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.

  • 1
    $\begingroup$ Right idea. But you can assume $y$ is an power series, and the same reasoning works. $\endgroup$ – 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 $a_3$.


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