2.7.2.1 Example 1 \(y^{\prime \prime }+xy^{\prime }+y=2x+x^{2}+x^{4}\)

Solved using Taylor series method.

\begin{align*} y^{\prime \prime }+xy^{\prime }+y & =2x+x^{2}+x^{4}\\ y^{\prime \prime } & =-xy^{\prime }-y+2x+x^{2}+x^{4}\\ y^{\prime \prime } & =f\left ( x,y,y^{\prime }\right ) \end{align*}

Hence

\[ y=y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+1\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0},y_{0}^{\prime }}\]

Where

\begin{align*} F_{0} & =f\left ( x,y,y^{\prime }\right ) \\ F_{n} & =\frac {d}{dx}\left ( F_{n-1}\right ) \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) F_{0}\end{align*}

Hence

\begin{align*} F_{1} & =\frac {\partial \left ( -xy^{\prime }-y+2x+x^{2}+x^{4}\right ) }{\partial x}+\frac {\partial \left ( -xy^{\prime }-y+2x+x^{2}+x^{4}\right ) }{\partial y}y^{\prime }+\frac {\partial \left ( -xy^{\prime }-y+2x+x^{2}+x^{4}\right ) }{\partial y^{\prime }}y^{\prime \prime }\\ & =\left ( 4x^{3}+2x-y^{\prime }+2\right ) -y^{\prime }-xy^{\prime \prime }\\ & =2x-2y^{\prime }-xy^{\prime \prime }+4x^{3}+2 \end{align*}

But \(y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \), the above becomes

\[ F_{1}=2x-2y^{\prime }+x^{2}y^{\prime }+xy-2x^{2}+3x^{3}-x^{5}+2 \]

And

\begin{align*} F_{2} & =\frac {d}{dx}\left ( F_{n-1}\right ) \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\\ & =\frac {\partial }{\partial x}\left ( 2x-2y^{\prime }+x^{2}y^{\prime }+xy-2x^{2}+3x^{3}-x^{5}+2\right ) +\\ & +\left ( \frac {\partial }{\partial y}\left ( 2x-2y^{\prime }+x^{2}y^{\prime }+xy-2x^{2}+3x^{3}-x^{5}+2\right ) \right ) y^{\prime }\\ & +\left ( \frac {\partial }{\partial y^{\prime }}\left ( 2x-2y^{\prime }+x^{2}y^{\prime }+xy-2x^{2}+3x^{3}-x^{5}+2\right ) \right ) y^{\prime \prime }\\ & =\left ( y-4x+2xy^{\prime }+9x^{2}-5x^{4}+2\right ) +xy^{\prime }+\left ( -2+x^{2}\right ) y^{\prime \prime }\\ & =y-4x-2y^{\prime \prime }+3xy^{\prime }+x^{2}y^{\prime \prime }+9x^{2}-5x^{4}+2 \end{align*}

But \(y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \), the above becomes

\begin{align*} F_{2} & =y-4x-2\left ( -xy^{\prime }-y+2x+x^{2}+x^{4}\right ) +3xy^{\prime }+x^{2}\left ( -xy^{\prime }-y+2x+x^{2}+x^{4}\right ) +9x^{2}-5x^{4}+2\\ & =3y-8x+5xy^{\prime }-x^{2}y-x^{3}y^{\prime }+7x^{2}+2x^{3}-6x^{4}+x^{6}+2 \end{align*}

And

\begin{align*} F_{3} & =\frac {d}{dx}\left ( F_{2}\right ) \\ & =\frac {\partial }{\partial x}F_{2}+\left ( \frac {\partial F_{2}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{2}}{\partial y^{\prime }}\right ) y^{\prime \prime }\\ & =\frac {\partial }{\partial x}\left ( 3y-8x+5xy^{\prime }-x^{2}y-x^{3}y^{\prime }+7x^{2}+2x^{3}-6x^{4}+x^{6}+2\right ) \\ & +\left ( \frac {\partial }{\partial y}\left ( 3y-8x+5xy^{\prime }-x^{2}y-x^{3}y^{\prime }+7x^{2}+2x^{3}-6x^{4}+x^{6}+2\right ) \right ) y^{\prime }\\ & +\left ( \frac {\partial }{\partial y^{\prime }}\left ( 3y-8x+5xy^{\prime }-x^{2}y-x^{3}y^{\prime }+7x^{2}+2x^{3}-6x^{4}+x^{6}+2\right ) \right ) y^{\prime \prime }\\ & =14x+5y^{\prime }-3x^{2}y^{\prime }-2xy+6x^{2}-24x^{3}+6x^{5}-8+\left ( 3-x^{2}\right ) y^{\prime }+\left ( 5x-x^{3}\right ) y^{\prime \prime }\end{align*}

But \(y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \), the above becomes

\begin{align*} F_{3} & =14x+5y^{\prime }-3x^{2}y^{\prime }-2xy+6x^{2}-24x^{3}+6x^{5}-8+\left ( 3-x^{2}\right ) y^{\prime }+\left ( 5x-x^{3}\right ) \left ( -xy^{\prime }-y+2x+x^{2}+x^{4}\right ) \\ & =14x+8y^{\prime }+x^{3}y-9x^{2}y^{\prime }+x^{4}y^{\prime }-7xy+16x^{2}-19x^{3}-2x^{4}+10x^{5}-x^{7}-8 \end{align*}

And so on. Evaluating each of the above at \(x=0,y=y_{0},y^{\prime }=y_{0}^{\prime }\) gives

\begin{align*} F_{0} & =\left ( -xy^{\prime }-y+2x+x^{2}+x^{4}\right ) _{x=0,y_{0},y_{0}^{\prime }}=-y_{0}\\ F_{1} & =\left ( 2x-2y^{\prime }+x^{2}y^{\prime }+xy-2x^{2}+3x^{3}-x^{5}+2\right ) _{x=0,y_{0},y_{0}^{\prime }}=\left ( -2y_{0}^{\prime }+2\right ) \\ F_{2} & =3y-8x+5xy^{\prime }-x^{2}y-x^{3}y^{\prime }+7x^{2}+2x^{3}-6x^{4}+x^{6}+2=3y_{0}+2\\ F_{3} & =14x+8y^{\prime }+x^{3}y-9x^{2}y^{\prime }+x^{4}y^{\prime }-7xy+16x^{2}-19x^{3}-2x^{4}+10x^{5}-x^{7}-8=8y_{0}^{\prime }-8 \end{align*}

Hence

\begin{align*} y\left ( x\right ) & =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0},y_{0}^{\prime }}\\ & =y_{0}+xy_{0}^{\prime }+\frac {x^{2}}{2}F_{0}+\frac {x^{3}}{6}F_{1}+\frac {x^{4}}{24}F_{2}+\frac {x^{5}}{5!}F_{3}+\cdots \\ & =y_{0}+xy_{0}^{\prime }+\frac {x^{2}}{2}\left ( -y_{0}\right ) +\frac {x^{3}}{6}\left ( -2y_{0}^{\prime }+2\right ) +\frac {x^{4}}{24}\left ( 3y_{0}+2\right ) +\frac {x^{5}}{5!}\left ( 8y_{0}^{\prime }-8\right ) +\cdots \\ & =y_{0}\left ( 1-\frac {x^{2}}{2}+\frac {1}{8}x^{4}+\cdots \right ) +y_{0}^{\prime }\left ( x-\frac {x^{3}}{3}+\frac {1}{15}x^{4}\cdots \right ) +\left ( \frac {1}{3}x^{3}+\frac {1}{12}x^{4}-\frac {1}{15}x^{4}\right ) \\ & =c_{1}\left ( 1-\frac {x^{2}}{2}+\frac {1}{8}x^{4}+\cdots \right ) +c_{2}\left ( x-\frac {x^{3}}{3}+\frac {1}{15}x^{4}\cdots \right ) +\left ( \frac {1}{3}x^{3}+\frac {1}{12}x^{4}-\frac {1}{15}x^{4}\right ) \end{align*}

Solved using power series method.

\[ y^{\prime \prime }+xy^{\prime }+y=2x+x^{2}+x^{4}\]

Comparing the homogenous ode to \(y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0\) shows that \(p\left ( x\right ) =x,q\left ( x\right ) =1\). These are defined everywhere. Let the expansion point be \(x_{0}=0\).  This is ordinary point since \(p\left ( x\right ) ,q\left ( x\right ) \) are defined at \(x_{0}\). Let \(y=\sum _{n=0}^{\infty }a_{n}x^{n}\). Hence \(y^{\prime }=\sum _{n=0}^{\infty }na_{n}x^{n-1}=\sum _{n=1}^{\infty }na_{n}x^{n-1}\) and \(y^{\prime \prime }=\sum _{n=1}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}=\sum _{n=2}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}\). The homogenous ode becomes

\begin{align} \sum _{n=2}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}+x\sum _{n=1}^{\infty }na_{n}x^{n-1}+\sum _{n=0}^{\infty }a_{n}x^{n} & =2x+x^{2}+x^{4}\nonumber \\ \sum _{n=2}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}+\sum _{n=1}^{\infty }na_{n}x^{n}+\sum _{n=0}^{\infty }a_{n}x^{n} & =2x+x^{2}+x^{4}\nonumber \end{align}

Adjust all sums to lowest power on \(x\) gives

\[ \sum _{n=2}^{\infty }\left ( n\right ) \left ( n-1\right ) a_{n}x^{n-2}+\sum _{n=3}^{\infty }\left ( n-2\right ) a_{n-2}x^{n-2}+\sum _{n=2}^{\infty }a_{n-2}x^{n-2}=2x+x^{2}+x^{4}\]

\(n=2\) gives \(x^{0}\) on the LHS with no match on the RHS. Hence

\begin{align*} 2a_{2}+a_{0} & =0\\ a_{2} & =-\frac {1}{2}a_{0}\end{align*}

\(n=3\) gives \(x^{1}\) on the LHS with one match on the RHS. Hence

\begin{align*} 6a_{3}+2a_{1} & =2\\ a_{3} & =\frac {2-2a_{1}}{6}\\ & =\frac {1}{3}-\frac {1}{3}a_{1}\end{align*}

\(n=4\) gives \(x^{2}\) on the LHS with one match on the RHS. Hence

\begin{align*} 12a_{4}+3a_{2} & =1\\ a_{4} & =\frac {1-3a_{2}}{12}\\ & =\frac {1-3\left ( -\frac {1}{2}a_{0}\right ) }{12}\\ & =\frac {1}{8}a_{0}+\frac {1}{12}\end{align*}

\(n=5\) gives \(x^{3}\) on the LHS with no match on the RHS. Hence

\begin{align*} 20a_{5}+4a_{3} & =0\\ a_{5} & =\frac {-4a_{3}}{20}\\ & =\frac {-4\left ( \frac {1}{3}-\frac {1}{3}a_{1}\right ) }{20}\\ & =\frac {1}{15}a_{1}-\frac {1}{15}\end{align*}

\(n=6\) gives \(x^{4}\) on the LHS with one match on the RHS. Hence

\begin{align*} 30a_{6}+5a_{4} & =1\\ a_{6} & =\frac {1-5a_{4}}{30}\\ & =\frac {1-5\left ( \frac {1}{8}a_{0}+\frac {1}{12}\right ) }{30}\\ & =\frac {7}{360}-\frac {1}{48}a_{0}\end{align*}

And for \(n\geq 7\) we have recurrence relation

\begin{align*} \left ( n\right ) \left ( n-1\right ) a_{n}+\left ( n-2\right ) a_{n-2}+a_{n-2} & =0\\ a_{n} & =-\frac {n-1}{n\left ( n-1\right ) }a_{n-2}\end{align*}

Hence for \(n=7\)

\begin{align*} a_{7} & =-\frac {6}{42}a_{5}\\ & =-\frac {6}{42}\left ( \frac {1}{15}a_{1}-\frac {1}{15}\right ) \\ & =\frac {1}{105}-\frac {1}{105}a_{1}\end{align*}

For \(n=8\)

\begin{align*} a_{8} & =-\frac {7}{\left ( 8\right ) \left ( 7\right ) }a_{6}\\ & =-\frac {7}{\left ( 8\right ) \left ( 7\right ) }\left ( \frac {7}{360}-\frac {1}{48}a_{0}\right ) \\ & =\frac {1}{384}a_{0}-\frac {7}{2880}\end{align*}

And so on. Hence

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots \\ & =a_{0}+a_{1}x-\frac {1}{2}a_{0}x^{2}+\left ( \frac {1}{3}-\frac {1}{3}a_{1}\right ) x^{3}+\left ( \frac {1}{8}a_{0}+\frac {1}{12}\right ) x^{4}+\left ( \frac {1}{15}a_{1}-\frac {1}{15}\right ) x^{5}+\left ( \frac {7}{360}-\frac {1}{48}a_{0}\right ) x^{6}+\left ( \frac {1}{105}-\frac {1}{105}a_{1}\right ) x^{7}+\cdots \\ & =a_{0}\left ( 1-\frac {1}{2}x^{2}+\frac {1}{8}x^{4}-\frac {1}{48}x^{6}+\cdots \right ) +a_{1}\left ( x-\frac {1}{3}x^{3}+\frac {1}{15}x^{5}-\frac {1}{105}x^{7}-\cdots \right ) +\left ( \frac {1}{3}x^{3}+\frac {1}{12}x^{4}-\frac {1}{15}x^{5}+\frac {7}{360}x^{6}+\frac {1}{105}x^{7}+\cdots \right ) \end{align*}

Which is the same answer given using the Taylor series method. We see that the Taylor series method is much simpler, but requires using the computer to calculate the derivatives as they become very complicated as more terms are needed.

Even though the expansion point is ordinary, we can also solve this using Frobenius series as follows. Comparing the ode \(y^{\prime \prime }+xy^{\prime }+y=0\) to

\[ y^{\prime \prime }+p\left ( x\right ) y^{\prime }+q\left ( x\right ) y=0 \]

Hence \(p\left ( x\right ) =x,q\left ( x\right ) =1\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}x^{2}=0\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}x^{2}=0\). Hence the indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) & =0\\ r & =1,0 \end{align*}

Hence \(r_{1}=1,r_{2}=0\). All ordinary points will have the same roots. Let

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ y^{\prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}\\ y^{\prime \prime } & =\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}\end{align*}

The ode becomes

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

Reindex to lowest powers gives

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+\sum _{n=2}^{\infty }\left ( n+r-2\right ) a_{n-2}x^{n+r-2}+\sum _{n=2}^{\infty }a_{n-2}x^{n+r-2}=0 \tag {1}\end{equation}

For \(n=0\)

\[ r\left ( r-1\right ) a_{0}x^{r-2}=0 \]

The homogenous ode therefore satisfies

\begin{equation} y^{\prime \prime }+xy^{\prime }+y=r\left ( r-1\right ) a_{0}x^{r} \tag {2}\end{equation}

For \(n=1\), Eq (1) gives

\[ \left ( 1+r\right ) \left ( r\right ) a_{1}=0 \]

For \(r=1\) we see that \(a_{1}=0\). But for \(r=0\) then the above gives \(0b_{1}=0\). This means \(b_{1}\) can be any value and we choose \(b_{1}=0\) in this case.

For \(n\geq 2\) we obtain the recurrence relation

\begin{align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r-2\right ) a_{n-2}+a_{n-2} & =0\nonumber \\ a_{n} & =\frac {-\left ( n+r-2\right ) a_{n-2}-a_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) }=\frac {-\left ( n+r-1\right ) a_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) } \tag {3}\end{align}

Now we find \(y_{1}\) which is associated with \(r=1\). From (3) and for \(r=1\) it becomes

\begin{equation} a_{n}=-\frac {n}{\left ( n+1\right ) n}a_{n-2}=-\frac {1}{n+1}a_{n-2} \tag {4}\end{equation}

For \(n=2\) and using \(a_{0}=1\)

\[ a_{2}=-\frac {1}{3}a_{0}=-\frac {1}{3}\]

For \(n=3\)

\[ a_{3}=-\frac {1}{4}a_{1}=0 \]

All odd \(a_{n}\) will be zero. For \(n=4\)

\[ a_{4}=-\frac {1}{5}a_{2}=-\frac {1}{5}\left ( -\frac {1}{3}\right ) =\frac {1}{15}\]

And so on. Hence

\begin{align*} y_{1} & =\sum a_{n}x^{n+r_{1}}\\ & =x\sum a_{n}x^{n}\\ & =x\left ( a_{0}+a_{1}x+a_{2}x^{2}+\cdots \right ) \\ & =x\left ( 1-\frac {1}{2}x^{2}+\frac {1}{10}x^{4}-\cdots \right ) \\ & =x-\frac {1}{3}x^{3}+\frac {1}{15}x^{5}-\cdots \end{align*}

Now we find \(y_{2}\) associated with \(r=0\). From (3) this becomes (using \(b\) instead of \(a\)) and \(r=0\)

\begin{align} b_{n} & =\frac {-\left ( n+r-1\right ) b_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) }\nonumber \\ & =\frac {-\left ( n-1\right ) b_{n-2}}{\left ( n\right ) \left ( n-1\right ) }\nonumber \\ & =-\frac {b_{n-2}}{n} \tag {5}\end{align}

From above, we found that \(b_{1}=0\). Now we use (5) to find all \(b_{n}\) for \(n\geq 2\). For \(n=2\)

\[ b_{2}=-\frac {b_{0}}{2}=-\frac {1}{2}\]

For \(n=3\)

\[ b_{3}=-\frac {b_{1}}{3}=0 \]

For \(n=4\)

\[ b_{4}=-\frac {b_{2}}{4}=\frac {1}{8}\]

And so on. Hence

\begin{align*} y_{2} & =\sum b_{n}x^{n+r_{2}}\\ & =\sum b_{n}x^{n}\\ & =\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =\left ( 1-\frac {1}{2}x^{2}+\frac {1}{8}x^{4}+\cdots \right ) \end{align*}

Hence the solution \(y_{h}\) is

\begin{align*} y & =c_{1}y_{1}+c_{2}y_{2}\\ & =c_{1}\left ( x-\frac {1}{3}x^{3}+\frac {1}{15}x^{5}-\cdots \right ) +c_{2}\left ( 1-\frac {1}{2}x^{2}+\frac {1}{8}x^{4}+\cdots \right ) \end{align*}

We see this is the same \(y_{h}\) obtained using standard power series. This shows that we can also use Frobenius series to solve for ordinary point. The roots will always be \(r_{1}=1,r_{2}=0\). But this requires more work than using standard power series. The main advantage of using Frobenius series for ordinary point comes in when the RHS has no series expansion at \(x=0\). For example, if the RHS in this ode was say \(\sqrt {x}\) then we must use Frobenius to be able to solve it as standard power series will fail, since\(\sqrt {x}\) has no series representation at \(x=0\). Examples below shows how to do this.