We now try the case of $$d=2$$. Since $$d=2$$, it means the polynomial $$d$$ is of degree two. Let $P=x^{2}+ax+b$ Substituting this in (8) using $$\omega =x-\frac {3}{2x}+\frac {1}{x^{2}}-1$$. This gives

\begin {align*} P^{\prime \prime }+2\left ( x-\frac {3}{2x}+\frac {1}{x^{2}}-1\right ) P^{\prime }+\left ( \left ( \frac {3}{2x^{2}}-\frac {2}{x^{3}}+1\right ) +\left ( x-\frac {3}{2x}+\frac {1}{x^{2}}-1\right ) ^{2}-\left ( \frac {4x^{6}-8x^{5}+12x^{4}+4x^{3}+7x^{2}-20x+4}{4x^{4}}\right ) \right ) P & =0\\ P^{\prime \prime }+2\left ( x-\frac {3}{2x}+\frac {1}{x^{2}}-1\right ) P^{\prime }+\left ( \frac {4}{x}-4\right ) P & =0 \end {align*}

Using $$P=x^{2}+ax+b$$ the above becomes\begin {align*} 2+2\left ( x-\frac {3}{2x}+\frac {1}{x^{2}}-1\right ) \left ( 2x+a\right ) +\left ( \frac {4}{x}-4\right ) \left ( x^{2}+ax+b\right ) & =0\\ 2a-4b-3\frac {a}{x}+2\frac {a}{x^{2}}+4\frac {b}{x}-2ax+\frac {4}{x}-4 & =0\\ \left ( 2a-4b-4\right ) +\frac {1}{x}\left ( -\frac {3}{4}a+4b+4\right ) +\frac {1}{x^{2}}\left ( 2a\right ) -2ax & =0 \end {align*}

Therefore\begin {align*} 2a-4b-4 & =0\\ -\frac {3}{4}a+4b+4 & =0\\ 2a & =0\\ 2a & =0 \end {align*}

hence $$a=0$$ from last equation. Using ﬁrst or second equation gives $$b=-1$$. Therefore a solution is found. Hence $p\left ( x\right ) =x^{2}-1$ Therefore the solution to $$y^{\prime \prime }=ry$$ is\begin {align*} y & =p\left ( x\right ) e^{\int \omega dx}\\ & =\left ( x^{2}-1\right ) e^{\int x-\frac {3}{2x}+\frac {1}{x^{2}}-1\ dx}\\ & =\left ( x^{2}-1\right ) e^{-\frac {1}{x}-\frac {3}{2}\ln x+\frac {x^{2}}{2}-x}\\ & =\left ( x^{2}-1\right ) x^{\frac {-3}{2}}e^{-\frac {1}{x}+\frac {x^{2}}{2}-x} \end {align*}

The second solution can be found by reduction of order.

##### 3.2.2 Example 2

Let $y^{\prime \prime }=\left ( \frac {2}{x^{2}}-1\right ) y$ Therefore \begin {align} r & =\frac {s}{t}\nonumber \\ & =\frac {2-x^{2}}{x^{2}} \tag {1} \end {align}

Step 1 In this we ﬁnd all $$\left [ \sqrt {r}\right ] _{c}$$ and associated $$\alpha _{c}^{\pm }$$ for each pole. There is one pole at $$x=0$$ of order 2. In this case, from the description of the algorithm earlier, we write\begin {align*} \left [ \sqrt {r}\right ] _{c} & =0\\ \alpha _{c}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end {align*}

Where $$b$$ is the coeﬃcient of $$\frac {1}{\left ( x-0\right ) ^{2}}$$ in the partial fraction decomposition of $$r$$ which is $$\frac {2}{x^{2}}-1$$. Hence $$b=2$$. Therefore\begin {align*} \left [ \sqrt {r}\right ] _{c} & =0\\ \alpha _{c}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+8}=\frac {1}{2}+\frac {3}{2}=2\\ \alpha _{c}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+8}=\frac {1}{2}-\frac {3}{2}=-1 \end {align*}

We are done with all the poles. Now we consider $$O\left ( \infty \right )$$ which is $$\deg \left ( t\right ) -\deg \left ( s\right ) =2-2=0$$. This falls in the case $$-2v\leq 0$$. Hence $v=0$ We need the Laurent series of $$\sqrt {r}$$ around $$\infty$$. Using the computer this is $$i-\frac {i}{x^{2}}-\frac {1}{2x^{4}}i+\cdots$$. Hence we need the coeﬃcient of $$x^{0}$$ in this series ($$0$$ because that is value of $$v$$).$\left [ \sqrt {r}\right ] _{\infty }=ix^{0}$ Recall that $$\left [ \sqrt {r}\right ] _{\infty }$$ is the sum of terms of $$x^{j}$$ for $$0\leq j\leq v$$ or for $$j=0$$ since $$v=0$$. Looking at the series above, we see that $a=i$ Which is the coeﬃcient of the term $$x^{0}$$. Now we need to ﬁnd $$\alpha _{\infty }^{\pm }$$ associated with $$\left [ \sqrt {r}\right ] _{\infty }$$. For this we need to ﬁrst ﬁnd $$b$$ which is the coeﬃcient of $$x^{v-1}=\frac {1}{x}$$ in $$r$$ minus the coeﬃcient of $$x^{v-1}=\frac {1}{x}$$ in $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}$$. But$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}=i^{2}=-1$ Hence the coeﬃcient of $$x^{-1}$$ in $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}$$ is $$0$$. To ﬁnd the coeﬃcient of $$x^{-1}$$ in $$r$$ long division is done\begin {align*} r & =\frac {s}{t}\\ & =\frac {2-x^{2}}{x^{2}}\\ & =Q+\frac {R}{x^{2}} \end {align*}

Where $$Q$$ is the quotient and $$R$$ is the remainder. This gives$r=-1+\frac {2}{x^{2}}$ For the other case of $$v=0$$ then the coeﬃcient of $$x^{-1}$$ is found by looking up the coeﬃcient in $$R$$ of $$x$$ to the degree of of $$t$$ then subtracting one and dividing result by $$lcoeff\left ( t\right )$$. But degree of $$t$$ is $$2$$. Therefore we want the coeﬃcient of $$x^{2-1}$$ or $$x$$ in $$R$$ which is zero. Hence $$b=0-0=0$$.

Now that we found $$a,b$$, then from the above section describing the algorithm, we see in this case that\begin {align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =\frac {1}{2}\left ( 0-0\right ) =0\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( -0-0\right ) =0 \end {align*}

This completes step 1 of the solution. We have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{\pm }$$ and found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{\pm }$$. Now we go to step 2 which is to ﬁnd the $$d^{\prime }s$$.

step 2 Since we have a pole at zero, and we have one $$O\left ( \infty \right )$$, each with $$\pm$$ signs, then we set up this table to make it easier to work with. This implements$d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{1}\alpha _{c_{i}}^{\pm }$ Therefore we obtain 4 possible $$d$$ values.

 pole $$c$$ $$\alpha _{c}$$ value $$O\left ( \infty \right )$$ value $$d$$ $$d$$ value $$x=0$$ $$\alpha _{c}^{+}=2$$ $$\alpha _{\infty }^{+}=0$$ $$\alpha _{\infty }^{+}-\left ( \alpha _{c}^{+}\right ) =0-\left ( 2\right )$$ $$-2$$ $$x=0$$ $$\alpha _{c}^{+}=2$$ $$\alpha _{\infty }^{-}=0$$ $$\alpha _{\infty }^{-}-\left ( \alpha _{c}^{+}\right ) =0-\left ( 2\right )$$ $$-2$$ $$x=0$$ $$\alpha _{c}^{-}=-1$$ $$\alpha _{\infty }^{+}=0$$ $$\alpha _{\infty }^{+}-\left ( \alpha _{c}^{-}\right ) =0-\left ( -1\right )$$ $$1$$ $$x=0$$ $$\alpha _{c}^{-}=-1$$ $$\alpha _{\infty }^{-}=0$$ $$\alpha _{\infty }^{-}-\left ( \alpha _{c}^{-}\right ) =0-\left ( -1\right )$$ $$1$$

We see from the above that we took each pole in this problem (there is only one pole here at $$x=0$$) and its associated $$\alpha _{c}^{\pm }$$ with each $$\alpha _{\infty }^{\pm }$$ and generated all possible $$d$$ values from all the combinations. Hence we obtain 4 possible $$d$$ values in this case. If we had 2 poles, then we would have 8 possible $$d$$ values. Hence the maximum possible $$d$$ values we can get is $$2^{p+1}$$ where $$p$$ is number of poles. Now we remove all negative $$d$$ values. Hence the trial $$d$$ values remaining is $d=\left \{ 1\right \}$ There is one $$d$$ value to try. We can pick any one of the two values of $$d$$ generated since there are both $$d=1$$. Both will give same solution. We generate $$\omega$$ using$\omega =\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$ To apply the above, we update the table above, now using only the ﬁrst $$d=1$$ value in the above table. (selecting the second $$d=1$$ row, will not change the ﬁnal solution). but we also add columns for $$\left [ \sqrt {r}\right ] _{c},\left [ \sqrt {r}\right ] _{\infty }$$ to make the computation easier. Here is the new table

pole $$c$$ $$\alpha _{c}$$ value $$s\left ( c\right )$$ $$\left [ \sqrt {r}\right ] _{c}$$ $$O\left ( \infty \right )$$ value $$s\left ( \infty \right )$$ $$\left [ \sqrt {r}\right ] _{\infty }$$ $$d$$ value
 $$\omega$$ value $$\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$$
$$x=0$$ $$\alpha _{c}^{-}=-1$$ $$-$$ $$0$$ $$\alpha _{\infty }^{+}=0$$ $$+$$ $$i$$ $$1$$ $$\left ( -\left ( 0\right ) +\frac {-1}{x-0}\right ) +\left ( +\right ) \left ( i\right ) =\frac {-1}{x}+i$$

The above gives one candidate $$\omega$$ value to try. For this $$\omega$$ we need to ﬁnd polynomial $$P$$ by solving \begin {equation} P^{\prime \prime }+2\omega P^{\prime }+\left ( \omega ^{\prime }+\omega ^{2}-r\right ) P=0 \tag {8} \end {equation} If we are able to ﬁnd $$P$$, then we stop and the ode $$y^{\prime \prime }=ry$$ is solved. If we try all candidate $$\omega$$ and can not ﬁnd $$P$$ then this case is not successful and we go to the next case.

step 3 Now for each candidate $$\omega$$ we solve the above Eq (8). Starting with $$\omega =\frac {-1}{x}+i$$ associated with ﬁrst $$d=1$$ in the table, then (8) becomes\begin {align*} P^{\prime \prime }+2\left ( \frac {-1}{x}+i\right ) P^{\prime }+\left ( \left ( \frac {-1}{x}+i\right ) ^{\prime }+\left ( \frac {-1}{x}+i\right ) ^{2}-\left ( \frac {2}{x^{2}}-1\right ) \right ) P & =0\\ P^{\prime \prime }+2\left ( \frac {-1}{x}+i\right ) P^{\prime }+\left ( \frac {-2i}{x}\right ) P & =0 \end {align*}

This needs to be solved for $$P$$. Since degree of $$p\left ( x\right )$$ is $$d=1$$. Let $$p=x+a$$. The above becomes\begin {align*} 2\left ( \frac {-1}{x}+i\right ) +\left ( \frac {-2i}{x}\right ) \left ( x+a\right ) & =0\\ \frac {-2}{x}+2i-2i-\frac {2ia}{x} & =0\\ \frac {-2}{x}-\frac {2ia}{x} & =0 \end {align*}

Which means$a=i$ Hence we found the polynomial $p\left ( x\right ) =x+i$ Therefore the solution to $$y^{\prime \prime }=ry$$ is\begin {align*} y & =pe^{\int \omega dx}\\ & =\left ( x+i\right ) e^{\int \frac {-1}{x}+i\ dx}\\ & =\left ( x+i\right ) \frac {1}{x}e^{ix}\\ & =\frac {x+i}{x}\left ( \cos x+i\sin x\right ) \end {align*}

The second solution can be found by reduction of order. The full general solution to $$y^{\prime \prime }=\left ( \frac {2}{x^{2}}-1\right ) y$$ is$y\left ( x\right ) =\frac {c_{1}}{x}\left ( x\cos x-\sin x\right ) +\frac {c_{2}}{x}\left ( \cos x+x\sin x\right )$

##### 3.2.3 Example 3

Let $y^{\prime \prime }=\left ( x^{2}+3\right ) y$ Therefore \begin {align} r & =\frac {s}{t}\nonumber \\ & =\frac {x^{2}+3}{1} \tag {1} \end {align}

Step 1 In this step we ﬁnd all $$\left [ \sqrt {r}\right ] _{c}$$ and associated $$\alpha _{c}^{\pm }$$ for each pole. There are no poles. In this case$$\ \alpha _{c}^{\pm }=0$$ (paper was not explicit in saying this, but from example 3 in paper this can be inferred). Hence the value of $$d$$ is decided by $$\alpha _{\infty }^{\pm }$$ only in this case.

Now we consider $$O\left ( \infty \right )$$ which is $$\deg \left ( t\right ) -\deg \left ( s\right ) =0-2=-2$$. This falls in the case $$-2v\leq 0$$. Hence $$2v=-2$$ or $v=1$ We need the Laurent series of $$\sqrt {r}$$ around $$\infty$$. Using the computer this is $x+\frac {3}{2x}-\frac {9}{8x^{3}}+\cdots$ Hence we need the coeﬃcient of $$x^{1}$$ in this series ($$1$$ because that is value of $$v$$). Recall that $$\left [ \sqrt {r}\right ] _{\infty }$$ is the sum of terms of $$x^{j}$$ for $$0\leq j\leq v$$ or for $$j=0,1$$ since $$v=1$$. Looking at the series above, we see that $a=1$ Which is the coeﬃcient of the term $$x$$. There is no term $$x^{0}$$. Hence $\left [ \sqrt {r}\right ] _{\infty }=x$ Now we need to ﬁnd $$\alpha _{\infty }^{\pm }$$ associated with $$\left [ \sqrt {r}\right ] _{\infty }$$. For this we need to ﬁrst ﬁnd $$b$$. Recall from above that $$b$$ is the coeﬃcient of $$x^{v-1}$$ or $$x^{0}$$ in $$r$$ minus the coeﬃcient of $$x^{v-1}=x^{0}$$ in $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}$$.  But $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}=x^{2}$$. Hence the coeﬃcient of $$x^{0}$$ is zero. To ﬁnd the coeﬃcient of $$x^{0}$$ in $$r$$ long division is done\begin {align*} r & =\frac {s}{t}\\ & =\frac {x^{2}+3}{1}\\ & =Q+\frac {R}{1} \end {align*}

Where $$Q$$ is the quotient and $$R$$ is the remainder. This gives$r=x^{2}+3+\frac {0}{1}$ For the case of $$v\neq 0$$ then the coeﬃcient is read from $$Q$$ above. Which is $$3$$. Hence\begin {align*} b & =3-0\\ & =3 \end {align*}

Now that we found $$a,b$$, then from the above section describing the algorithm, we see in this case that\begin {align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =\frac {1}{2}\left ( 3-1\right ) =1\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( -3-1\right ) =-2 \end {align*}

This completes step 1 of the solution. We have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{\pm }$$ (these are zero, in this example, since there are no poles) and found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{\pm }$$. Now we go to step 2 which is to ﬁnd the $$d^{\prime }s$$.

step 2 We set up this table to make it easier to work with. This implements$d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{0}\alpha _{c_{i}}^{\pm }$ Therefore we obtain 2 possible $$d$$ values.

 pole $$c$$ $$\alpha _{c}$$ values (all zero) $$O\left ( \infty \right )$$ value $$d$$ $$d$$ value $$x=\$$N/A $$\alpha _{c}=0$$ $$\alpha _{\infty }^{+}=1$$ $$\alpha _{\infty }^{+}=1$$ $$1$$ $$x=\$$N/A $$\alpha _{c}=0$$ $$\alpha _{\infty }^{-}=-2$$ $$\alpha _{\infty }^{-}=-2$$ $$-2$$

Picking the positive $$d$$ integers, this gives$d=\left \{ 1\right \}$ There is one $$d$$ value to try. We can pick any one of the two values of $$d$$ generated since there are both $$d=1$$. Both will give same solution. We generate $$\omega$$ using$\omega =\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$ To apply the above, we update the table above, now using only the ﬁrst $$d=1$$ value in the above table. (selecting the ﬁrst $$d=1$$ row). but we also add columns for $$\left [ \sqrt {r}\right ] _{c},\left [ \sqrt {r}\right ] _{\infty }$$ to make the computation easier. Here is the new table

pole $$c$$ $$\alpha _{c}$$ value $$s\left ( c\right )$$ $$\left [ \sqrt {r}\right ] _{c}$$ $$O\left ( \infty \right )$$ value $$s\left ( \infty \right )$$ $$\left [ \sqrt {r}\right ] _{\infty }$$ $$d$$ value
 $$\omega$$ value $$\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$$
$$x=0$$ $$\alpha _{c}=0$$ $$+$$ $$0$$ $$\alpha _{\infty }^{+}=1$$ $$+$$ $$x$$ $$1$$ $$\left ( +\left ( 0\right ) +0\right ) +\left ( +\right ) \left ( x\right ) =x$$

The above gives one candidate $$\omega$$ value to try. For this $$\omega$$ we need to ﬁnd polynomial $$P$$ by solving \begin {equation} P^{\prime \prime }+2\omega P^{\prime }+\left ( \omega ^{\prime }+\omega ^{2}-r\right ) P=0 \tag {8} \end {equation} If we are able to ﬁnd $$P$$, then we stop and the ode $$y^{\prime \prime }=ry$$ is solved. If we try all candidate $$\omega$$ and can not ﬁnd $$P$$ then this case is not successful and we go to the next case.

step 3 Now for each candidate $$\omega$$ (there is only one in this example) we solve the above Eq (8). Starting with $$\omega =x$$ associated with ﬁrst $$d=1$$ in the table, then (8) becomes\begin {align*} P^{\prime \prime }+2\left ( x\right ) P^{\prime }+\left ( \left ( x\right ) ^{\prime }+\left ( x\right ) ^{2}-\left ( x^{2}+3\right ) \right ) P & =0\\ P^{\prime \prime }+2xP^{\prime }+\left ( 1+x^{2}-x^{2}-3\right ) P & =0\\ P^{\prime \prime }+2xP^{\prime }-2P & =0 \end {align*}

This needs to be solved for $$P$$. Since degree of $$p\left ( x\right )$$ is $$d=1$$. Let $$p=x+a$$. The above becomes\begin {align*} 2x-2\left ( x+a\right ) & =0\\ 2x-2x-2a & =0\\ 2a & =0 \end {align*}

Which means$a=0$ Hence we found the polynomial $p\left ( x\right ) =x$ Therefore the solution to $$y^{\prime \prime }=ry$$ is\begin {align*} y & =pe^{\int \omega dx}\\ & =xe^{\int x\ dx}\\ & =xe^{\frac {x^{2}}{2}} \end {align*}

The second solution can be found by reduction of order. The full general solution to $$y^{\prime \prime }=\left ( x^{2}+3\right ) y$$ is$y\left ( x\right ) =c_{1}xe^{\frac {x^{2}}{2}}+c_{2}\left ( xe^{\frac {x^{2}}{2}}\sqrt {\pi }\operatorname {erf}\left ( x\right ) +e^{\frac {-x^{2}}{2}}\right )$

##### 3.2.4 Example 4

Let $y^{\prime \prime }=\frac {2}{x^{2}}y$ Therefore \begin {align} r & =\frac {s}{t}\nonumber \\ & =\frac {2}{x^{2}} \tag {1} \end {align}

Step 1 In this we ﬁnd all $$\left [ \sqrt {r}\right ] _{c}$$ and associated $$\alpha _{c}^{\pm }$$ for each pole. There is one pole at $$x=0$$ of order 2. In this case, from the description of the algorithm earlier, we write\begin {align*} \left [ \sqrt {r}\right ] _{c} & =0\\ \alpha _{c}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end {align*}

Where $$b$$ is the coeﬃcient of $$\frac {1}{\left ( x-0\right ) ^{2}}$$ in the partial fraction decomposition of $$r$$ which is $$\frac {2}{x^{2}}$$. Hence $$b=2$$. Therefore\begin {align*} \left [ \sqrt {r}\right ] _{c} & =0\\ \alpha _{c}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+8}=\frac {1}{2}+\frac {3}{2}=2\\ \alpha _{c}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+8}=\frac {1}{2}-\frac {3}{2}=-1 \end {align*}

We are done with all the poles. Now we consider $$O\left ( \infty \right )$$ which is $$\deg \left ( t\right ) -\deg \left ( s\right ) =2-0=2$$. Since $$O\left ( \infty \right ) =2$$ then from the algorithm above $\left [ \sqrt {r}\right ] _{\infty }=0$ Now we calculate $$b$$ for this case. This is given by the leading coeﬃcient of $$s$$ divided by the leading coeﬃcient of $$t$$ when $$\gcd \left ( s,t\right ) =1$$. In this case $$r=\frac {2}{x^{2}}$$ , hence $$b=\frac {2}{1}=2$$. Therefore\begin {align*} \left [ \sqrt {r}\right ] _{\infty } & =0\\ \alpha _{\infty }^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}=\frac {1}{2}+\frac {1}{2}\sqrt {1+8}=2\\ \alpha _{\infty }^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b}=\frac {1}{2}-\frac {1}{2}\sqrt {1+8}=-1 \end {align*}

This completes step 1 of the solution. We have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{\pm }$$ and found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{\pm }$$. Now we go to step 2 which is to ﬁnd the $$d^{\prime }s$$.

step 2 Since we have a pole at zero, and we have one $$O\left ( \infty \right )$$, each with $$\pm$$ signs, then we set up this table to make it easier to work with. This implements$d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{1}\alpha _{c_{i}}^{\pm }$ Therefore we obtain 4 possible $$d$$ values.

 pole $$c$$ $$\alpha _{c}$$ value $$O\left ( \infty \right )$$ value $$d$$ $$d$$ value $$x=0$$ $$\alpha _{c}^{+}=2$$ $$\alpha _{\infty }^{+}=2$$ $$\alpha _{\infty }^{+}-\left ( \alpha _{c}^{+}\right ) =2-\left ( 2\right )$$ $$0$$ $$x=0$$ $$\alpha _{c}^{+}=2$$ $$\alpha _{\infty }^{-}=-1$$ $$\alpha _{\infty }^{-}-\left ( \alpha _{c}^{+}\right ) =-1-\left ( 2\right )$$ $$-3$$ $$x=0$$ $$\alpha _{c}^{-}=-1$$ $$\alpha _{\infty }^{+}=2$$ $$\alpha _{\infty }^{+}-\left ( \alpha _{c}^{-}\right ) =2-\left ( -1\right )$$ $$3$$ $$x=0$$ $$\alpha _{c}^{-}=-1$$ $$\alpha _{\infty }^{-}=-1$$ $$\alpha _{\infty }^{-}-\left ( \alpha _{c}^{-}\right ) =-1-\left ( -1\right )$$ $$0$$

Hence the trial $$d$$ values which are not negative are $d=\left \{ 0,3\right \}$ For $$d=0\,$$, since it shows in two rows, we take the ﬁrst row. Now we generate $$\omega$$ for each $$d$$ using$\omega =\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$ To apply the above, we update the table above, but we also add columns for $$\left [ \sqrt {r}\right ] _{c},\left [ \sqrt {r}\right ] _{\infty }$$ to make the computation easier. Here is the new table

pole $$c$$ $$\alpha _{c}$$ value $$s\left ( c\right )$$ $$\left [ \sqrt {r}\right ] _{c}$$ $$O\left ( \infty \right )$$ value $$s\left ( \infty \right )$$ $$\left [ \sqrt {r}\right ] _{\infty }$$ $$d$$ value
 $$\omega$$ value $$\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$$
$$x=0$$ $$\alpha _{c}^{+}=2$$ $$+$$ $$0$$ $$\alpha _{\infty }^{+}=2$$ $$+$$ $$0$$ $$0$$ $$\left ( +\left ( 0\right ) +\frac {2}{x-0}\right ) +\left ( +\right ) \left ( 0\right ) =\frac {2}{x}$$
$$x=0$$ $$\alpha _{c}^{-}=-1$$ $$-$$ $$0$$ $$\alpha _{\infty }^{+}=2$$ $$+$$ $$0$$ $$3$$ $$\left ( -\left ( 0\right ) +\frac {-1}{x-0}\right ) +\left ( +\right ) \left ( 0\right ) =\frac {-1}{x}$$

The above gives two candidate $$\omega =\left \{ \frac {2}{x},\frac {-1}{x}\right \}$$ value to try. For this $$\omega$$ we need to ﬁnd polynomial $$P$$ by solving \begin {equation} P^{\prime \prime }+2\omega P^{\prime }+\left ( \omega ^{\prime }+\omega ^{2}-r\right ) P=0 \tag {8} \end {equation} If we are able to ﬁnd $$P$$, then we stop and the ode $$y^{\prime \prime }=ry$$ is solved. If we try all candidate $$\omega$$ and can not ﬁnd $$P$$ then this case is not successful and we go to the next case.

step 3 Now for each candidate $$\omega$$ we solve the above Eq (8). Starting with $$\omega =\frac {2}{x}$$ associated with ﬁrst $$d=0$$ in the table, then (8) becomes\begin {align*} P^{\prime \prime }+2\left ( \frac {2}{x}\right ) P^{\prime }+\left ( \left ( \frac {2}{x}\right ) ^{\prime }+\left ( \frac {2}{x}\right ) ^{2}-\left ( \frac {2}{x^{2}}\right ) \right ) P & =0\\ P^{\prime \prime }+\frac {4}{x}P^{\prime }+\left ( -\frac {2}{x^{2}}+\frac {4}{x^{2}}-\frac {2}{x^{2}}\right ) P & =0\\ P^{\prime \prime }+\frac {4}{x}P^{\prime } & =0 \end {align*}

This needs to be solved for $$P$$. Since degree of $$p\left ( x\right )$$ is $$d=0$$. Let $$p=a$$. The above becomes$0=0$ No unique solution. Hence $$d=0$$ did not work. Now we try the second $$\omega =\frac {-1}{x}$$ associated with $$d=3$$. Substituting in 8 gives\begin {align*} P^{\prime \prime }+2\left ( \frac {-1}{x}\right ) P^{\prime }+\left ( \left ( \frac {-1}{x}\right ) ^{\prime }+\left ( \frac {-1}{x}\right ) ^{2}-\left ( \frac {2}{x^{2}}\right ) \right ) P & =0\\ P^{\prime \prime }+\frac {-2}{x}P^{\prime }+\left ( \frac {1}{x^{2}}+\frac {1}{x^{2}}-\frac {2}{x^{2}}\right ) P & =0\\ P^{\prime \prime }-\frac {2}{x}P^{\prime } & =0 \end {align*}

Since $$d=3$$, let $p\left ( x\right ) =x^{3}+ax^{2}+bx+c$ Then $$P^{\prime \prime }-\frac {2}{x}P^{\prime }=0$$ becomes\begin {align*} \left ( 6x+2a\right ) -\frac {2}{x}\left ( 3x^{2}+2ax+b\right ) & =0\\ -2a-2\frac {b}{x} & =0 \end {align*}

Hence $$a=0,b=0$$ is solution. $$c$$ is arbitrary. Taking $$c=0$$ then$p\left ( x\right ) =x^{3}$ Therefore the solution to $$y^{\prime \prime }=ry$$ is\begin {align*} y & =p\left ( x\right ) e^{\int \omega dx}\\ & =x^{3}e^{\int \frac {-1}{x}\ dx}\\ & =x^{3}e^{-\ln x}\\ & =x^{2} \end {align*}

The second solution can be found by reduction of order. The full general solution to $$y^{\prime \prime }=\frac {2}{x^{2}}y$$ is$y\left ( x\right ) =c_{1}\frac {1}{x}+c_{2}x^{2}$

##### 3.2.5 Example 5

Let $y^{\prime \prime }=\frac {32x^{2}-27x+27}{144x^{4}-288x^{3}+144x^{2}}y$ Therefore \begin {align} r & =\frac {s}{t}\nonumber \\ & =\frac {32x^{2}-27x+27}{144x^{4}-288x^{3}+144x^{2}}\tag {1}\\ & =\frac {3}{16x}-\frac {3}{16}\frac {1}{\left ( x-1\right ) }+\frac {3}{16}\frac {1}{x^{2}}+\frac {2}{9}\frac {1}{\left ( x-1\right ) ^{2}} \tag {2} \end {align}

The necessary conditions for case 1 are met.

Step 1 In this we ﬁnd all $$\left [ \sqrt {r}\right ] _{c}$$ and associated $$\alpha _{c}^{\pm }$$ for each pole. There is one pole at $$x=0$$ of order 2 and one pole at $$x=1$$ of order 2. For the pole at $$x=0$$ since order is $$2$$ then\begin {align*} \left [ \sqrt {r}\right ] _{c=0} & =0\\ \alpha _{c=0}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c=0}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end {align*}

Where $$b$$ is the coeﬃcient of $$\frac {1}{\left ( x-0\right ) ^{2}}$$ in the partial fraction decomposition of $$r$$ given in Eq (2) which is $$\frac {3}{16}$$. Hence $$b=\frac {3}{16}$$. Therefore\begin {align*} \left [ \sqrt {r}\right ] _{c=0} & =0\\ \alpha _{c=0}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {3}{16}\right ) }=\frac {1}{4}\sqrt {7}+\frac {1}{2}\\ \alpha _{c=0}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {3}{16}\right ) }=\frac {1}{2}-\frac {1}{4}\sqrt {7} \end {align*}

And for the pole at $$x=1$$ which is order 2, \begin {align*} \left [ \sqrt {r}\right ] _{c=1} & =0\\ \alpha _{c=1}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c=1}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end {align*}

Where $$b$$ is the coeﬃcient of $$\frac {1}{\left ( x-1\right ) ^{2}}$$ in the partial fraction decomposition of $$r$$ given in Eq (2) which is $$\frac {2}{9}$$. Hence $$b=\frac {2}{9}$$. Therefore the above becomes\begin {align*} \left [ \sqrt {r}\right ] _{c=1} & =0\\ \alpha _{c=1}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {2}{9}\right ) }=\frac {1}{6}\sqrt {17}+\frac {1}{2}\\ \alpha _{c=1}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {2}{9}\right ) }=\frac {1}{2}-\frac {1}{6}\sqrt {17} \end {align*}

We are done with all the poles. Now we consider $$O\left ( \infty \right )$$ which is $$\deg \left ( t\right ) -\deg \left ( s\right ) =4-2=2$$. Since $$O\left ( \infty \right ) =2$$ then from the algorithm above $\left [ \sqrt {r}\right ] _{\infty }=0$ Now we calculate $$b$$ for this case. This is given by the leading coeﬃcient of $$s$$ divided by the leading coeﬃcient of $$t$$ when $$\gcd \left ( s,t\right ) =1$$. In this case $$r=\frac {32x^{2}-27x+27}{144x^{4}-288x^{3}+144x^{2}}$$ from Eq (1) , hence $$b=\frac {32}{144}=\frac {2}{9}$$. Therefore\begin {align*} \left [ \sqrt {r}\right ] _{\infty } & =0\\ \alpha _{\infty }^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}=\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {2}{9}\right ) }=\frac {1}{6}\sqrt {17}+\frac {1}{2}\\ \alpha _{\infty }^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b}=\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {2}{9}\right ) }=\frac {1}{2}-\frac {1}{6}\sqrt {17} \end {align*}

This completes step 1 of the solution. We have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{\pm }$$ and found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{\pm }$$. Now we go to step 2 which is to ﬁnd the $$d^{\prime }s$$.

step 2 Since we have a pole at $$x=c_{1}=0$$ and pole at $$x=c_{2}=1$$, and we have one $$O\left ( \infty \right )$$, each with $$\pm$$ signs. The following now implements $d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{2}\alpha _{c_{i}}^{\pm }$ By trying all possible combinations. There are 8 possible $$d$$ values. This gives\begin {align*} d_{1} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) -\left ( \frac {1}{4}\sqrt {7}+\frac {1}{2}\right ) -\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) =-\frac {1}{4}\sqrt {7}-\frac {1}{2}\\ d_{2} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) -\left ( \frac {1}{4}\sqrt {7}+\frac {1}{2}\right ) -\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) =\frac {1}{3}\sqrt {17}-\frac {1}{4}\sqrt {7}-\frac {1}{2}\\ d_{3} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) -\left ( \frac {1}{2}-\frac {1}{4}\sqrt {7}\right ) -\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) =\frac {1}{4}\sqrt {7}-\frac {1}{2}\\ d_{4} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) -\left ( \frac {1}{2}-\frac {1}{4}\sqrt {7}\right ) -\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) =\frac {1}{4}\sqrt {7}+\frac {1}{3}\sqrt {17}-\frac {1}{2}\\ d_{5} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) -\left ( \frac {1}{4}\sqrt {7}+\frac {1}{2}\right ) -\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) =-\frac {1}{4}\sqrt {7}-\frac {1}{3}\sqrt {17}-\frac {1}{2}\\ d_{6} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) -\left ( \frac {1}{4}\sqrt {7}+\frac {1}{2}\right ) -\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) =-\frac {1}{4}\sqrt {7}-\frac {1}{2}\\ d_{7} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) -\left ( \frac {1}{2}-\frac {1}{4}\sqrt {7}\right ) -\left ( \frac {1}{6}\sqrt {17}+\frac {1}{2}\right ) =\frac {1}{4}\sqrt {7}-\frac {1}{3}\sqrt {17}-\frac {1}{2}\\ d_{8} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) -\left ( \frac {1}{2}-\frac {1}{4}\sqrt {7}\right ) -\left ( \frac {1}{2}-\frac {1}{6}\sqrt {17}\right ) =\frac {1}{4}\sqrt {7}-\frac {1}{2} \end {align*}

None of the $$d$$ found are integer. Hence case 1 did not work we need to try case 2 and if that also fail, try case 3. We will ﬁnd all three cases fail on this ode..

##### 3.2.6 Example 6

Let $y^{\prime \prime }=y$ Therefore $r=\frac {s}{t}=1$ The necessary conditions for case 1 are met since zero order pole and $$O\left ( \infty \right ) =0$$.

Step 1 In this we ﬁnd all $$\left [ \sqrt {r}\right ] _{c}$$ and associated $$\alpha _{c}^{\pm }$$ for each pole. There are no poles. In this case$$\ \left [ \sqrt {r}\right ] _{c}=0$$ and $$\alpha _{c}^{\pm }=0$$. Since $$O\left ( \infty \right ) =0$$, we are in case $$2v\leq 0$$. Hence $$v=0$$. Then now $$\left [ \sqrt {r}\right ] _{\infty }$$ is the sum of all terms $$x^{i}$$ for $$0\leq i\leq v$$ in the Laurent series expansion of $$\sqrt {r}$$ at $$\infty$$.$\left [ \sqrt {r}\right ] _{\infty }=1$ Hence $$a=1$$.  And $$b$$ is the coeﬃcient of $$x^{v-1}=x^{-1}$$ in $$r$$ minus the coeﬃcient of $$x^{v-1}=x^{-1}$$ in $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}$$. Hence $$b=0$$. Then \begin {align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =0\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =0 \end {align*}

This completes step 1 of the solution. We have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{\pm }$$ and found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{\pm }$$. Now we go to step 2 which is to ﬁnd the $$d^{\prime }s$$.

step 2 Since we have a pole at zero and pole at $$x=1$$, and we have one $$O\left ( \infty \right )$$, each with $$\pm$$ signs, then we set up this table to make it easier to work with. This implements$d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{0}\alpha _{c_{i}}^{\pm }$ Therefore we obtain 2 possible $$d$$ values.

 pole $$c$$ $$\alpha _{c}$$ value $$O\left ( \infty \right )$$ value $$d$$ $$d$$ value $$x=NA$$ $$\alpha _{c}=0$$ $$\alpha _{\infty }^{+}=0$$ $$\alpha _{\infty }^{+}=0$$ $$0$$ $$x=NA$$ $$\alpha _{c}=0$$ $$\alpha _{\infty }^{-}=0$$ $$\alpha _{\infty }^{+}=0$$ $$0$$

Hence the trial $$d$$ values which are not negative integers are $d=\left \{ 0\right \}$ For $$d=0\,$$, since it shows in two rows, we take the ﬁrst row. Now we generate $$\omega$$ for each $$d$$ using$\omega =\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$ To apply the above, we update the table above, but we also add columns for $$\left [ \sqrt {r}\right ] _{c},\left [ \sqrt {r}\right ] _{\infty }$$ to make the computation easier. Here is the new table

pole $$c$$ $$\alpha _{c}$$ value $$s\left ( c\right )$$ $$\left [ \sqrt {r}\right ] _{c}$$ $$O\left ( \infty \right )$$ value $$s\left ( \infty \right )$$ $$\left [ \sqrt {r}\right ] _{\infty }$$ $$d$$ value
 $$\omega$$ value $$\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$$
$$x=NA$$ $$\alpha _{c}=0$$ $$+$$ $$0$$ $$\alpha _{\infty }^{+}=0$$ $$+$$ $$0$$ $$0$$ $$\left ( +\left ( 0\right ) +0\right ) +\left ( +\right ) \left ( 1\right ) =1$$

The above gives candidate $$\omega =1$$ value to try. For this $$\omega$$ we need to ﬁnd polynomial $$P$$ by solving \begin {equation} P^{\prime \prime }+2\omega P^{\prime }+\left ( \omega ^{\prime }+\omega ^{2}-r\right ) P=0 \tag {8} \end {equation} If we are able to ﬁnd $$P$$, then we stop and the ode $$y^{\prime \prime }=ry$$ is solved.

step 3 Now for each candidate $$\omega$$ we solve the above Eq (8). Starting with $$\omega =1$$ associated with $$d=0$$ in the table. Let $$p\left ( x\right ) =1$$ since degree is zero, then (8) becomes\begin {align*} P^{\prime \prime }+2\left ( 1\right ) P^{\prime }+\left ( \left ( 1\right ) ^{\prime }+\left ( 1\right ) ^{2}-\left ( 1\right ) \right ) P & =0\\ \left ( 0+1-1\right ) & =0\\ 0 & =0 \end {align*}

Hence $$p\left ( x\right ) =1$$ is valid solution. Therefore the solution to $$y^{\prime \prime }=y$$ is\begin {align*} y & =p\left ( x\right ) e^{\int \omega dx}\\ & =e^{\int 1\ dx}\\ & =e^{x} \end {align*}

The second solution can be found by reduction of order. The full general solution to $$y^{\prime \prime }=y$$ is$y\left ( x\right ) =c_{1}e^{x}+c_{2}e^{-x}$

##### 3.2.7 Example 7

Let $\left ( x^{2}-2x\right ) y^{\prime \prime }+\left ( 2-x^{2}\right ) y^{\prime }+\left ( 2x-2\right ) y=0$ Normalizing so that coeﬃcient of $$y^{\prime \prime }$$ is one gives\begin {align} y^{\prime \prime }+\frac {\left ( 2-x^{2}\right ) }{\left ( x^{2}-2x\right ) }y^{\prime }+\frac {\left ( 2x-2\right ) }{\left ( x^{2}-2x\right ) }y & =0\nonumber \\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0 \tag {1} \end {align}

Hence \begin {align*} a & =\frac {\left ( 2-x^{2}\right ) }{\left ( x^{2}-2x\right ) }\\ b & =\frac {\left ( 2x-2\right ) }{\left ( x^{2}-2x\right ) } \end {align*}

It is ﬁrst transformed to the following ode by eliminating the ﬁrst derivative \begin {equation} z^{\prime \prime }=rz \tag {2} \end {equation} Using what is known as the Liouville transformation given by\begin {equation} y=ze^{\frac {-1}{2}\int adx} \tag {3} \end {equation} Where it can be found that $$r$$ in (2) is given by \begin {align} r & =\frac {1}{4}a^{2}+\frac {1}{2}a^{\prime }-b\nonumber \\ & =\frac {1}{4}\left ( \frac {\left ( 2-x^{2}\right ) }{\left ( x^{2}-2x\right ) }\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {\left ( 2-x^{2}\right ) }{\left ( x^{2}-2x\right ) }\right ) -\frac {\left ( 2x-2\right ) }{\left ( x^{2}-2x\right ) }\nonumber \\ & =\frac {\left ( x^{4}-8x^{3}+24x^{2}-24x+12\right ) }{4x^{2}\left ( x-2\right ) ^{2}} \tag {4} \end {align}

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is \begin {equation} z^{\prime \prime }=\frac {\left ( x^{4}-8x^{3}+24x^{2}-24x+12\right ) }{4x^{2}\left ( x-2\right ) ^{2}}z \tag {5} \end {equation} Therefore \begin {align*} r & =\frac {s}{t}\\ & =\frac {\left ( x^{4}-8x^{3}+24x^{2}-24x+12\right ) }{4x^{2}\left ( x-2\right ) ^{2}} \end {align*}

The necessary conditions for case 1 are met.

Step 1 In this we ﬁnd all $$\left [ \sqrt {r}\right ] _{c}$$ and associated $$\alpha _{c}^{\pm }$$ for each pole. There is one pole at $$x=0$$ of order 2 and one pole at $$x=2$$ of order 2. For the pole at $$x=0$$ since order is $$2$$ then\begin {align*} \left [ \sqrt {r}\right ] _{c=0} & =0\\ \alpha _{c=0}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c=0}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end {align*}

Where $$b$$ is the coeﬃcient of $$\frac {1}{\left ( x-0\right ) ^{2}}$$ in the partial fraction decomposition of $$r$$ which is \begin {equation} \frac {x^{4}-8x^{3}+24x^{2}-24x+12}{4x^{2}\left ( x-2\right ) ^{2}}=\frac {1}{4}-\frac {3}{4}\frac {1}{x}-\frac {1}{4}\frac {1}{\left ( x-2\right ) }+\frac {3}{4}\frac {1}{\left ( x-2\right ) ^{2}}+\frac {3}{4}\frac {1}{x^{2}}\tag {6} \end {equation} Hence $$b=\frac {3}{4}$$. Therefore\begin {align*} \left [ \sqrt {r}\right ] _{c=0} & =0\\ \alpha _{c=0}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {3}{4}\right ) }=\frac {3}{2}\\ \alpha _{c=0}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {3}{4}\right ) }=-\frac {1}{2} \end {align*}

And for the pole at $$x=2$$ which is order 2, \begin {align*} \left [ \sqrt {r}\right ] _{c=2} & =0\\ \alpha _{c=2}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c=2}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end {align*}

Where $$b$$ is the coeﬃcient of $$\frac {1}{\left ( x-2\right ) ^{2}}$$ in the partial fraction decomposition of $$r$$ given in Eq (6). Hence $$b=\frac {3}{4}$$. Therefore the above becomes\begin {align*} \left [ \sqrt {r}\right ] _{c=2} & =0\\ \alpha _{c=2}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {3}{4}\right ) }=\frac {3}{2}\\ \alpha _{c=2}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {3}{4}\right ) }=-\frac {1}{2} \end {align*}

We are done with all the poles. Now we consider $$O\left ( \infty \right )$$ which is $$\deg \left ( t\right ) -\deg \left ( s\right ) =4-4=0$$. Since $$O\left ( \infty \right ) =0$$, we are in case $$2v\leq 0$$. Hence $$v=0$$. Then now $$\left [ \sqrt {r}\right ] _{\infty }$$ is the sum of all terms $$x^{i}$$ for $$0\leq i\leq v$$ in the Laurent series expansion of $$\sqrt {r}$$ at $$\infty$$ which is \begin {equation} \left [ \sqrt {r}\right ] _{\infty }=\frac {1}{2}-\frac {1}{x}+\frac {2}{x^{3}}+\frac {11}{x^{4}}+\cdots \tag {7} \end {equation} We want only terms for $$0\leq i\leq v$$ but $v=0$ Therefore only the constant term. Hence\begin {equation} \left [ \sqrt {r}\right ] _{\infty }=\frac {1}{2}\tag {8} \end {equation} Which means $a=\frac {1}{2}$ As it is the the term that matches $$\left [ \sqrt {r}\right ] _{\infty }=ax^{v}+\cdots$$. Hence $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}=\frac {1}{4}$$ and the coeﬃcient of $$\frac {1}{x}$$ is zero. To ﬁnd the coeﬃcient of $$\frac {1}{x}$$ in $$r$$ long division is done\begin {align*} r & =\frac {s}{t}\\ & =\frac {x^{4}-8x^{3}+24x^{2}-24x+12}{4x^{4}-16x^{3}+16x^{2}}\\ & =Q+\frac {R}{4x^{4}-16x^{3}+16x^{2}} \end {align*}

Where $$Q$$ is the quotient and $$R$$ is the remainder. This gives$r=\frac {1}{4}+\frac {-4x^{3}+20x^{2}-24x+12}{4x^{4}-16x^{3}+16x^{2}}$ Since $$v=0$$ then the coeﬃcient of $$x^{-1}$$ in $$r$$ is found using $$\frac {lcoeff\left ( R\right ) }{lcoeff\left ( t\right ) }$$. This gives $$-1$$ as seen from above. Hence $$b=-1-0=-1$$. Therefore\begin {align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =\frac {1}{2}\left ( \frac {-1}{\frac {1}{2}}-0\right ) =-1\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( -\frac {-1}{\frac {1}{2}}-0\right ) =1 \end {align*}

This completes step 1 of the solution. We have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{\pm }$$ and found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{\pm }$$. Now we go to step 2 which is to ﬁnd the $$d^{\prime }s$$.

step 2 Since we have a pole at $$x=c_{1}=0$$ and pole at $$x=c_{2}=1$$, and we have one $$O\left ( \infty \right )$$, each with $$\pm$$ signs. The following now implements $d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{2}\alpha _{c_{i}}^{\pm }$ By trying all possible combinations. There are 8 possible $$d$$ values. This gives\begin {align*} d_{1} & =1-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =1-\left ( \frac {3}{2}+\frac {3}{2}\right ) =-2\\ d_{2} & =1-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =1-\left ( \frac {3}{2}-\frac {1}{2}\right ) =0\\ d_{3} & =1-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =1-\left ( -\frac {1}{2}+\frac {3}{2}\right ) =0\\ d_{4} & =1-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =1-\left ( -\frac {1}{2}-\frac {1}{2}\right ) =2\\ d_{5} & =-1-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =-1-\left ( \frac {3}{2}+\frac {3}{2}\right ) =-4\\ d_{6} & =-1-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =-1-\left ( \frac {3}{2}-\frac {1}{2}\right ) =-2\\ d_{7} & =-1-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =-1-\left ( -\frac {1}{2}+\frac {3}{2}\right ) =-2\\ d_{8} & =-1-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =-1-\left ( -\frac {1}{2}-\frac {1}{2}\right ) =0 \end {align*}

Need to complete the solution next.

##### 3.2.8 Example 8

Let $\left ( x^{2}+1\right ) y^{\prime \prime }+2xy^{\prime }-2y=0$ Normalizing so that coeﬃcient of $$y^{\prime \prime }$$ is one gives\begin {align} y^{\prime \prime }+\frac {2x}{\left ( x^{2}+1\right ) }y^{\prime }-\frac {2}{\left ( x^{2}+1\right ) }y & =0\nonumber \\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0 \tag {1} \end {align}

Hence \begin {align*} a & =\frac {2x}{\left ( x^{2}+1\right ) }\\ b & =-\frac {2}{\left ( x^{2}+1\right ) } \end {align*}

It is ﬁrst transformed to the following ode by eliminating the ﬁrst derivative \begin {equation} z^{\prime \prime }=rz \tag {2} \end {equation} Using what is known as the Liouville transformation given by\begin {equation} y=ze^{\frac {-1}{2}\int adx} \tag {3} \end {equation} Where it can be found that $$r$$ in (2) is given by \begin {align} r & =\frac {1}{4}a^{2}+\frac {1}{2}a^{\prime }-b\nonumber \\ & =\frac {1}{4}\left ( \frac {2x}{\left ( x^{2}+1\right ) }\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {2x}{\left ( x^{2}+1\right ) }\right ) -\left ( -\frac {2}{\left ( x^{2}+1\right ) }\right ) \nonumber \\ & =\frac {2x^{2}+3}{\left ( x^{2}+1\right ) ^{2}} \tag {4} \end {align}

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is \begin {equation} z^{\prime \prime }=\frac {2x^{2}+3}{\left ( x^{2}+1\right ) ^{2}}z \tag {5} \end {equation} Therefore \begin {align} r & =\frac {s}{t}\nonumber \\ & =\frac {2x^{2}+3}{\left ( x^{2}+1\right ) ^{2}}\nonumber \\ & =\frac {2x^{2}+3}{x^{4}+2x^{2}+1} \tag {5A} \end {align}

The necessary conditions for case 1 are met.

Step 1 In this we ﬁnd all $$\left [ \sqrt {r}\right ] _{c}$$ and associated $$\alpha _{c}^{\pm }$$ for each pole. There is one pole at $$x=-i$$ of order 2 and one pole at $$x=i$$ of order 2. For the pole at $$x=-i$$ since order is $$2$$ then\begin {align*} \left [ \sqrt {r}\right ] _{c_{1}} & =0\\ \alpha _{c_{1}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c_{1}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end {align*}

Where $$b$$ is the coeﬃcient of $$\frac {1}{\left ( x+i\right ) ^{2}}$$ in the partial fraction decomposition of $$r$$ which is (in Maple this can be found using fullparfrac. \begin {equation} \frac {2x^{2}+3}{\left ( x^{2}+1\right ) ^{2}}=-\frac {1}{4}\frac {1}{\left ( x-i\right ) ^{2}}-\frac {1}{4}\frac {1}{\left ( x+i\right ) ^{2}}-\frac {5i}{4}\frac {1}{\left ( x-i\right ) }+\frac {5i}{4}\frac {1}{x+i}\tag {6} \end {equation} Hence $$b=-\frac {1}{4}$$. Therefore\begin {align*} \left [ \sqrt {r}\right ] _{c=0} & =0\\ \alpha _{c_{1}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( -\frac {1}{4}\right ) }=\frac {1}{2}\\ \alpha _{c_{1}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( -\frac {1}{4}\right ) }=\frac {1}{2} \end {align*}

And for the pole at $$x=+i$$ which is order 2, \begin {align*} \left [ \sqrt {r}\right ] _{c_{2}} & =0\\ \alpha _{c_{2}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c_{2}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end {align*}

Where $$b$$ is the coeﬃcient of $$\frac {1}{\left ( x-i\right ) ^{2}}$$ in the partial fraction decomposition of $$r$$ given in Eq (6). Hence $$b=-\frac {1}{4}$$. Therefore the above becomes\begin {align*} \left [ \sqrt {r}\right ] _{c=2} & =0\\ \alpha _{c=2}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( -\frac {1}{4}\right ) }=\frac {1}{2}\\ \alpha _{c=2}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( -\frac {1}{4}\right ) }=\frac {1}{2} \end {align*}

We are done with all the poles. Now we consider $$O\left ( \infty \right )$$ which is $$\deg \left ( t\right ) -\deg \left ( s\right ) =4-2=2$$. Since $$O\left ( \infty \right ) =2$$, then $$\left [ \sqrt {r}\right ] _{\infty }=0$$. Now $$b$$ is the coeﬃcient of $$\frac {1}{x^{2}}$$ in $$r$$ minus coeﬃcient of $$\frac {1}{x^{2}}$$ in $$\left [ \sqrt {r}\right ] _{\infty }^{2}$$ which is zero. the coeﬃcient of $$\frac {1}{x^{2}}$$ in $$r$$ is found from $$\frac {lcoeff\left ( s\right ) }{lcoeff\left ( t\right ) }$$ which from Eq (5A) above is $$\frac {2}{1}=2$$. Hence $$b=2-0=2.$$\begin {align*} \alpha _{\infty }^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}=\frac {1}{2}+\frac {1}{2}\sqrt {1+8}=2\\ \alpha _{\infty }^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b}=\frac {1}{2}-\frac {1}{2}\sqrt {1+8}=-1 \end {align*}

This completes step 1 of the solution. We have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{\pm }$$ and found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{\pm }$$. Now we go to step 2 which is to ﬁnd the $$d^{\prime }s$$.

step 2 Since we have a pole at $$x=-i$$ and pole at $$x=+i$$ each of order 2, and we have one $$O\left ( \infty \right )$$, each with $$\pm$$ signs. The following now implements $d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{2}\alpha _{c_{i}}^{\pm }$ By trying all possible combinations. There are 8 possible $$d$$ values. This gives\begin {align*} d_{1} & =2-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =2-\left ( \frac {1}{2}+\frac {1}{2}\right ) =1\\ d_{2} & =2-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =2-\left ( \frac {1}{2}+\frac {1}{2}\right ) =1\\ d_{3} & =2-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =2-\left ( \frac {1}{2}+\frac {1}{2}\right ) =1\\ d_{4} & =2-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =2-\left ( \frac {1}{2}+\frac {1}{2}\right ) =1\\ d_{5} & =-1-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =-1-\left ( \frac {1}{2}+\frac {1}{2}\right ) =-2\\ d_{6} & =-1-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =-1-\left ( \frac {1}{2}+\frac {1}{2}\right ) =-2\\ d_{7} & =-1-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =-1-\left ( \frac {1}{2}+\frac {1}{2}\right ) =-2\\ d_{8} & =-1-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =-1-\left ( \frac {1}{2}+\frac {1}{2}\right ) =-2 \end {align*}

Need to complete the solution next.

##### 3.2.9 Example 9

Let $\left ( 1-x\right ) y^{\prime \prime }+xy^{\prime }-y=0$ Normalizing so that coeﬃcient of $$y^{\prime \prime }$$ is one gives\begin {align} y^{\prime \prime }+\frac {x}{\left ( 1-x\right ) }y^{\prime }-\frac {1}{\left ( 1-x\right ) }y & =0\nonumber \\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0 \tag {1} \end {align}

Hence \begin {align*} a & =\frac {x}{\left ( 1-x\right ) }\\ b & =-\frac {1}{\left ( 1-x\right ) } \end {align*}

It is ﬁrst transformed to the following ode by eliminating the ﬁrst derivative \begin {equation} z^{\prime \prime }=rz \tag {2} \end {equation} Using what is known as the Liouville transformation given by\begin {equation} y=ze^{\frac {-1}{2}\int adx} \tag {3} \end {equation} Where it can be found that $$r$$ in (2) is given by \begin {align} r & =\frac {1}{4}a+\frac {1}{2}a^{\prime }-b\nonumber \\ & =\frac {1}{4}\left ( \frac {x}{\left ( 1-x\right ) }\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {x}{\left ( 1-x\right ) }\right ) -\left ( -\frac {1}{\left ( 1-x\right ) }\right ) \nonumber \\ & =\frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}} \tag {4} \end {align}

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is \begin {equation} z^{\prime \prime }=\frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}z \tag {5} \end {equation} Therefore \begin {align} r & =\frac {s}{t}\nonumber \\ & =\frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}\nonumber \\ & =\frac {x^{2}-4x+6}{4x^{2}-8x+4} \tag {5A} \end {align}

The necessary conditions for case 1 are met.

Step 1 In this we ﬁnd all $$\left [ \sqrt {r}\right ] _{c}$$ and associated $$\alpha _{c}^{\pm }$$ for each pole. There is one pole at $$x=1$$ of order 2. Hence\begin {align*} \left [ \sqrt {r}\right ] _{c_{1}} & =0\\ \alpha _{c_{1}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c_{1}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end {align*}

Where $$b$$ is the coeﬃcient of $$\frac {1}{\left ( x-1\right ) ^{2}}$$ in the partial fraction decomposition of $$r$$ which is (in Maple this can be found using fullparfrac). \begin {equation} \frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}=\frac {1}{4}+\frac {3}{4\left ( x-1\right ) ^{2}}-\frac {1}{2}\frac {1}{x-1}\tag {6} \end {equation} Hence $$b=\frac {3}{4}$$. Therefore\begin {align*} \left [ \sqrt {r}\right ] _{c_{1}} & =0\\ \alpha _{c_{1}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {3}{4}\right ) }=\frac {3}{2}\\ \alpha _{c_{1}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {3}{4}\right ) }=-\frac {1}{2} \end {align*}

We are done with all the poles. Now we consider $$O\left ( \infty \right )$$ which is $$\deg \left ( t\right ) -\deg \left ( s\right ) =2-2=0$$. Since $$O\left ( \infty \right ) =0$$, we are in case $$2v\leq 0$$. Hence $v=0$ Then now $$\left [ \sqrt {r}\right ] _{\infty }$$ is the sum of all terms $$x^{i}$$ for $$0\leq i\leq v$$ in the Laurent series expansion of $$\sqrt {r}$$ at $$\infty$$ which is \begin {equation} \left [ \sqrt {r}\right ] _{\infty }=\frac {1}{2}-\frac {1}{2x}+\frac {1}{x^{3}}+\frac {11}{4x^{4}}+\cdots \tag {7} \end {equation} We want only terms for $$0\leq i\leq v$$ but $$v=0$$. Therefore only the constant term. Hence\begin {equation} \left [ \sqrt {r}\right ] _{\infty }=\frac {1}{2}\tag {8} \end {equation} Which means $a=\frac {1}{2}$ As it is the term that matches $$\left [ \sqrt {r}\right ] _{\infty }=ax^{v}+\cdots$$. $$\$$ Now we need to ﬁnd $$b$$. This will be the coeﬃcient of $$x^{v-1}=\frac {1}{x}$$ in $$r-\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}$$. But $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}=\frac {1}{4}$$. So coeﬃcient of $$\frac {1}{x}$$ is zero in $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}$$.  To ﬁnd the coeﬃcient of $$\frac {1}{x}$$ in $$r$$ long division is done\begin {align*} r & =\frac {s}{t}\\ & =\frac {x^{2}-4x+6}{4x^{2}-8x+4}\\ & =Q+\frac {R}{4x^{2}-8x+4} \end {align*}

Where $$Q$$ is the quotient and $$R$$ is the remainder. This gives$r=\frac {1}{4}+\frac {-2x+5}{4x^{2}-8x+4}$ Hence the coeﬃcient of $$\frac {1}{x}$$ is $$\frac {lcoeff\left ( R\right ) }{lcoeff\left ( t\right ) }=\frac {-2}{4}=-\frac {1}{2}$$. Therefore $$b=-\frac {1}{2}-0=-\frac {1}{2}$$. Hence\begin {align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =\frac {1}{2}\left ( \frac {-\frac {1}{2}}{\frac {1}{2}}-0\right ) =-\frac {1}{2}\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( -\frac {-\frac {1}{2}}{\frac {1}{2}}-0\right ) =\frac {1}{2} \end {align*}

This completes step 1 of the solution. We have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{\pm }$$ and found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{\pm }$$. Now we go to step 2 which is to ﬁnd the $$d^{\prime }s$$.

step 2 Now $$d$$ is found using$d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{1}\alpha _{c_{i}}^{\pm }$ By trying all possible combinations. There are 4 possible $$d$$ values. This gives\begin {align*} d_{1} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{+}\right ) =-\frac {1}{2}-\frac {3}{2}=-2\\ d_{2} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{-}\right ) =-\frac {1}{2}-\left ( -\frac {1}{2}\right ) =0\\ d_{3} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{+}\right ) =\frac {1}{2}-\frac {3}{2}=-1\\ d_{4} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{-}\right ) =\frac {1}{2}-\left ( -\frac {1}{2}\right ) =1 \end {align*}

Using entry $$d=1$$ entry above now we ﬁnd $$\omega$$ using$\omega =\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}^{s\left ( c\right ) }}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$ Hence\begin {align*} \omega & =\left ( 0+\frac {\alpha _{c_{1}}^{-}}{x-c_{1}}\right ) +\left ( -\right ) \frac {1}{2}\\ & =\frac {-\frac {1}{2}}{x-1}-\frac {1}{2}\\ & =-\frac {1}{2}\frac {1}{x-1}-\frac {1}{2} \end {align*}

If this $$\omega$$ fails to ﬁnd $$p\left ( x\right )$$, then we will try the entry $$d=0$$. Which will give$\omega =-\frac {1}{2}\frac {1}{x-1}+\frac {1}{2}$ Will ﬁnish the solution later.

##### 3.2.10 Example 10

Let \begin {align} y^{\prime \prime }-x^{2}y^{\prime }-3xy & =0\tag {1}\\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end {align}

Hence \begin {align*} a & =-x^{2}\\ b & =-3x \end {align*}

It is ﬁrst transformed to the following ode by eliminating the ﬁrst derivative \begin {equation} z^{\prime \prime }=rz \tag {2} \end {equation} Using what is known as the Liouville transformation given by\begin {equation} y=ze^{\frac {-1}{2}\int adx} \tag {3} \end {equation} Where it can be found that $$r$$ in (2) is given by \begin {align} r & =\frac {1}{4}a^{2}+\frac {1}{2}a^{\prime }-b\nonumber \\ & =\frac {1}{4}\left ( -x^{2}\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( -x^{2}\right ) -\left ( -3x\right ) \nonumber \\ & =\frac {1}{4}x^{4}+2x \tag {4} \end {align}

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is \begin {equation} z^{\prime \prime }=\left ( \frac {1}{4}x^{4}+2x\right ) z \tag {5} \end {equation} Therefore \begin {align} r & =\frac {s}{t}\nonumber \\ & =\frac {1}{4}x^{4}+2x\nonumber \end {align}

The necessary conditions for case 1 are met.

Step 1 In this we ﬁnd all $$\left [ \sqrt {r}\right ] _{c}$$ and associated $$\alpha _{c}^{\pm }$$ for each pole. There are no poles. Hence set $$\Gamma$$ of poles is empty. Now we consider $$O\left ( \infty \right )$$ which is $$\deg \left ( t\right ) -\deg \left ( s\right ) =0-4=-4$$. We are in case $$2v\leq 0$$. Hence $$-2v=-4$$ or $v=2$ Then now $$\left [ \sqrt {r}\right ] _{\infty }$$ is the sum of all terms $$x^{i}$$ for $$0\leq i\leq v$$ in the Laurent series expansion of $$\sqrt {r}$$ at $$\infty$$ which is \begin {equation} \left [ \sqrt {r}\right ] _{\infty }=\frac {x^{2}}{2}+\frac {2}{x}-\frac {4}{x^{4}}+\frac {16}{x^{7}}-\cdots \tag {7} \end {equation} We want only terms for $$0\leq i\leq v$$ but $$v=2$$. Therefore need to sum terms $$x^{0},x^{1},x^{2}$$. From the above we see that\begin {align*} \left [ \sqrt {r}\right ] _{\infty } & =\frac {x^{2}}{2}+0x^{1}+0x^{0}\\ & =\frac {x^{2}}{2} \end {align*}

Which means $a=\frac {1}{2}$ As that is the term which matches $$\left [ \sqrt {r}\right ] _{\infty }=ax^{2}+\cdots$$. $$\$$ Now we need to ﬁnd $$b$$. This will be the coeﬃcient of $$x^{v-1}=x$$ in $$r$$ minus coeﬃcient of $$x$$ in$$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}$$. But $\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}=\frac {x^{4}}{4}$ Hence the coeﬃcient is zero here. Now we ﬁnd coeﬃcient of $$x$$ in $$r$$. But $$r=$$ $$\frac {1}{4}x^{4}+2x$$ hence the coeﬃcient of $$x$$ is $$2$$. Therefore\begin {align*} b & =2-0\\ & =2 \end {align*}

Hence\begin {align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =\frac {1}{2}\left ( \frac {2}{\frac {1}{2}}-2\right ) =1\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( -\frac {2}{\frac {1}{2}}-2\right ) =-3 \end {align*}

This completes step 1 of the solution. We have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{\pm }$$ and found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{\pm }$$. Now we go to step 2 which is to ﬁnd the $$d^{\prime }s$$.

step 2 Now $$d$$ is found using$d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{0}\alpha _{c_{i}}^{\pm }$ By trying all possible combinations. There are 4 possible $$d$$ values. This gives\begin {align*} d_{1} & =\alpha _{\infty }^{+}=1\\ d_{2} & =\alpha _{\infty }^{-}=-3 \end {align*}

Using $$d=1$$ entry above now we ﬁnd $$\omega$$ using$\omega =\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}^{s\left ( c\right ) }}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$ Hence, since there are no poles, only last term above survives giving$\omega =s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }=\left ( +\right ) \frac {x^{2}}{2}=\frac {x^{2}}{2}$ Will ﬁnish the solution next.

##### 3.2.11 Example 11

Let \begin {align} y^{\prime \prime }-\frac {2}{5}xy^{\prime }+2y & =0\tag {1}\\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end {align}

Hence \begin {align*} a & =-\frac {2}{5}x\\ b & =2 \end {align*}

It is ﬁrst transformed to the following ode by eliminating the ﬁrst derivative \begin {equation} z^{\prime \prime }=rz \tag {2} \end {equation} Using what is known as the Liouville transformation given by\begin {equation} y=ze^{\frac {-1}{2}\int adx} \tag {3} \end {equation} Where it can be found that $$r$$ in (2) is given by \begin {align} r & =\frac {1}{4}a^{2}+\frac {1}{2}a^{\prime }-b\nonumber \\ & =\frac {1}{4}\left ( -\frac {2}{5}x\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( -\frac {2}{5}x\right ) -\left ( 2\right ) \nonumber \\ & =\frac {x^{2}-55}{25} \tag {4} \end {align}

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is \begin {equation} z^{\prime \prime }=\left ( \frac {x^{2}-55}{25}\right ) z \tag {5} \end {equation} Therefore \begin {align} r & =\frac {s}{t}\nonumber \\ & =\frac {x^{2}-55}{25}\nonumber \end {align}

The necessary conditions for case 1 are met.

Step 1 In this we ﬁnd all $$\left [ \sqrt {r}\right ] _{c}$$ and associated $$\alpha _{c}^{\pm }$$ for each pole. There are no poles. Hence set $$\Gamma$$ of poles is empty. Now we consider $$O\left ( \infty \right )$$ which is $$\deg \left ( t\right ) -\deg \left ( s\right ) =0-2=-2$$. We are in case $$2v\leq 0$$. Hence $$-2v=-2$$ or $v=1$ Then now $$\left [ \sqrt {r}\right ] _{\infty }$$ is the sum of all terms $$x^{i}$$ for $$0\leq i\leq v$$ in the Laurent series expansion of $$\sqrt {r}$$ at $$\infty$$ which is \begin {equation} \left [ \sqrt {r}\right ] _{\infty }=\frac {x}{5}-\frac {11}{2}\frac {1}{x}-\cdots \tag {7} \end {equation} We want only terms for $$0\leq i\leq v$$ but $$v=1$$. Therefore need to sum terms $$x^{0},x^{1}$$. From the above we see that$\left [ \sqrt {r}\right ] _{\infty }=\frac {x}{5}$ Which means $a=\frac {1}{5}$ As it is the the term that matches $$\left [ \sqrt {r}\right ] _{\infty }=ax$$. $$\$$ Now we need to ﬁnd $$b$$. This will be the coeﬃcient of $$x^{v-1}=x^{0}$$ in $$r$$ minus the coeﬃcient of $$x^{0}$$ in $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}$$. But \begin {align*} \left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2} & =\left ( \frac {x}{5}\right ) ^{2}\\ & =\frac {x^{2}}{25} \end {align*}

Hence the coeﬃcient of $$x^{0}$$ is zero. Now we ﬁnd coeﬃcient of $$x^{0}$$ in $$r$$. Since $$r=\frac {x^{2}}{25}-\frac {55}{25}$$ then coeﬃcient of $$x^{0}$$ is $$\frac {-55}{25}=-\frac {11}{5}$$. Hence $$b=-\frac {11}{5}-0=-\frac {11}{5}$$. Therefore\begin {align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =\frac {1}{2}\left ( \frac {-\frac {11}{5}}{\frac {1}{5}}-1\right ) =-6\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( -\frac {-\frac {11}{5}}{\frac {1}{5}}-1\right ) =5 \end {align*}

This completes step 1 of the solution. We have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{\pm }$$ and found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{\pm }$$. Now we go to step 2 which is to ﬁnd the $$d^{\prime }s$$.

step 2 Now $$d$$ is found using$d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{0}\alpha _{c_{i}}^{\pm }$ By trying all possible combinations. There are 2 possible $$d$$ values since no poles. \begin {align*} d_{1} & =\alpha _{\infty }^{+}=-6\\ d_{2} & =\alpha _{\infty }^{-}=5 \end {align*}

Using $$d=5$$ entry above now we ﬁnd $$\omega$$ using$\omega =\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}^{s\left ( c\right ) }}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$ Hence, since there are no poles, only last term above survives giving$\omega =s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }=\left ( -\right ) \frac {x}{5}=-\frac {x}{5}$ Will ﬁnish the solution next.

##### 3.2.12 Example 12

Let \begin {align} y^{\prime \prime }+\frac {x^{2}-1}{x}y^{\prime }+x^{2}y & =0\tag {1}\\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end {align}

Hence \begin {align*} a & =\frac {x^{2}-1}{x}\\ b & =x^{2} \end {align*}

It is ﬁrst transformed to the following ode by eliminating the ﬁrst derivative \begin {equation} z^{\prime \prime }=rz \tag {2} \end {equation} Using what is known as the Liouville transformation given by\begin {equation} y=ze^{\frac {-1}{2}\int adx} \tag {3} \end {equation} Where it can be found that $$r$$ in (2) is given by \begin {align} r & =\frac {1}{4}a^{2}+\frac {1}{2}a^{\prime }-b\nonumber \\ & =\frac {1}{4}\left ( \frac {x^{2}-1}{x}\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {x^{2}-1}{x}\right ) -x^{2}\nonumber \\ & =-\frac {3\left ( x^{4}-1\right ) }{4x^{2}} \tag {4} \end {align}

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is \begin {equation} z^{\prime \prime }=-\frac {3\left ( x^{4}-1\right ) }{4x^{2}}z \tag {5} \end {equation} Therefore \begin {align} r & =\frac {s}{t}\nonumber \\ & =\frac {-3\left ( x^{4}-1\right ) }{4x^{2}}\nonumber \end {align}

The necessary conditions for case 1 are met.

Step 1 In this we ﬁnd all $$\left [ \sqrt {r}\right ] _{c}$$ and associated $$\alpha _{c}^{\pm }$$ for each pole. There is one pole at $$x=0$$ of order 2. Hence

\begin {align*} \left [ \sqrt {r}\right ] _{c_{1}} & =0\\ \alpha _{c_{1}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c_{1}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end {align*}

Where $$b$$ is the coeﬃcient of $$\frac {1}{\left ( x-0\right ) ^{2}}$$ in the partial fraction decomposition of $$r$$ which is (in Maple this can be found using fullparfrac). \begin {equation} \frac {-3\left ( x^{4}-1\right ) }{4x^{2}}=-\frac {3}{4}x^{2}+\frac {3}{4}\frac {1}{x^{2}}\tag {6} \end {equation} Hence $$b=\frac {3}{4}$$. Therefore\begin {align*} \left [ \sqrt {r}\right ] _{c_{1}} & =0\\ \alpha _{c_{1}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {3}{4}\right ) }=\frac {3}{2}\\ \alpha _{c_{1}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {3}{4}\right ) }=-\frac {1}{2} \end {align*}

Now we consider $$O\left ( \infty \right )$$ which is $$\deg \left ( t\right ) -\deg \left ( s\right ) =2-4=-2$$. We are in case $$2v\leq 0$$. Hence $$-2v=-2$$ or $v=1$ Then now $$\left [ \sqrt {r}\right ] _{\infty }$$ is the sum of all terms $$x^{i}$$ for $$0\leq i\leq v$$ in the Laurent series expansion of $$\sqrt {r}$$ at $$\infty$$ which is \begin {equation} \left [ \sqrt {r}\right ] _{\infty }=\frac {i\sqrt {3}}{2}x-\frac {i\sqrt {3}}{4}\frac {1}{x^{3}}+\cdots \tag {7} \end {equation} We want only terms for $$0\leq i\leq v$$ but $$v=1$$. Therefore need to sum terms $$x^{0},x^{1}$$. From the above we see that$\left [ \sqrt {r}\right ] _{\infty }=\frac {i\sqrt {3}}{2}x$ Which means $a=\frac {i\sqrt {3}}{2}$ As it is the the term that matches $$\left [ \sqrt {r}\right ] _{\infty }=ax$$. $$\$$ Now we need to ﬁnd $$b$$. This will be the coeﬃcient of $$x^{v-1}=x^{0}$$ in $$r$$ minus the coeﬃcient of $$x^{0}$$ in $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}$$. But \begin {align*} \left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2} & =\left ( \frac {i\sqrt {3}}{2}x\right ) ^{2}\\ & =-\frac {3}{4}x^{2} \end {align*}

Hence the coeﬃcient is zero here. Now we ﬁnd coeﬃcient of $$x^{0}$$ in $$r$$. Since $$r=-\frac {3}{4}x^{2}+\frac {3}{4}\frac {1}{x^{2}}$$ then coeﬃcient of $$x^{0}$$ is zero also. Hence $$b=0-0=0$$. Therefore\begin {align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =\frac {1}{2}\left ( 0-1\right ) =-\frac {1}{2}\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( 0-1\right ) =-\frac {1}{2} \end {align*}

This completes step 1 of the solution. We have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{\pm }$$ and found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{\pm }$$. Now we go to step 2 which is to ﬁnd the $$d^{\prime }s$$.

step 2 Now $$d$$ is found using$d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{0}\alpha _{c_{i}}^{\pm }$ By trying all possible combinations. There are 4 possible $$d$$ values. This gives\begin {align*} d_{1} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{+}\right ) =-\frac {1}{2}-\frac {3}{2}=-1\\ d_{2} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{-}\right ) =-\frac {1}{2}-\left ( -\frac {1}{2}\right ) =0\\ d_{3} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{+}\right ) =-\frac {1}{2}-\frac {3}{2}=-2\\ d_{4} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{-}\right ) =-\frac {1}{2}-\left ( -\frac {1}{2}\right ) =0 \end {align*}

Using ﬁrst $$d=0$$ entry above now we ﬁnd $$\omega$$ using$\omega =\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}^{s\left ( c\right ) }}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$ Hence$\omega =\left ( -1\right ) \left ( 0\right ) +\frac {-\frac {1}{2}}{x-0}+\left ( +\right ) \left ( \frac {i\sqrt {3}}{2}x\right ) =\frac {-1}{2x}+\frac {i\sqrt {3}}{2}x$ Notice that if have taken the last $$d=0$$ entry, we will get$\omega =\left ( -1\right ) \left ( 0\right ) +\frac {-\frac {1}{2}}{x-0}+\left ( -\right ) \left ( \frac {i\sqrt {3}}{2}x\right ) =\frac {-1}{2x}-\frac {i\sqrt {3}}{2}x$ In practice, we will try the second one if the ﬁrst fails. Will ﬁnish the solution next.

##### 3.2.13 Example 13

Let \begin {align} \left ( 1-x\right ) y^{\prime \prime }+xy^{\prime }-y & =0\tag {1}\\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end {align}

Hence \begin {align*} a & =\frac {x}{1-x}\\ b & =-\frac {1}{1-x} \end {align*}

It is ﬁrst transformed to the following ode by eliminating the ﬁrst derivative \begin {equation} z^{\prime \prime }=rz \tag {2} \end {equation} Using what is known as the Liouville transformation given by\begin {equation} y=ze^{\frac {-1}{2}\int adx} \tag {3} \end {equation} Where it can be found that $$r$$ in (2) is given by \begin {align} r & =\frac {1}{4}a^{2}+\frac {1}{2}a^{\prime }-b\nonumber \\ & =\frac {1}{4}\left ( \frac {x}{1-x}\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {x}{1-x}\right ) -\left ( -\frac {1}{1-x}\right ) \nonumber \\ & =\frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}} \tag {4} \end {align}

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is \begin {equation} z^{\prime \prime }=\frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}z \tag {5} \end {equation} Therefore \begin {align} r & =\frac {s}{t}\nonumber \\ & =\frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}\nonumber \end {align}

The necessary conditions for case 1 are met.

Step 1 In this we ﬁnd all $$\left [ \sqrt {r}\right ] _{c}$$ and associated $$\alpha _{c}^{\pm }$$ for each pole. There is one pole at $$x=1$$ of order 2. Hence

\begin {align*} \left [ \sqrt {r}\right ] _{c_{1}} & =0\\ \alpha _{c_{1}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c_{1}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end {align*}

Where $$b$$ is the coeﬃcient of $$\frac {1}{\left ( x-1\right ) ^{2}}$$ in the partial fraction decomposition of $$r$$ which is \begin {equation} \frac {x^{2}-4x+6}{4\left ( x-1\right ) ^{2}}=\frac {1}{4}+\frac {3}{4\left ( x-1\right ) ^{2}}-\frac {1}{2}\frac {1}{x-1}\tag {6} \end {equation} Hence $$b=\frac {3}{4}$$. Therefore\begin {align*} \left [ \sqrt {r}\right ] _{c_{1}} & =0\\ \alpha _{c_{1}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {3}{4}\right ) }=\frac {3}{2}\\ \alpha _{c_{1}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {3}{4}\right ) }=-\frac {1}{2} \end {align*}

Now we consider $$O\left ( \infty \right )$$ which is $$\deg \left ( t\right ) -\deg \left ( s\right ) =2-2=0$$. We are in case $$2v\leq 0$$. Hence $$-2v=0$$ or $v=0$ Then now $$\left [ \sqrt {r}\right ] _{\infty }$$ is the sum of all terms $$x^{i}$$ for $$0\leq i\leq v$$ in the Laurent series expansion of $$\sqrt {r}$$ at $$\infty$$ which is \begin {equation} \left [ \sqrt {r}\right ] _{\infty }=\frac {1}{2}-\frac {1}{2x}+\frac {1}{x^{3}}+\cdots \tag {7} \end {equation} But we want only terms for $$0\leq i\leq v$$ but $$v=0$$. Therefore need to sum terms $$x^{0}$$. Which is the constant term\begin {equation} \left [ \sqrt {r}\right ] _{\infty }=\frac {1}{2}\tag {8} \end {equation} Which means $a=\frac {1}{2}$ Now we need to ﬁnd $$b$$. Which is given by the coeﬃcient of $$\frac {1}{x}$$ in $$r$$ minus coeﬃcient of $$\frac {1}{x}$$ in $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}$$. But $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}=\frac {1}{4}$$ Hence the coeﬃcient is zero here.  To ﬁnd the coeﬃcient of $$\frac {1}{x}$$ in $$r$$ long division is done (here paper is not clear at all what it means by coeﬃcient of $$x^{v-1}$$ in $$r$$ as that depends on the form of $$r$$ and how it is represented). This method of using long division to ﬁnd the coeﬃcient works to obtain the correct result. But it is still not clear what the paper actually means by this.\begin {align*} r & =\frac {s}{t}\\ & =\frac {x^{2}-4x+6}{4x^{2}-8x+4}\\ & =Q+\frac {R}{4x^{2}-8x+4} \end {align*}

Where $$Q$$ is the quotient and $$R$$ is the remainder. This gives$r=\frac {1}{4}+\frac {-2x+5}{4x^{2}-8x+4}$ For the case of $$v=0$$ then the coeﬃcient of $$x^{-1}$$ is  $$\frac {lcoeff\left ( R\right ) }{lcoeff\left ( t\right ) }=\frac {-2}{4}=-\frac {1}{2}$$. Notice that if we just expanded $$r$$ it will give $$\frac {x^{2}}{4\left ( x-1\right ) ^{2}}-\frac {x}{\left ( x-1\right ) ^{2}}+\frac {3}{2\left ( x-1\right ) ^{2}}$$ and we see there is no coeﬃcient of $$\frac {1}{x}$$ in this representation. So we would have obtain wrong value of $$b$$ if we just used what the paper said. Now $$b=-\frac {1}{2}-0=-\frac {1}{2}$$. Therefore \begin {align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =\frac {1}{2}\left ( \frac {\frac {1}{2}}{\frac {1}{2}}-0\right ) =\frac {1}{2}\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( -\frac {\frac {1}{2}}{\frac {1}{2}}-0\right ) =-\frac {1}{2} \end {align*}

This completes step 1 of the solution. We have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{\pm }$$ and found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{\pm }$$. Now we go to step 2 which is to ﬁnd the $$d^{\prime }s$$.

step 2 Now $$d$$ is found using$d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{1}\alpha _{c_{i}}^{\pm }$ By trying all possible combinations. There are 4 possible $$d$$ values. This gives\begin {align*} d_{1} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{+}\right ) =\frac {1}{2}-\left ( \frac {3}{2}\right ) =-1\\ d_{2} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{-}\right ) =\frac {1}{2}-\left ( -\frac {1}{2}\right ) =1\\ d_{3} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{+}\right ) =-\frac {1}{2}-\left ( \frac {3}{2}\right ) =-2\\ d_{4} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{-}\right ) =-\frac {1}{2}-\left ( -\frac {1}{2}\right ) =0 \end {align*}

Using ﬁrst $$d=1$$ entry from above we ﬁnd $$\omega$$ using$\omega =\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}^{s\left ( c\right ) }}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$ Hence$\omega =\left ( \left ( -1\right ) \left ( 0\right ) +\frac {-\frac {1}{2}}{x-1}\right ) +\left ( +\right ) \left ( \frac {1}{2}\right ) =\frac {-1}{2(x-1)}+\frac {1}{2}$ And if use the last entry $$d=0$$ then$\omega \left ( =\left ( -1\right ) \left ( 0\right ) +\frac {-\frac {1}{2}}{x-1}\right ) +\left ( -\right ) \left ( \frac {1}{2}x\right ) =\frac {-1}{2(x-1)}-\frac {1}{2}$ In practice, we will try the second one if the ﬁrst fails. Will ﬁnish the solution next.

##### 3.2.14 Example 14

Let \begin {align} 3y^{\prime \prime }+xy^{\prime }-4y & =0\tag {1}\\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end {align}

Hence \begin {align*} a & =\frac {x}{3}\\ b & =-\frac {4}{3} \end {align*}

It is ﬁrst transformed to the following ode by eliminating the ﬁrst derivative \begin {equation} z^{\prime \prime }=rz \tag {2} \end {equation} Using what is known as the Liouville transformation given by\begin {equation} y=ze^{\frac {-1}{2}\int adx} \tag {3} \end {equation} Where it can be found that $$r$$ in (2) is given by \begin {align} r & =\frac {1}{4}a^{2}+\frac {1}{2}a^{\prime }-b\nonumber \\ & =\frac {1}{4}\left ( \frac {x}{3}\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {x}{3}\right ) -\left ( -\frac {4}{3}\right ) \nonumber \\ & =\frac {x^{2}}{36}+\frac {3}{2} \tag {4} \end {align}

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is \begin {equation} z^{\prime \prime }=\frac {x^{2}+54}{36}z \tag {5} \end {equation} Therefore \begin {align} r & =\frac {s}{t}\nonumber \\ & =\frac {x^{2}+54}{36}\nonumber \end {align}

The necessary conditions for case 1 are met.

Step 1 In this we ﬁnd all $$\left [ \sqrt {r}\right ] _{c}$$ and associated $$\alpha _{c}^{\pm }$$ for each pole. There are no poles. Hence $$\Gamma$$ is empty.

Now we consider $$O\left ( \infty \right )$$ which is $$\deg \left ( t\right ) -\deg \left ( s\right ) =0-2=-2$$. We are in case $$2v\leq 0$$. Hence $$-2v=-2$$ or $v=1$ Then now $$\left [ \sqrt {r}\right ] _{\infty }$$ is the sum of all terms $$x^{i}$$ for $$0\leq i\leq v$$ in the Laurent series expansion of $$\sqrt {r}$$ at $$\infty$$ which is \begin {equation} \left [ \sqrt {r}\right ] _{\infty }=\frac {x}{6}+\frac {9}{2}\frac {1}{x}+\cdots \tag {7} \end {equation} But we want only terms for $$0\leq i\leq v$$ but $$v=1$$. Therefore need to sum terms $$x^{0},x^{1}$$. Therefore\begin {equation} \left [ \sqrt {r}\right ] _{\infty }=\frac {x}{6}\tag {8} \end {equation} Which means $a=\frac {1}{6}$ Now we need to ﬁnd $$b$$. Which is given by the coeﬃcient of $$x^{v-1}=x^{0}$$ or the constant term in $$r$$ minus coeﬃcient of $$x^{0}$$ in $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}$$. But $$\left ( \left [ \sqrt {r}\right ] _{\infty }\right ) ^{2}=\frac {x^{2}}{36}$$. Hence the coeﬃcient of $$x^{0}$$ is zero here. Now we ﬁnd coeﬃcient of $$x^{0}$$ in $$r$$. Since $$r=\frac {x^{2}}{36}+\frac {54}{36}$$ then the coeﬃcient of $$x^{0}$$ is $$\frac {54}{36}=\frac {3}{2}$$. Hence $$b=\frac {3}{2}-0=\frac {3}{2}$$. Therefore\begin {align*} \alpha _{\infty }^{+} & =\frac {1}{2}\left ( \frac {b}{a}-v\right ) =\frac {1}{2}\left ( \frac {\frac {3}{2}}{\frac {1}{6}}-1\right ) =4\\ \alpha _{\infty }^{-} & =\frac {1}{2}\left ( -\frac {b}{a}-v\right ) =\frac {1}{2}\left ( -\frac {\frac {3}{2}}{\frac {1}{6}}-1\right ) =-5 \end {align*}

This completes step 1 of the solution. We have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{\pm }$$ and found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{\pm }$$. Now we go to step 2 which is to ﬁnd the $$d^{\prime }s$$.

step 2 Now $$d$$ is found using$d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{0}\alpha _{c_{i}}^{\pm }$ By trying all possible combinations. There are 2 possible $$d$$ values (since no poles). This gives\begin {align*} d_{1} & =\alpha _{\infty }^{+}=4\\ d_{2} & =\alpha _{\infty }^{-}=-5 \end {align*}

Using $$d=4$$ entry from above we ﬁnd $$\omega$$ using$\omega =\left ( \sum _{c}s\left ( c\right ) \left [ \sqrt {r}\right ] _{c}+\frac {\alpha _{c}^{s\left ( c\right ) }}{x-c}\right ) +s\left ( \infty \right ) \left [ \sqrt {r}\right ] _{\infty }$ Hence$\omega =\left ( 0\right ) +\left ( +\right ) \left ( \frac {x}{6}\right ) =\frac {x}{6}$

##### 3.2.15 Example 15

Let \begin {align} \left ( 4-x^{2}\right ) y^{\prime \prime }+xy^{\prime }+2y & =0\tag {1}\\ y^{\prime \prime }+ay^{\prime }\left ( x\right ) +by & =0\nonumber \end {align}

Hence \begin {align*} a & =\frac {x}{\left ( 4-x^{2}\right ) }\\ b & =\frac {2}{\left ( 4-x^{2}\right ) } \end {align*}

It is ﬁrst transformed to the following ode by eliminating the ﬁrst derivative \begin {equation} z^{\prime \prime }=rz \tag {2} \end {equation} Using what is known as the Liouville transformation given by\begin {equation} y=ze^{\frac {-1}{2}\int adx} \tag {3} \end {equation} Where it can be found that $$r$$ in (2) is given by \begin {align} r & =\frac {1}{4}a^{2}+\frac {1}{2}a^{\prime }-b\nonumber \\ & =\frac {1}{4}\left ( \frac {x}{\left ( 4-x^{2}\right ) }\right ) ^{2}+\frac {1}{2}\frac {d}{dx}\left ( \frac {x}{\left ( 4-x^{2}\right ) }\right ) -\left ( \frac {2}{\left ( 4-x^{2}\right ) }\right ) \nonumber \\ & =\frac {11x^{2}-24}{4\left ( x^{2}-4\right ) ^{2}} \tag {4} \end {align}

Hence the DE we will solve using Kovacic algorithm is Eq (2) which is \begin {equation} z^{\prime \prime }=\frac {11x^{2}-24}{4\left ( x^{2}-4\right ) ^{2}}z \tag {5} \end {equation} Therefore \begin {align} r & =\frac {s}{t}\nonumber \\ & =\frac {11x^{2}-24}{4\left ( x^{2}-4\right ) ^{2}}=\frac {11x^{2}-24}{4x^{4}-32x^{2}+64} \tag {5A} \end {align}

The necessary conditions for case 1 are met.

Step 1 In this we ﬁnd all $$\left [ \sqrt {r}\right ] _{c}$$ and associated $$\alpha _{c}^{\pm }$$ for each pole. There are two poles at $$\pm 2$$ each of order 2. For pole at $$x=2=c_{1}$$

\begin {align*} \left [ \sqrt {r}\right ] _{c_{1}} & =0\\ \alpha _{c_{1}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4b}\\ \alpha _{c_{1}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4b} \end {align*}

Where $$b$$ is the coeﬃcient of $$\frac {1}{\left ( x-2\right ) ^{2}}$$ in the partial fraction decomposition of $$r$$ which is \begin {equation} \frac {11x^{2}-24}{4\left ( x^{2}-4\right ) ^{2}}=\frac {5}{16}\frac {1}{\left ( x+2\right ) ^{2}}+\frac {5}{16}\frac {1}{\left ( x-2\right ) ^{2}}-\frac {17}{32}\frac {1}{\left ( x+2\right ) }+\frac {17}{32}\frac {1}{\left ( x-2\right ) }\tag {6} \end {equation} Hence $$b=\frac {5}{16}$$. Therefore\begin {align*} \left [ \sqrt {r}\right ] _{c_{1}} & =0\\ \alpha _{c_{1}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {5}{16}\right ) }=\frac {5}{4}\\ \alpha _{c_{1}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {5}{16}\right ) }=-\frac {1}{4} \end {align*}

And for pole at $$x=-2=c_{2}$$\begin {align*} \left [ \sqrt {r}\right ] _{c_{2}} & =0\\ \alpha _{c_{2}}^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {5}{16}\right ) }=\frac {5}{4}\\ \alpha _{c_{2}}^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {5}{16}\right ) }=-\frac {1}{4} \end {align*}

Now we consider $$O\left ( \infty \right )$$ which is $$\deg \left ( t\right ) -\deg \left ( s\right ) =4-2=2$$. Therefore $$v=1$$. In this case \begin {equation} \left [ \sqrt {r}\right ] _{\infty }=0\tag {7} \end {equation} And $\alpha _{\infty }^{\pm }=\frac {1}{2}\pm \frac {1}{2}\sqrt {1+4b}$ The coeﬃcient of $$x^{v-1}=x^{0}$$ is zero in $$\left [ \sqrt {r}\right ] _{\infty }^{2}$$. To ﬁnd coeﬃcient of $$x^{0}$$ in $$r=\frac {11x^{2}-24}{4\left ( x^{2}-4\right ) ^{2}}$$ and since $$v=0$$ then using $$b=\frac {lcoeff\left ( s\right ) }{lcoeff\left ( t\right ) }$$ gives $$\frac {11}{4}$$. Hence $$b=\frac {11}{4}-0=\frac {11}{4}$$. Therefore\begin {align*} \alpha _{\infty }^{+} & =\frac {1}{2}+\frac {1}{2}\sqrt {1+4\left ( \frac {11}{4}\right ) }=\frac {1}{2}+\sqrt {3}\\ \alpha _{\infty }^{-} & =\frac {1}{2}-\frac {1}{2}\sqrt {1+4\left ( \frac {11}{4}\right ) }=\frac {1}{2}-\sqrt {3} \end {align*}

This completes step 1 of the solution. We have found $$\left [ \sqrt {r}\right ] _{c}$$ and its associated $$\alpha _{c}^{\pm }$$ and found $$\left [ \sqrt {r}\right ] _{\infty }$$ and its associated $$\alpha _{\infty }^{\pm }$$. Now we go to step 2 which is to ﬁnd the $$d^{\prime }s$$.

step 2 Now $$d$$ is found using$d=\alpha _{\infty }^{\pm }-\sum _{i=1}^{0}\alpha _{c_{i}}^{\pm }$ By trying all possible combinations. There are 8 possible $$d$$ values. These are\begin {align*} d_{1} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =\frac {1}{2}+\sqrt {3}-\left ( \frac {5}{4}+\frac {5}{4}\right ) =\sqrt {3}-2\\ d_{2} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =\frac {1}{2}+\sqrt {3}-\left ( \frac {5}{4}-\frac {1}{4}\right ) =\sqrt {3}-\frac {1}{2}\\ d_{3} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =\frac {1}{2}+\sqrt {3}-\left ( -\frac {1}{4}+\frac {5}{4}\right ) =\sqrt {3}-\frac {1}{2}\\ d_{4} & =\alpha _{\infty }^{+}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =\frac {1}{2}+\sqrt {3}-\left ( -\frac {1}{4}-\frac {1}{4}\right ) =\sqrt {3}+1\\ d_{5} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{+}\right ) =\frac {1}{2}-\sqrt {3}-\left ( \frac {5}{4}+\frac {5}{4}\right ) =-\sqrt {3}-2\\ d_{6} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{+}+\alpha _{c_{2}}^{-}\right ) =\frac {1}{2}-\sqrt {3}-\left ( \frac {5}{4}-\frac {1}{4}\right ) =-\sqrt {3}-\frac {1}{2}\\ d_{7} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{+}\right ) =\frac {1}{2}-\sqrt {3}-\left ( -\frac {1}{4}+\frac {5}{4}\right ) =-\sqrt {3}-\frac {1}{2}\\ d_{8} & =\alpha _{\infty }^{-}-\left ( \alpha _{c_{1}}^{-}+\alpha _{c_{2}}^{-}\right ) =\frac {1}{2}-\sqrt {3}-\left ( -\frac {1}{4}-\frac {1}{4}\right ) =1-\sqrt {3} \end {align*}

There are no $$d\geq 0$$ integers. This means case 1 does not apply. We need to try case 2 now.

#### 3.3 Case two Kovacic algorithm

The following diagram shows the algorithm for case two.