# Easy ODE to solve using frobenius method help?

1. Jun 29, 2009

### 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.

2. Jun 29, 2009

### 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
$$y= \sum_{n=0}^\infty a_nx^{n+c}[/itex] differentiate term by term to find y' and y'' and substitute into that equation. Step 4: Look just at the terms having lowest power of x. That should be a quadratic equation in c times a0. Assume that $a_0\ne 0$ so that the coefficient must be 0 and solve that "indicial equation" for two values of c. Step 5: Put each of those values of c into your equation in turn to find two independent solutions to the homogeneous equation. Setting each coefficient of x to a power to 0, you will get a recursive equation for an. 3. Jun 29, 2009 ### ice109 i thought frobenius method was to look at the asymptotic solutions? halls what explained was simply how find a series solution? 4. Jun 29, 2009 ### HallsofIvy 5. Jun 30, 2009 ### razoribbon thank you for such a clear response! these are my steps: [tex] y = \sum_{n=0}^{\infty} a_{n}x^{n + c}$$
$$\frac{dy}{dx} = \sum_{n=0}^\infty (n + c)a_{n}x^{n + c -1}$$
$$\frac{d^{2}y}{dx^{2}} = \sum_{n=0}^{\infty} (n + c)(n + c -1)a_{n}x^{n + c - 2}$$

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

$$\sum_{n=0}^{\infty} (n + c)(n + c -1)a_{n}x^{n + c -1} + (n + c)a_{n}x^{n + c -1} - {\alpha}a_{n}x^{n + c + 1} = 0$$

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

$$\sum_{n=0}^{\infty} (n + c)^{2}a_{n}x^{n + c - 1} - {\alpha}a_{n}x^{n + c + 1} = 0$$

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