# Differential Equation problem

1. May 6, 2009

### cse63146

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

Compute the general solution of y'' + 6y' + 8y = 2t + et

2. Relevant equations

3. The attempt at a solution

after using determinants, I found the general solution to be y(t) = k1e-2 + k2e-4 + yp

To find yp, I would make it equal to yp = aet find its first and second derivates, and substitute them back into y'' + 6y' + 8y , but what do I do about the 2t?

2. May 6, 2009

### Cyosis

You can do some clever guess work. You can make sure that y'' is 0 by taking some linear function. This simplifies your problem to 6y'+8y=2t. Since we took a linear function y' will be a constant so this will reduce your problem to constant+8y=2t. A function of the form b+ct should suffice.

3. May 6, 2009

### cse63146

would yp = t be a good guess?

since y'p = 1 and y''p = 0

4. May 6, 2009

### Cyosis

But $6+8t \neq 2t$, so no. Try b+ct as suggested in my previous post.

5. May 6, 2009

### cse63146

how about $$\frac{1}{4}t - \frac{3}{16}$$

so 6(1/4) + 8($$\frac{1}{4}t - \frac{3}{16}$$) = 3/2 + 2t - 3/2 = 2t

so when considering et, I would let yp =$$\frac{1}{4}t - \frac{3}{16} + e^t$$ ?

6. May 6, 2009

### Cyosis

The linear term is correct now, however if we plug in e^t we get, e^t+6e^t+8e^t=15e^t!=e^t. So you have to divide by 15.

7. May 6, 2009

### cse63146

oops

it should be $$\frac{1}{4}t - \frac{3}{16} + ae^t$$

and a would be 1/15

8. May 6, 2009

Correct.

9. May 6, 2009

### cse63146

Thank you for all your help.

10. May 6, 2009

### Cyosis

You're welcome.

11. May 6, 2009

### Staff: Mentor

Pretty sure you meant to write y(t) = k1e-2t + k2e-4t + yp here.

12. May 6, 2009

### cse63146

Oops (again).

Yeah, I did. Thanks.