1
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $a_3$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.