Series Solutions

[Idealistic]

Solve y" - xy = 0

I have:

y" = sum(from n = 0 to inf){(n + 2)(n + 1)c[sub:3e2pos4t]n+2[/sub:3e2pos4t]x[sup:3e2pos4t]n[/sup:3e2pos4t]}

y = sum(from n = 0 to inf){c[sub:3e2pos4t]n[/sub:3e2pos4t]x[sup:3e2pos4t]n[/sup:3e2pos4t]}

but xy = y = sum(from n = 0 to inf){c[sub:3e2pos4t]n[/sub:3e2pos4t]x[sup:3e2pos4t]n+1[/sup:3e2pos4t]}

So when I try to solve:

x[sup:3e2pos4t]n[/sup:3e2pos4t][c[sub:3e2pos4t]n+2[/sub:3e2pos4t](n + 2)(n + 1) - xc[sub:3e2pos4t]n[/sub:3e2pos4t]] = 0


I have an "x" term in my equation. Can I still solve for my coefficients, c[sub:3e2pos4t]2n[/sub:3e2pos4t] and c[sub:3e2pos4t]2n+1[/sub:3e2pos4t], with the x variable in the equation

?

 

[Deleted member 4993]

Solve y" - xy = 0

I have:

y" = sum(from n = 0 to inf){(n + 2)(n + 1)c[sub:20lqzzf0]n+2[/sub:20lqzzf0]x[sup:20lqzzf0]n[/sup:20lqzzf0]}

y = sum(from n = 0 to inf){c[sub:20lqzzf0]n[/sub:20lqzzf0]x[sup:20lqzzf0]n[/sup:20lqzzf0]}

but xy = y = sum(from n = 0 to inf){c[sub:20lqzzf0]n[/sub:20lqzzf0]x[sup:20lqzzf0]n+1[/sup:20lqzzf0]}

So when I try to solve:

x[sup:20lqzzf0]n[/sup:20lqzzf0][c[sub:20lqzzf0]n+2[/sub:20lqzzf0](n + 2)(n + 1) - xc[sub:20lqzzf0]n[/sub:20lqzzf0]] = 0


I have an "x" term in my equation. Can I still solve for my coefficients, c[sub:20lqzzf0]2n[/sub:20lqzzf0] and c[sub:20lqzzf0]2n+1[/sub:20lqzzf0], with the x variable in the equation

?

Expand y" ~3 terms - do the same with "xy" - then look.

 

[Idealistic]

I'm not sure if I know what you mean, but this is what I think you're getting at:

first 3 "n" terms in y": 2*1c[sub:3fqxexjx]2[/sub:3fqxexjx] + 3*2c[sub:3fqxexjx]3[/sub:3fqxexjx]x + 4*3c[sub:3fqxexjx]4[/sub:3fqxexjx]x[sup:3fqxexjx]2[/sup:3fqxexjx]

first 3 "n" terms in xy: c[sub:3fqxexjx]0[/sub:3fqxexjx]x + c[sub:3fqxexjx]1[/sub:3fqxexjx]x[sup:3fqxexjx]2[/sup:3fqxexjx] + c[sub:3fqxexjx]2[/sub:3fqxexjx]x[sup:3fqxexjx]3[/sup:3fqxexjx]

these are the first three terms of each sum, y" and xy respectively, I'm not sure what I'm supposed to do really.

 

[Deleted member 4993]

I'm not sure if I know what you mean, but this is what I think you're getting at:

first 3 "n" terms in y": 2*1c[sub:3rapic6z]2[/sub:3rapic6z] + 3*2c[sub:3rapic6z]3[/sub:3rapic6z]x + 4*3c[sub:3rapic6z]4[/sub:3rapic6z]x[sup:3rapic6z]2[/sup:3rapic6z]

first 3 "n" terms in xy: c[sub:3rapic6z]0[/sub:3rapic6z]x + c[sub:3rapic6z]1[/sub:3rapic6z]x[sup:3rapic6z]2[/sup:3rapic6z] + c[sub:3rapic6z]2[/sub:3rapic6z]x[sup:3rapic6z]3[/sup:3rapic6z]

these are the first three terms of each sum, y" and xy respectively, I'm not sure what I'm supposed to do really.

y" - xy = 0

\(\displaystyle \sum_{n=1}[(n+1)(n+2)C_{n-2}-C_{n-1}]*x^n = - 2C_2\)