So I've been given this definition of function $F(x)$

$$F(x) = \sum_{k=0}^{\infty} \frac{x^{2k}}{2^kk!}$$

And the question is given as:

Show that $y=F(x)$ is a solution of $y''=xy'+y$ with $y(0) = 1$ and $y'(0)=0$.

So far I have gotten the fact that this is has a raidus of convergence of $\infty$ though that isn't much use here. I also know that it is a second order differential equation (linear).

I have only been able to so far solve first order differential equations, so I have no idea how to actively solve this.

Going from previous things I learned, I know that I can group together the terms in the differential like so and proceed to solve.

$$y''=xy'+y$$ $$\dashrightarrow y''-y = xy'$$ $$\dashrightarrow \frac{y''-y}{y'} = x$$

And from this point on I generally just integrate if it's first order, but the $y''$ has kinda lost me.

So is this the right method, if so, what next? if not, what am I doing wrong, and what should I be doing instead to solve it?

Note: This is not homework per se, it is simply from an exam review sheet that I was working on for a Calculus II class.

  • 4
    $\begingroup$ Isn't $F(x)=e^{x^2/2}$? $\endgroup$ – Thomas Andrews Dec 12 '13 at 23:07
  • $\begingroup$ Differentiate the series twice and check that it is a solution. I didn't check, but it should work. You can differentiate term by term, (why?). $\endgroup$ – Git Gud Dec 12 '13 at 23:08
  • $\begingroup$ @ThomasAndrews, Er, what do you mean? $\endgroup$ – Rivasa Dec 12 '13 at 23:10
  • $\begingroup$ @GitGud, When and if I do differentiate it twice, is it with respect to x? $\endgroup$ – Rivasa Dec 12 '13 at 23:10
  • $\begingroup$ @Link What I meant is'differentiate $F$', so yes, with respect to $x$. $\endgroup$ – Git Gud Dec 12 '13 at 23:11

HINT: Work directly with the series:

$$\begin{align*} xF'(x)+F(x)&=x\sum_{k\ge 0}\frac{2kx^{2k-1}}{2^kk!}+\sum_{k\ge 0}\frac{x^{2k}}{2^kk!}\\\\ &=\sum_{k\ge 0}\frac{2kx^{2k}}{2^kk!}+\sum_{k\ge 0}\frac{x^{2k}}{2^kk!}\\\\ &=\sum_{k\ge 0}\frac{(2k+1)x^{2k}}{2^kk!}\;. \end{align*}$$

Now calculate $F''(x)$ similarly and do a little simplification to show that it’s the same series.

  • $\begingroup$ Just a question, since I'm a bit fuzzy on this, when I differentiate with respect to $x$,I treat $k$ like just another integer right? $\endgroup$ – Rivasa Dec 12 '13 at 23:15
  • $\begingroup$ @Link: That’s right. It may help you to see what’s going on if you write out the first few terms of the series and follow the differentiation(s) term by term, as well as working with the summation notation. $\endgroup$ – Brian M. Scott Dec 12 '13 at 23:16
  • $\begingroup$ Okay then! But that's fine, I got the differentiation fine, just needed a small refresher. But the summation notation makes sense. Thanks for the help! Accepted! (As soon as I can) $\endgroup$ – Rivasa Dec 12 '13 at 23:18
  • $\begingroup$ @Link: You’re welcome! $\endgroup$ – Brian M. Scott Dec 12 '13 at 23:19

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