Question

Determine the first three nonzero terms of the Taylor polynomial approximations for the given initial value problem. ttps://tex.z-dn.net/?f=y%27%20%3D%209%20sin%28y%29%2B2%20e%5E%7Bx%7D%20" id="TexFormula1" title="y' = 9 sin(y)+2 e^{x} " alt="y' = 9 sin(y)+2 e^{x} " align="absmiddle" class="latex-formula"> and y(0)=0

Answer 1

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$\sin x=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{(2n+1)!}x^{2n+1}$
$\implies\sin y=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{(2n+1)!}\left(\sum_{m\ge0}a_mx^m\right)^{2n+1}$

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

$\displaystyle\sum_{n\ge1}na_nx^{n-1}=9\sum_{n\ge0}\frac{(-1)^n}{(2n+1)!}\left(\sum_{m\ge1}a_mx^m\right)^{2n+1}+2\sum_{n\ge0}\frac1{n!}x^n$
$a_1+2a_2x+3a_3x^2+4a_4x^3+\cdots=9\mathbf S+2\left(1+x+\frac12x^2+\frac16x^3+\cdots\right)$

where we can expand $\mathbf S$ as

$\mathbf S=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{(2n+1)!}(a_1x+a_2x^2+a_3x^3+\cdots)^{2n+1}$
$=a_1x+a_2x^2+a_3x^3+\cdots$
$\,\,-\dfrac1{3!}(a_1x+a_2x^2+a_3x^3+\cdots)^3+\cdots$
$=a_1x+a_2x^2+\left(a_3-\dfrac{{a_1}^3}{3!}\right)x^3+\cdots$

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

$a_1+2a_2x+3a_3x^2=9\left(a_1x+a_2x^2\right)x^3\right)+2\left(1+x+\dfrac12x^2\right)$
$a_1+2a_2x+3a_3x^2=2+\left(9a_1+2\right)x+\left(9a_2+1\right)x^2$
$\implies\begin{cases}a_1=2\\2a_2=9a_1+2\\3a_3=9a_2+1\end{cases}\implies a_1=2,a_2=10,a_3=\dfrac{91}3$

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

$y(x)\approx2x+10x^2+\dfrac{91}3x^3$

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 $-\dfrac\pi6$.