Easy ODE to solve using frobenius method help?

[razoribbon]

I can't seem to figure this one out:

y'' + (1/x)*y' -c*y = -c*y0

where c is just a constant.

if someone could please go through the steps of the frobenius method, i would really appreciate it!

thanks.

 

[HallsofIvy]

Step 1: Since the constant solution, y= cy0 for all x, is clearly a solution you can solve the homogeneous equation y''+ (1/x)y'- cy= 0 and add cy0.

Step 2: Multiply that equation by x to get xy''+ y'- cxy= 0.

Step 3: Write
[tex]y= \sum_{n=0}^\infty a_nx^{n+c}[/itex]
differentiate term by term to find y' and y'' and substitute into that equation.

Step 4: Look just at the terms having lowest power of x. That should be a quadratic equation in c times a₀. Assume that [itex]a_0\ne 0[/itex] so that the coefficient must be 0 and solve that "indicial equation" for two values of c.

Step 5: Put each of those values of c into your equation in turn to find two independent solutions to the homogeneous equation. Setting each coefficient of x to a power to 0, you will get a recursive equation for a_n.

 

[ice109]

i thought frobenius method was to look at the asymptotic solutions? halls what explained was simply how find a series solution?

 

[HallsofIvy]

No, Frobenius' method is finding series solution to a differential equation at a regular singular point.
http://en.wikipedia.org/wiki/Frobenius_method

 

[razoribbon]

thank you for such a clear response!

these are my steps:
[tex]
y = \sum_{n=0}^{\infty} a_{n}x^{n + c}
[/tex]
[tex]
\frac{dy}{dx} = \sum_{n=0}^\infty (n + c)a_{n}x^{n + c -1}
[/tex]
[tex]
\frac{d^{2}y}{dx^{2}} = \sum_{n=0}^{\infty} (n + c)(n + c -1)a_{n}x^{n + c - 2}
[/tex]

now i substitute them into the original equation multiplied by x:

[tex]
\sum_{n=0}^{\infty} (n + c)(n + c -1)a_{n}x^{n + c -1} + (n + c)a_{n}x^{n + c -1} - {\alpha}a_{n}x^{n + c + 1} = 0
[/tex]

the first two terms have the same power of x, so the equation can be simplified:

[tex]
\sum_{n=0}^{\infty} (n + c)^{2}a_{n}x^{n + c - 1} - {\alpha}a_{n}x^{n + c + 1} = 0
[/tex]

does this mean that the only solution is c = 0? have i done the right things so far?

thanks for your help!

 

[HallsofIvy]

Yes, that is correct. Now, since the indicial equation has a double root, c= 0, you might want to look at
http://en.wikipedia.org/wiki/Frobenius_method#Double_Roots