Coefficient of $x^3$ if $y''-y'\sin x+ xy = 0$ and $y(0),y'(0) = 0,1$ I'm a BC student studying DE. there is a multiple choice question that I can't answer it but I have some ideas, here it is: > what is the coefficient of $x^3$ in $y$ if $$y''-y'\sin x+ xy = 0$$ and $y(0) = 0$ and $y'(0) = 1$ > > 1. $\frac{1}{3}$ > 2. $\frac{-1}{3}$ > 3. $\frac{1}{6}$ > 4. $\frac{-1}{6}$ > MY IDEA: if I can be sure that $y$ is just a polynomial then I can easily see that $y'''(0) = 1$ so the coefficient of $x^3$ must be $\frac{1}{6}$ that is in the choices. but I don't know how to be sure about that and even doubt my sureness. any help would be appreciated.

Assuming a series solution of $f(x)$ exists around $a=0$, then

$$ f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n $$

So to find the coefficient of $x^3$, you just need to find $f'''(0)$.

First, observe from the equation that $f''(0)=0$. Differentiate throughout to get

$$ y''' - y''\sin x - y'\cos x + xy' + y = 0 $$

which gives

$$ y'''(0) = y'(0)-y(0) = 1 $$

Hence the answer is $\dfrac{1}{3!} = \dfrac16$