#### 2.30   ODE No. 30

$x^{-a-1} y(x)^2-x^a+y'(x)=0$ Mathematica : cpu = 0.152642 (sec), leaf count = 230

$\left \{\left \{y(x)\to \frac {x^{a+1} \left (-\frac {1}{2} (-1)^{-a} a x^{-\frac {a}{2}-1} \Gamma (1-a) I_{-a}\left (2 \sqrt {x}\right )+\frac {1}{2} (-1)^{-a} x^{-\frac {a}{2}-\frac {1}{2}} \Gamma (1-a) \left (I_{-a-1}\left (2 \sqrt {x}\right )+I_{1-a}\left (2 \sqrt {x}\right )\right )+c_1 \left (\frac {1}{2} x^{-\frac {a}{2}-\frac {1}{2}} \Gamma (a+1) \left (I_{a-1}\left (2 \sqrt {x}\right )+I_{a+1}\left (2 \sqrt {x}\right )\right )-\frac {1}{2} a x^{-\frac {a}{2}-1} \Gamma (a+1) I_a\left (2 \sqrt {x}\right )\right )\right )}{(-1)^{-a} x^{-a/2} \Gamma (1-a) I_{-a}\left (2 \sqrt {x}\right )+c_1 x^{-a/2} \Gamma (a+1) I_a\left (2 \sqrt {x}\right )}\right \}\right \}$ Maple : cpu = 0.061 (sec), leaf count = 54

$\left \{ y \left ( x \right ) ={{x}^{a+1} \left ( -{{\sl K}_{a+1}\left (2\,\sqrt {x}\right )}{\it \_C1}+{{\sl I}_{a+1}\left (2\,\sqrt {x}\right )} \right ) {\frac {1}{\sqrt {x}}} \left ( {{\sl K}_{a}\left (2\,\sqrt {x}\right )}{\it \_C1}+{{\sl I}_{a}\left (2\,\sqrt {x}\right )} \right ) ^{-1}} \right \}$

Hand solution

\begin {align} y^{\prime }+x^{-a-1}y^{2}-x^{a} & =0\nonumber \\ y^{\prime } & =x^{a}-x^{-a-1}y^{2}\nonumber \\ & =P\left ( x\right ) +Q\left ( x\right ) y+R\left ( x\right ) y^{2} \tag {1} \end {align}

This is Ricatti ﬁrst order non-linear ODE. Using standard transformation$y=-\frac {u^{\prime }}{uR\left ( x\right ) }=x^{a+1}\frac {u^{\prime }}{u}$

Hence

$y^{\prime }=\left ( a+1\right ) x^{a}\frac {u^{\prime }}{u}+x^{a+1}\frac {u^{\prime \prime }}{u}-x^{a+1}\frac {\left ( u^{\prime }\right ) ^{2}}{u^{2}}$

Comparing to (1) gives

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

In standard form $$u^{\prime \prime }+\frac {1}{x}\left ( 1+a\right ) u^{\prime }-\frac {1}{x}u=0$$ or $$u^{\prime \prime }+p\left ( x\right ) \left ( 1+a\right ) u^{\prime }+q\left ( x\right ) u=0$$. We see that $$p\left ( x\right )$$ is not analytic at $$x=0$$ (the expansion point). So we can’t use power series solution, and will use Forbenius series. Power series, which is $$u=\sum _{n=0}^{\infty }c_{n}x^{n}$$ is used when the expansion point is not singular point. (i.e. $$p\left ( x\right )$$ and $$q\left ( x\right )$$ are analytic there). Forbenius series $$u=x^{r}\sum _{n=0}^{\infty }c_{n}x^{n}$$ is used when there is a removable singular point (called also regular singular point), as in this case. Starting with$u=x^{r}\sum _{n=0}^{\infty }c_{n}x^{n}=\sum _{n=0}^{\infty }c_{n}x^{n+r}$ Hence \begin {align*} u^{\prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) c_{n}x^{n+r-1}\\ u^{\prime \prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n+r-2} \end {align*}

Substituting in (2) gives\begin {align*} x\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n+r-2}+\left ( 1+a\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) c_{n}x^{n+r-1}-\sum _{n=0}^{\infty }c_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n+r-1}+\left ( 1+a\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) c_{n}x^{n+r-1}-\sum _{n=0}^{\infty }c_{n}x^{n+r} & =0 \end {align*}

Dividing out $$x^{r}$$$\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n-1}+\left ( 1+a\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) c_{n}x^{n-1}-\sum _{n=0}^{\infty }c_{n}x^{n}=0$ Each term should have $$x^{n-1}$$ in it. So we adjust the last term$\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n-1}+\left ( 1+a\right ) \sum _{n=0}^{\infty }\left ( n+r\right ) c_{n}x^{n-1}-\sum _{n=1}^{\infty }c_{n-1}x^{n-1}=0$ Expanding the second term$\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n-1}+\sum _{n=0}^{\infty }\left ( n+r\right ) c_{n}x^{n-1}+\sum _{n=0}^{\infty }a\left ( n+r\right ) c_{n}x^{n-1}-\sum _{n=1}^{\infty }c_{n-1}x^{n-1}=0$ Hence for $$n=0$$\begin {align*} \left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n-1}+\left ( n+r\right ) c_{n}x^{n-1}+a\left ( n+r\right ) c_{n}x^{n-1} & =0\\ r\left ( r-1\right ) c_{0}+rc_{0}+arc_{0} & =0 \end {align*}

Since $$c_{0}\neq 0$$ then$r\left ( r-1\right ) +r+ar=0$ Hence $$r=$$ $$-a$$ or $$r=0$$. Now for $$n\geq 1$$\begin {align*} \ \left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n-1}+\ \left ( n+r\right ) c_{n}x^{n-1}+\ a\left ( n+r\right ) c_{n}x^{n-1}-\ c_{n-1}x^{n-1} & =0\\ \left ( n+r\right ) \left ( n+r-1\right ) c_{n}\ +\ \left ( n+r\right ) c_{n}\ +\ a\left ( n+r\right ) c_{n}\ -\ c_{n-1}\ & =0\\ \left ( \left ( n+r\right ) \left ( n+r-1\right ) \ +\ \left ( n+r\right ) \ +\ a\left ( n+r\right ) \right ) c_{n}\ \ & =c_{n-1}\\ c_{n}\ \ & =\frac {c_{n-1}}{\left ( n+r\right ) \left ( n+r-1\right ) \ +\ \left ( n+r\right ) \ +\ a\left ( n+r\right ) } \end {align*}

For $$r=0$$, we obtain\begin {equation} c_{n}\ \ =\frac {c_{n-1}}{n\left ( n-1\right ) \ +\ n\ +\ an}\tag {3} \end {equation} For $$r=-a$$\begin {equation} c_{n}\ \ =\frac {c_{n-1}}{\left ( n-a\right ) \left ( n-a-1\right ) \ +\ \left ( n-a\right ) \ +\ a\left ( n-a\right ) }\tag {4} \end {equation} There are two solutions. Looking at (3) for now, for $$n=1$$$c_{1}=\frac {c_{0}}{\ 1\ +\ a}$ For $$n=2$$$c_{2}\ \ =\frac {c_{1}}{4\ +2a}=\frac {c_{0}}{\ 1\ +\ a}\frac {1}{2\left ( 2\ +a\right ) }$ For $$n=3$$$c_{3}=\frac {c_{2}}{3\left ( 2\right ) \ +\ 3\ +3a}=\frac {c_{2}}{3\left ( 3\ +a\right ) }=\frac {c_{0}}{\ 1\ +\ a}\frac {1}{2\left ( 2\ +a\right ) }\frac {1}{3\left ( 3\ +a\right ) }$ And so on. Since the solution is assumed to be $$x^{r}\sum _{n=0}^{\infty }c_{n}x^{n}$$ and we are looking at case $$r=0$$ then\begin {align} u_{r=0}\left ( x\right ) & =\sum _{n=1}^{\infty }c_{n}x^{n}\nonumber \\ & =c_{0}+c_{1}x+c_{2}x^{2}+\cdots \nonumber \\ & =c_{0}x^{0}+\frac {c_{0}}{\ 1\ +\ a}x+\frac {c_{0}}{\ 1\ +\ a}\frac {1}{2\left ( 2\ +a\right ) }x^{2}+\frac {c_{0}}{\ 1\ +\ a}\frac {1}{2\left ( 2\ +a\right ) }\frac {1}{3\left ( 3\ +a\right ) }x^{3}+\cdots \nonumber \\ & =c_{0}\left ( x^{0}+\frac {1}{\ 1\ +\ a}x+\frac {1}{\ \left ( 1\ +\ a\right ) }\frac {1}{2\left ( 2\ +a\right ) }x^{2}+\frac {1}{\ \left ( 1\ +\ a\right ) }\frac {1}{2\left ( 2\ +a\right ) }\frac {1}{3\left ( 3\ +a\right ) }x^{3}+\cdots \right ) \tag {5} \end {align}

Since $\Gamma \left ( n\right ) =\left ( n-1\right ) !$ and $a\left ( 1+a\right ) \left ( 2+a\right ) \cdots \left ( n+a\right ) =\frac {\Gamma \left ( a+n+1\right ) }{\Gamma \left ( a\right ) }$ Then$\left ( 1+a\right ) \left ( 2+a\right ) \cdots \left ( n+a\right ) =\frac {\Gamma \left ( a+n+1\right ) }{a\Gamma \left ( a\right ) }$ And (5) can now be written as\begin {equation} y_{r=0}\left ( x\right ) =c_{0}\sum _{n=1}^{\infty }\frac {1}{n!}\frac {a\Gamma \left ( a\right ) }{\Gamma \left ( a+n+1\right ) }x^{n}\tag {6} \end {equation} But modiﬁed Bessel function of ﬁrst kind is $\operatorname {BesselI}\left ( a,z\right ) =\sum _{n=0}^{\infty }\frac {1}{n!}\frac {1}{\Gamma \left ( a+n+1\right ) }\left ( \frac {z}{2}\right ) ^{2n+a}$ So if we let $$z=2\sqrt {x}$$ we obtain\begin {align*} \operatorname {BesselI}\left ( a,2\sqrt {x}\right ) & =\sum _{n=0}^{\infty }\frac {1}{n!}\frac {1}{\Gamma \left ( a+n+1\right ) }\left ( \frac {2\sqrt {x}}{2}\right ) ^{2n+a}\\ & =\sum _{n=0}^{\infty }\frac {1}{n!}\frac {1}{\Gamma \left ( a+n+1\right ) }\left ( \sqrt {x}\right ) ^{2n}\left ( \sqrt {x}\right ) ^{a}\\ & =\sum _{n=0}^{\infty }\frac {1}{n!}\frac {1}{\Gamma \left ( a+n+1\right ) }x^{n}\left ( \sqrt {x}\right ) ^{a} \end {align*}

Hence\begin {equation} \frac {1}{\sqrt {x^{a}}}\operatorname {BesselI}\left ( a,2\sqrt {x}\right ) =\sum _{n=0}^{\infty }\frac {1}{n!}\frac {1}{\Gamma \left ( a+n+1\right ) }x^{n}\tag {7} \end {equation} If we now compare (6) and (7), we see that if we set $$c_{0}$$, which is arbitrary, to be $$c_{0}=\frac {1}{a\Gamma \left ( a\right ) }$$, then we obtain\begin {align*} u_{r=0}\left ( x\right ) & =\frac {1}{a\Gamma \left ( a\right ) }\sum _{n=0}^{\infty }\frac {1}{n!}\frac {a\Gamma \left ( a\right ) }{\Gamma \left ( a+n+1\right ) }x^{n}\\ & =\sum _{n=0}^{\infty }\frac {1}{n!}\frac {1}{\Gamma \left ( a+n+1\right ) }x^{n} \end {align*}

But this is (7). Hence we found the ﬁrst solution, which is \begin {equation} u_{r=0}\left ( x\right ) =\frac {1}{\sqrt {x^{a}}}\operatorname {BesselI}\left ( a,2\sqrt {x}\right ) \tag {8} \end {equation}

The above was for  $$r=0$$. Now we ﬁnd the second solution for $$r=-a$$. From (4)

$c_{n}\ \ =\frac {c_{n-1}}{\left ( n-a\right ) \left ( n-a-1\right ) \ +\ \left ( n-a\right ) \ +\ a\left ( n-a\right ) }$

For $$n=1$$

$c_{1}\ \ =\frac {c_{0}}{-a\left ( 1-a\right ) +\ \left ( 1-a\right ) \ +\ a\left ( 1-a\right ) }=\frac {c_{0}}{\ \left ( 1-a\right ) \ }$

For $$n=2$$

$c_{2}\ \ =\frac {c_{1}}{\left ( 2-a\right ) \left ( 1-a\right ) \ +\ \left ( 2-a\right ) \ +\ a\left ( 2-a\right ) }=\frac {c_{1}}{4-2a}=\frac {c_{0}}{\ \left ( 1-a\right ) \ }\frac {1}{2\left ( 2-a\right ) }$

For $$n=3$$

$c_{3}\ \ =\frac {c_{2}}{\left ( 3-a\right ) \left ( 2-a\right ) \ +\ \left ( 3-a\right ) \ +\ a\left ( 3-a\right ) }=\frac {c_{2}}{3\left ( 3-a\right ) }=\frac {c_{0}}{\ \left ( 1-a\right ) \ }\frac {1}{2\left ( 2-a\right ) }\frac {1}{3\left ( 3-a\right ) }$

And so on. Since the solution is assumed to be $$x^{r}\sum _{n=0}^{\infty }c_{n}x^{n}$$ then

\begin {align*} u_{r=-a} & =x^{-a}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =\sum _{n=0}^{\infty }c_{n}x^{n-a}\\ & =c_{0}x^{-a}\sum _{n=0}^{\infty }\frac {1}{\ n!}\left ( \frac {1}{\left ( 1-a\right ) }\frac {1}{\left ( 2-a\right ) }\frac {1}{\left ( 3-a\right ) }\cdots \frac {1}{\left ( n-a\right ) }\right ) x^{n-a} \end {align*}

But as we found above, we obtain that $$\left ( 1-a\right ) \left ( 2-a\right ) \cdots \left ( n-a\right ) =\frac {\Gamma \left ( -a+n+1\right ) }{-a\Gamma \left ( -a\right ) }$$, therefore

$u_{r=-a}=c_{0}\sum _{n=0}^{\infty }\frac {1}{\ n!}\frac {-a\Gamma \left ( -a\right ) }{\Gamma \left ( -a+n+1\right ) }x^{n-a}$

Modiﬁed Bessel function of second kind is $$\operatorname {BesselK}\left ( a,z\right ) =\frac {\pi }{2}\frac {1}{\sin \left ( a\pi \right ) }\left ( \operatorname {BesselI}\left ( -a,z\right ) -\operatorname {BesselI}\left ( a,z\right ) \right )$$. The above should result in $$\frac {1}{\sqrt {x^{a}}}\operatorname {BesselK}\left ( a,2\sqrt {x}\right )$$ for $$z=2\sqrt {x}$$ by setting $$c_{0}$$ to appropriate arbitrary value. I need to work out this ﬁnal manipulation later. Hence we ﬁnd $$u_{r=-a}\left ( x\right ) =\frac {1}{\sqrt {x^{a}}}\operatorname {BesselK}\left ( a,2\sqrt {x}\right )$$. Therefore, the solution is$u=C_{1}\frac {1}{\sqrt {x^{a}}}\operatorname {BesselI}\left ( a,2\sqrt {x}\right ) +C_{2}\frac {1}{\sqrt {x^{a}}}\operatorname {BesselK}\left ( a,2\sqrt {x}\right )$ But \begin {align*} \frac {d}{dx}\frac {1}{\sqrt {x^{a}}}\operatorname {BesselI}\left ( a,2\sqrt {x}\right ) & =\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselI}\left ( 1+a,2\sqrt {x}\right ) \\ \frac {d}{dx}\frac {1}{\sqrt {x^{a}}}\operatorname {BesselI}\left ( a,2\sqrt {x}\right ) & =-\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselK}\left ( 1+a,2\sqrt {x}\right ) \end {align*}

Hence$u^{\prime }=C_{1}\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselI}\left ( 1+a,2\sqrt {x}\right ) -C_{2}\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselK}\left ( 1+a,2\sqrt {x}\right )$ And from $$y=x^{a+1}\frac {u^{\prime }}{u}$$$y=x^{1+a}\frac {C_{1}\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselI}\left ( 1+a,2\sqrt {x}\right ) -C_{2}\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselK}\left ( 1+a,2\sqrt {x}\right ) }{C_{1}\frac {1}{\sqrt {x^{a}}}\operatorname {BesselI}\left ( a,2\sqrt {x}\right ) +C_{2}\frac {1}{\sqrt {x^{a}}}\operatorname {BesselK}\left ( a,2\sqrt {x}\right ) }$ Let $$C=\frac {C_{2}}{C_{1}}$$ hence$y=x^{1+a}\frac {\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselI}\left ( 1+a,2\sqrt {x}\right ) -C\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselK}\left ( 1+a,2\sqrt {x}\right ) }{\frac {1}{\sqrt {x^{a}}}\operatorname {BesselI}\left ( a,2\sqrt {x}\right ) +C\frac {1}{\sqrt {x^{a}}}\operatorname {BesselK}\left ( a,2\sqrt {x}\right ) }$ Or\begin {align*} y & =x^{1+a}\frac {x^{-\frac {1}{2}}\operatorname {BesselI}\left ( 1+a,2\sqrt {x}\right ) -Cx^{-\frac {1}{2}}\operatorname {BesselK}\left ( 1+a,2\sqrt {x}\right ) }{\operatorname {BesselI}\left ( a,2\sqrt {x}\right ) +C\operatorname {BesselK}\left ( a,2\sqrt {x}\right ) }\\ & =\frac {x^{\frac {1}{2}+a}\operatorname {BesselI}\left ( 1+a,2\sqrt {x}\right ) -Cx^{\frac {1}{2}+a}\operatorname {BesselK}\left ( 1+a,2\sqrt {x}\right ) }{\operatorname {BesselI}\left ( a,2\sqrt {x}\right ) +C\operatorname {BesselK}\left ( a,2\sqrt {x}\right ) } \end {align*}

Veriﬁcation

eq:=diff(y(x),x)+x^(-a-1)*y(x)^2-x^a = 0;
num:=x^(1/2+a)*BesselI(1+a,2*sqrt(x))-_C1*x^(1/2+a)*BesselK(1+a,2*sqrt(x));
den:=BesselI(a,2*sqrt(x))+_C1*BesselK(a,2*sqrt(x));
my_sol:=num/den;
odetest(y(x)=my_sol,eq);
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