- #1

jazznaz

- 23

- 0

Solve:

[tex]xy'' - 2y' + xy = -2\cos x[/tex]

Using the method of Laplace transforms...

So I do some jiggling and get to:

[tex](1+s^{2})\frac{dF}{ds}+4sF(s) = \frac{2s}{s^{2} + 1}[/tex]

To which I find the following solution:

[tex]F(s) = \frac{C}{(1 + s^{2})^{2}} + \frac{s^{2}}{(1 + s^{2})^{2}}[/tex]

But I'm supposed to get to:

[tex]F(s) = \frac{C}{(1 + s^{2})^{2}} + \frac{s^{2}-1}{(1 + s^{2})^{2}}[/tex]

But the method I'm using to solve the first order ODE (variation of parameters/constants) won't seem to lead to the answer I need. It almost looks as though I should include an aribtrary constant where they're expecting me to get -1, but it seems that any constant value will be a solution to the first order DE obtained after applying the transform. So I don't really understand how they've deduced that it should be -1...

We've been given the answer to the problem,

[tex]y = \frac{C}{2}(\sin x - x\cos x) + x\cos x[/tex]

Where the result is obtained easily using a table of Laplace Transforms.