Write the ODE as
[tex]\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}[/tex]
and truncate as many terms as needed to obtain only the terms up to order 4:
[tex](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[/tex]
[tex]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[/tex]
We only care about the coefficients up to [tex]a_4[/tex], so we take the system
[tex]\begin{cases}2a_2=1\\6a_3-a_0=0\\12a_4-a_1=-\frac12\end{cases}[/tex]
Now, we're given initial values [tex]y(0)=-3[/tex] and [tex]y'(0)=7[/tex], so that
[tex]y(0)=\displaystyle a_0+\sum_{n\ge1}a_n0^n=a_0=-3[/tex]
[tex]y'(0)=\displaystyle a_1+\sum_{n\ge2}na_n0^{n-1}=a_1=7[/tex]
which gives
[tex]a_0=-3,a_1=7,a_2=\dfrac12,a_3=-\dfrac12,a_4=\dfrac{13}{24}[/tex]
