# [Differential Equations] Verify by direct substitution that a given power series is a solution

1. Nov 16, 2014

### 24karatbear

1. The problem statement, all variables and given/known data

2. Relevant equations
All relevant equations are given above.

3. The attempt at a solution
I decided to make a substitution for x2 so that x2 = u. By doing that, I could make the k=n+1 and k=n-1 substitutions so I can keep the series in phase. I took y' and just plugged everything into the DE. In the 2xy part, I just pulled the x into y so that I get the x2n+1 term. However, I can't seem to combine the series in one without having a loose term included (the first term of the y' series). I get the combined series with the loose term, all equal to zero - and I don't know where to go from there.

2. Nov 16, 2014

### PeroK

I'm not sure why you would need $u = x^2$. What did you get for $y'$?

3. Nov 16, 2014

### Zondrina

The question is simply asking you to find $\frac{d}{dx} [ \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} x^{2n} ]$ and then use it along with $y$ to show the equation is satisfied.

4. Nov 16, 2014

### 24karatbear

@PeroK

I thought making that substitution would make the series easier to work with in terms of getting each series in phase. With that substitution, I get:

y' = sum_n=1..inf 2n(-1)nx2n-1 / n! = sum_n=1..inf 2(-1)nun-1 / (n-1)!
(canceled out the n's)

and for 2xy (using the same substitution) I get:

2xy = sum_n=0..inf 2(-1)nun+1 / n!

But if I let k=n-1 for y' and k=n+1 for y, I can make both indices of summation start at the same number, k=1 if I shift the y' series up so that it starts at n=2. I get the following:

y'+2xy=0
→ -2 + sum_k=1..inf 2(-1)k+1uk / (k)! + sum_k=1..inf 2(-1)k-1uk / (k-1)! = 0

I can combine the two series, but after that, I'm not sure what to do.

@Zondrina

Yeah, that's why I'm not sure why this problem is causing difficulties for me. I was hoping it would be an easy substitute, but I keep running into the same issue above.

5. Nov 16, 2014

### PeroK

You got $y'$ right, but the you've gone astray with the pointless substitution $u=x^2$.

$u^{n-1} = x^{2n-2} \ne x^{2n-1}$

6. Nov 16, 2014

### 24karatbear

I see. Yeah, I tried to make that substitution because I needed a way to get the series in phase with each other / make their indices of summation the same and couldn't think of anything else. I see where I went wrong though.

Anyway, I just decided to write out the terms and verify it that way. Thanks for the help!