# Showing that a series is a solution of a second order linear differential equation.

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.

• Isn't $F(x)=e^{x^2/2}$? – Thomas Andrews Dec 12 '13 at 23:07
• 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?). – Git Gud Dec 12 '13 at 23:08
• @ThomasAndrews, Er, what do you mean? – Rivasa Dec 12 '13 at 23:10
• @GitGud, When and if I do differentiate it twice, is it with respect to x? – Rivasa Dec 12 '13 at 23:10
• @Link What I meant is'differentiate $F$', so yes, with respect to $x$. – Git Gud Dec 12 '13 at 23:11

\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.
• 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? – Rivasa Dec 12 '13 at 23:15