Question

Determine the first three nonzero terms of the Taylor polynomial approximations for the given initial value problem.
`y' = 9 sin(y)+2 e^(x)` and y(0)=0

Answer 1

`y=\displaystyle\sum_(n\ge0)a_nx^n`

`\sin x=\displaystyle\sum_(n\ge0)((-1)^n)/((2n+1)!)x^(2n+1)`

`\implies\sin y=\displaystyle\sum_(n\ge0)((-1)^n)/((2n+1)!)\left(\sum_(m\ge0)a_mx^m\right)^(2n+1)`

Note that with
`y(0)=0`, the first sum reduces immediately (after taking out the first term in the sum) to
`a_0`, which implies that
`a_0=0`.

So we have


`\displaystyle\sum_(n\ge1)na_nx^(n-1)=9\sum_(n\ge0)((-1)^n)/((2n+1)!)\left(\sum_(m\ge1)a_mx^m\right)^(2n+1)+2\sum_(n\ge0)\frac1{n!}x^n`

`a_1+2a_2x+3a_3x^2+4a_4x^3+\cdots=9\mathbf S+2\left(1+x+\frac12x^2+\frac16x^3+\cdots\right)`

where we can expand
`\mathbf S` as


`\mathbf S=\displaystyle\sum_(n\ge0)((-1)^n)/((2n+1)!)(a_1x+a_2x^2+a_3x^3+\cdots)^(2n+1)`

`=a_1x+a_2x^2+a_3x^3+\cdots`

`\,\,-\frac1{3!}(a_1x+a_2x^2+a_3x^3+\cdots)^3+\cdots`

`=a_1x+a_2x^2+\left(a_3-\frac{{a_1}^3}{3!}\right)x^3+\cdots`

Dropping all terms with order greater than 2 (because this is enough to generate 3 nonzero terms) reduces the equation to


`a_1+2a_2x+3a_3x^2=9\left(a_1x+a_2x^2\right)x^3\right)+2\left(1+x+\frac12x^2\right)`

`a_1+2a_2x+3a_3x^2=2+\left(9a_1+2\right)x+\left(9a_2+1\right)x^2`

`\implies\begin{cases}a_1=2\\2a_2=9a_1+2\\3a_3=9a_2+1\end{cases}\implies a_1=2,a_2=10,a_3=\frac{91}3`

so that the Taylor polynomial approximating the solution
`y` is


`y(x)\approx2x+10x^2+\frac{91}3x^3`

Just to demonstrate that the result is reasonable, I've attached a plot of an approximate solution with higher accuracy (blue) and the one we found (orange) over the interval
`-\frac\pi6<x<\frac\pi6`.