2.7.2.3 Example 3 \(\frac {1}{x^{2}}y^{\prime \prime }+y^{\prime }+y=\sin x\)

\[ \frac {1}{x^{2}}y^{\prime \prime }+y^{\prime }+y=\sin x \]

Expansion around \(x=0\). This is ordinary point. Since RHS is not zero, do not find recurrence relation. 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 ode becomes

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

Hence

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

Reindex so all powers to start from \(n\). This results in

\[ \sum _{n=0}^{\infty }\left ( n+2\right ) \left ( n+1\right ) a_{n+2}x^{n}+\sum _{n=2}^{\infty }\left ( n-1\right ) a_{n-1}x^{n}+\sum _{n=2}^{\infty }a_{n-2}x^{n}=x^{2}\sin x \]

To be able to continue, we have to expand \(\sin x\) as Taylor series around \(x\). The above becomes

\begin{align*} \sum _{n=0}^{\infty }\left ( n+2\right ) \left ( n+1\right ) a_{n+2}x^{n}+\sum _{n=2}^{\infty }\left ( n-1\right ) a_{n-1}x^{n}+\sum _{n=2}^{\infty }a_{n-2}x^{n} & =x^{2}\left ( x-\frac {1}{6}x^{3}+\frac {1}{120}x^{5}-\frac {1}{5040}x^{7}+\cdots \right ) \\ \sum _{n=0}^{\infty }\left ( n+2\right ) \left ( n+1\right ) a_{n+2}x^{n}+\sum _{n=2}^{\infty }\left ( n-1\right ) a_{n-1}x^{n}+\sum _{n=2}^{\infty }a_{n-2}x^{n} & =x^{3}-\frac {1}{6}x^{5}+\frac {1}{120}x^{7}-\frac {1}{5040}x^{9}+\cdots \end{align*}

For \(n=0\)

\begin{align*} 2a_{2} & =0\\ a_{2} & =0 \end{align*}

For \(n=1\)

\begin{align*} \left ( 3\right ) \left ( 2\right ) a_{3} & =0\\ a_{3} & =0 \end{align*}

For \(n=2\)

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

For \(n=3\) (now we pick one term from the RHS which match on \(x^{3}\))

\begin{align*} 20a_{5}+2a_{2}+a_{1} & =1\\ a_{5} & =\frac {1-a_{1}}{20}\end{align*}

For \(n=4\)

\begin{align*} 30a_{6}+3a_{3}+a_{2} & =0\\ a_{6} & =0 \end{align*}

For \(n=5\)

\begin{align*} 42a_{7}+4a_{4}+a_{3} & =-\frac {1}{6}\\ a_{7} & =\frac {-\frac {1}{6}-4a_{4}}{42}=\frac {-\frac {1}{6}-4\left ( \frac {-a_{1}-a_{0}}{12}\right ) }{42}=\frac {1}{126}a_{0}+\frac {1}{126}a_{1}-\frac {1}{252}\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+\left ( \frac {-a_{1}-a_{0}}{12}\right ) x^{4}+\left ( \frac {1-a_{1}}{20}\right ) x^{5}+\left ( \frac {1}{126}a_{0}+\frac {1}{126}a_{1}-\frac {1}{252}\right ) x^{7}+\cdots \\ & =a_{0}\left ( 1-\frac {1}{12}x^{4}+\frac {1}{126}x^{7}+\cdots \right ) +a_{1}\left ( x-\frac {1}{12}x^{4}-\frac {1}{20}x^{5}+\frac {1}{126}x^{7}+\cdots \right ) +\left ( \frac {1}{20}x^{5}-\frac {1}{252}x^{7}+\cdots \right ) \end{align*}