Solve the Differential Equation by method of Series

[Sol-chan]

Solve by the method of series

y''-xy'+y=0

From what I understand, I need to substitute in
y(x)=SUM(C_n*xⁿ) from n=0 to infinity
I get that:
y'(x)=SUM(n*C_n*x^n-1) from n=1 to infinity
and
y''(x)=SUM(n*(n-1)*C_n*x^n-2) from n=2 to infinity

When I substitute in and collect the terms, I get:
2C₂ + C₀ + SUM[(C_n+2*(n+2)*(n+1)-C_n*n+C_n)*xⁿ] from n=1 to infinity = 0

So then I get that
C₂ = -C₀/2
and
C_n+2 = C_n/(n+2)

And I don't know what to do after that. Any hints on where to go from there would be much appreciated...

 

[tiny-tim]

Hi Sol-chan !

C₂ = -C₀/2

hint:

C₄ = C₀ *… ?

C₆ = C₀ *… ?

C₈ = C₀ *… ? ​

 

[SammyS]

So then I get that
C₂ = -C₀/2
and
C_n+2 = C_n/(n+2)

I get (for n even):

C_{\,n+2}=\frac{(n-1)C_n}{(n+1)(n+2)}

Plug in y = C₁x to find out about C₁.

Take the derivative of both sides of y''-xy'+y=0, to get the recurrence relation for C_n for odd n > 1.

 

[Sol-chan]

Ah, right, I solved that wrong. So, with

C_n+2 = (n-1)C_n/(n+1)(n+2)

I get that
C₄ = -C₀/24
C₆ = -C₀/240
C₈ = -C₀/2688

What am I plugging y=C₁x into?

 

[SammyS]

y''-xy'+y=0

 

[Sol-chan]

Doesn't that just give 0 = 0?
Because if y=C₁x then y'=C₁ and y''=0, then you get
0-x*C₁ + x*C₁ = 0
0 = 0

And I don't really understand the recurrence relation, either. I'm not sure if I'm doing this right, but taking the derivative of both sides I get
y''' - xy'' +2y' = 0
And I'm not sure what that's supposed to be telling me...

 

[SammyS]

Right, so C₁x is a solution. Right?

C₁ is an arbitrary coefficient. So is C₀ arbitrary.

 

[Sol-chan]

Okay, arbitrary... meaning I can just set them equal to whatever I want, right? But I still don't see how that helps me...

 

[SammyS]

No, it means you have to set them to something general, like ... C₀ & C₁.

Try to come up with a general expression for C_n, where n is even. It will be of the form, C₀*f(n).

The expression: C_{\,n+2}=\frac{(n-1)C_n}{(n+1)(n+2)} tells you what about C_n for n=3 and odd n > 3?

 

[SammyS]

Ah, right, I solved that wrong. So, with

C_n+2 = (n-1)C_n/(n+1)(n+2)

I get that
C₄ = -C₀/24
C₆ = -C₀/240
C₈ = -C₀/2688

Write these without canceling or combining factors, so that you can the general form for C_n :

C₄ = -C₀ (1) / [(1)(2)(3)(4)]
C₆ = -C₀ (1)(3) / [(1)(2)(3)(4)(5)(6)]
C₈ = -C₀ (1)(3)(5) / [(1)(2)(3)(4)(5)(6)(7)(8)]
...​

Note: For even n, (n-1)(n-3)...(5)(3)(1) = [(n-1)(n-3)...(5)(3)(1)][(n)(n-2)(n-4)...(4)(2)]/[(n)(n-2)(n-4)...(4)(2)] = n!/[2^(n/2)(n/2)!]