Hi, I have an initial value problem to solve for a second order differential equation. Currently in my class, we have learned undetermined coefficients, characteristic equations, variation of parameters

The questions is:

Okay, so first I need to find the general solution. The only method we've learned for finding the general solution of a non-homogeneous, second order differential equation is variation of parameters. But to use that method, we first need to find two solutions for the homogeneous form.

If we use the characteristic equation method, we end up with

Now I'm not sure how to proceed. The only way I can see to get two linearly independent solutions is by setting one of the terms to ln(x). So we could do the following:

Then when I try to use variation of parameters, solving for u and v in y=u'*y

And it's not clear to me how to integrate that function. So I'm wondering if 1) I'm approaching this wrong at some point, 2) I just made a mistake somewhere along the line, or 3) if I'm missing something really obvious and over-complicating the problem.

Thanks for any help.

The questions is:

Code:

```
Solve the initial value problem:
y'' + y = x
y(0)=1, y'(0)=1
```

If we use the characteristic equation method, we end up with

Code:

```
[FONT=arial]y = C[SUB]1[/SUB]*e[SUP]-x[/SUP] + C[SUB]2[/SUB]*e[SUP]-x[/SUP] = e[SUP]-x[/SUP] (C[SUB]1[/SUB]+C[SUB]2[/SUB])
[/FONT]
```

Now I'm not sure how to proceed. The only way I can see to get two linearly independent solutions is by setting one of the terms to ln(x). So we could do the following:

Code:

```
C[SUB]1[/SUB]=1, C[SUB]2[/SUB]=0
y[SUB]1[/SUB] = e[SUP]-x[/SUP]
C[SUB]1[/SUB]=ln(x), C[SUB]2[/SUB]=0
y[SUB]2[/SUB] = -1
```

_{1}+ v'*y_{2}, I end up with the integral:
Code:

`[TEX]$u = \int \frac{x}{e^{-x}}$[/TEX]`

Thanks for any help.

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