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)((-1)^n)/((2n+1)!)x^(2n+1)\sum_(n\ge0)a_nx^n=\sum_(n\ge0)((-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-\frac16x^3\right)(a_0+a_1x+a_2x^2+a_3x^3)=1-\frac12x^2+\frac1{24}x^4`

`2a_2+(6a_3-a_0)x+(12a_4-a_1)x^2+\left(20a_5+\frac16a_0-a_2\right)x^3+\left(30a_6-a_3+\frac16a_1\right)x^4=1-\frac12x^2+\frac1{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=\frac12,a_3=-\frac12,a_4=(13)/(24)`