Question

Find the indicated coefficients of the power series solution about x=0 of the differential equation

Answer 1

Write the ODE as

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge0}\frac{(-1)^n}{(2n+1)!}x^{2n+1}\sum_{n\ge0}a_nx^n=\sum_{n\ge0}\frac{(-1)^n}{(2n)!}x^{2n}$

and truncate as many terms as needed to obtain only the terms up to order 4:

$(2a_2+6a_3x+12a_4x^2+20a_5x^3+30a_6x^4)-\left(x-\dfrac16x^3\right)(a_0+a_1x+a_2x^2+a_3x^3)=1-\dfrac12x^2+\dfrac1{24}x^4$
$2a_2+(6a_3-a_0)x+(12a_4-a_1)x^2+\left(20a_5+\dfrac16a_0-a_2\right)x^3+\left(30a_6-a_3+\dfrac16a_1\right)x^4=1-\dfrac12x^2+\dfrac1{24}x^4$

We only care about the coefficients up to $a_4$, so we take the system

$\begin{cases}2a_2=1\\6a_3-a_0=0\\12a_4-a_1=-\frac12\end{cases}$

Now, we're given initial values $y(0)=-3$ and $y'(0)=7$, so that

$y(0)=\displaystyle a_0+\sum_{n\ge1}a_n0^n=a_0=-3$
$y'(0)=\displaystyle a_1+\sum_{n\ge2}na_n0^{n-1}=a_1=7$

which gives

$a_0=-3,a_1=7,a_2=\dfrac12,a_3=-\dfrac12,a_4=\dfrac{13}{24}$