# 2nd order de

1. Dec 9, 2009

### manenbu

1. The problem statement, all variables and given/known data

Solve:
$(x^2-1)y'' + 4xy' + 2y = 6x$, given that $y_1=\frac{1}{x-1}$ and $y_2=\frac{1}{x+1}$.

2. Relevant equations

3. The attempt at a solution

Since both solutions are given, the solution to the homogenous system is:
$$y_h=C_1\frac{1}{x-1} + C_2\frac{1}{x+1}$$
And the solution to the original equation would be:
$$y=C_1(x)\frac{1}{x-1} + C_2(x)\frac{1}{x+1}$$

To solve I use this system of equations:

$$C_1'(x)\frac{1}{x-1} + C_2'(x)\frac{1}{x+1} = 0$$
$$-C_1'(x)\frac{1}{(x-1)^2} - C_2'(x)\frac{1}{(x+1)^2} = 6x$$

Somehow, all of this should end up being: $y=\frac{C_1}{x-1} + \frac{C_2}{x+1} + x$, according to the answers, but I just can't get there. Was there anything wrong in the systems of equations? Or is it me solving for the constants (I didn't write it here)?

2. Dec 9, 2009

### Staff: Mentor

It's not clear in your OP, but apparently y1 and y2 are solutions to the homogeneous equation. If that's the case, then you should try yp = Ax + B for your particular solution.

3. Dec 11, 2009

### manenbu

Yes, those are solutions to the homogenous equation. I'm trying to solve it using the variation method, using y = c1(x) y1 + c2(x) y2.

4. Dec 11, 2009

### HallsofIvy

Staff Emeritus
You could do that, but Mark44's suggestion for "undetermined coefficients" is much simpler.

5. Dec 13, 2009

### manenbu

Ok, so let's assume I still want to solve it using the variation method, just to see that I'm doing it ok (because obviously I went wrong there).

Is my system ok? Any recommendations on a preferred way to solve it?