# 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.

• Right idea. But you can assume $y$ is an power series, and the same reasoning works. – B. Goddard Mar 19 at 20:20

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 $$\dfrac{1}{3!} = \dfrac16$$

Consider a series solution $$y(x) = x + a_2 x^2 + a_3 x^3 + \ldots$$. Substitute this into the differential equation, using the Taylor series of $$\sin(x)$$. The coefficients of $$x$$ and $$x^2$$ should be the same on both sides, and solving for this will tell you $$a_2$$ and $$a_3$$.