Problem source

\begin{questionparts}
\item
Let $\displaystyle y= \sum_{n=0}^\infty a_n x^n\,$, where the coefficients $a_n$ are independent of $x$ and are such that this series and all others in this question converge.
Show that 
\[
 \displaystyle y'= \sum_{n=1}^\infty na_n x^{n-1}\,,
\]
and write down a similar expression for $y''$.
Write out explicitly  each of the three
series as far as the term containing $a_3$.
\item
It is given that $y$ satisfies the differential
equation
\[
xy''-y'+4x^3y =0\,.
\]
By substituting the series of part (i) into the differential
equation and comparing coefficients, show that $a_1=0$. 
 Show that, for $n\ge4$, 
\[ a_n =- \frac{4}{n(n-2)}\, a_{n-4}\,,
\]
and  that, if $a_0=1$ and $a_2=0$, then $ y=\cos (x^2)\,$.
Find the corresponding result when $a_0=0$ and $a_2=1$.
\end{questionparts}

Solution source

\begin{questionparts}
\item Let $\displaystyle y= \sum_{n=0}^\infty a_n x^n\,$ then

\begin{align*}
y' &= \frac{\d}{\d x}  \l \sum_{n=0}^\infty a_n x^n \r \\
&=  \sum_{n=0}^\infty \frac{\d}{\d x}  \l a_n x^n \r \\
&= \sum_{n=0}^\infty  n a_n x^{n-1} \\
&= \sum_{n=1}^\infty  n a_n x^{n-1} \\
\\
y'' &=  \frac{\d}{\d x}  \l\sum_{n=1}^\infty  n a_n x^{n-1} \r \\
&= \sum_{n=1}^\infty  \frac{\d}{\d x}  \l n a_n x^{n-1} \r \\
&= \sum_{n=1}^\infty  n(n-1) a_n x^{n-2} \\
&= \sum_{n=2}^\infty  n(n-1) a_n x^{n-2} \\
\\
y &= a_0 + a_1 x+ a_2x^2 + a_3x^3 + \cdots \\
y'&= a_1 + 2a_2x+3a_3x^2 + \cdots \\
y'' &= 2a_2 + 6a_3x + \cdots
\end{align*}

\item 
\begin{align*}
&& 0 &= xy''-y'+4x^3y \\
&&&= x\sum_{n=2}^\infty  n(n-1) a_n x^{n-2}  -  \sum_{n=1}^\infty  n a_n x^{n-1} + 4x^3  \sum_{n=0}^\infty a_n x^n \\
&&&= \sum_{n=2}^\infty  n(n-1) a_n x^{n-1}  -  \sum_{n=1}^\infty  n a_n x^{n-1} + \sum_{n=0}^\infty 4a_n x^{n+3} \\
&&&= \sum_{n=2}^\infty  n(n-1) a_n x^{n-1}  -  \sum_{n=1}^\infty  n a_n x^{n-1} + \sum_{n=4}^\infty 4a_{n-4} x^{n-1} \\
&&&= \sum_{n=4}^{\infty} \l n(n-1) a_n-  n a_n +4a_{n-4}  \r x^{n-1} + 2a_2x + 6a_3x^2-a_1-2a_2x-3a_3x^2 \\
&&&= \sum_{n=4}^{\infty} \l n(n-2) a_n +4a_{n-4}  \r x^{n-1}+ 3a_3x^2-a_1 \\
\end{align*}

Therefore since all coefficients are $0$, $a_1 = 0$, $a_3 = 0$ and $\displaystyle a_n = -\frac{4}{n(n-2)}a_{n-4}$.

If $a_0 = 1, a_2 = 0$, and since $a_1 = 0, a_3 = 0$ the only values which will take non-zero value are $a_{4k}$. We can compute these values as: $a_{4k} = -\frac{4}{(4k)(4k-2)} a_{4k-4} = \frac{1}{2k(2k-1)}a_{4k-r}$ so $a_{4k} = \frac{(-1)^k}{(2k)!}$, which are precisely the coefficients in the expansion $\cos x^2$.

If $a_0 = 0, a_2 = 1$ then since $a_1 = 0, a_3 = 0$ the only values which take non-zero values are $a_{4k+2}$ we can compute these values as:

$a_{4k+2} = -\frac{4}{(4k+2)(4k)}a_{4k-2} = -\frac{1}{(2k+1)2k}a_{4k-2}$ so we can see that $a_{4k+2}= \frac{(-1)^k}{(2k+1)!}$ precisely the coefficients of $\sin x^2$
\end{questionparts}