Solving a Difficult Homework Equation

[24karatbear]

Homework Statement

Homework Equations

All relevant equations are given above.

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.

 

[PeroK]

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

 

[STEMucator]

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.

 

[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)ⁿx^2n-1 / n! = sum_n=1..inf 2(-1)ⁿu^n-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+1u^k / (k)! + sum_k=1..inf 2(-1)^k-1u^k / (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.

 

[PeroK]

@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)ⁿx^2n-1 / n! = sum_n=1..inf 2(-1)ⁿu^n-1 / (n-1)!
(canceled out the n's)

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

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

 

[24karatbear]

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

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

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!