# Simple Diff. Eq.

1. Aug 29, 2011

### ƒ(x)

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

2. Relevant equations

you can see my answer in the picture.

3. The attempt at a solution

y*e^(-2t) = (-1/2)t*e^(-2t) + (-1/4)e^(-2t) + (1/3)e^(3t) + C
y = (-1/2)t + (-1/4) + (1/3)e^(5t) + C/(e^(-2t))

1/3 = 0 - 1/4 + 1/3 + C
C = 1/4 (i tried 1/4 and -1/4 in the answer box)

I've even checked my answer by taking the derivative. Where'd I make a simple mistake?

2. Aug 29, 2011

### ƒ(x)

and i tried it with C = 0

3. Aug 29, 2011

### Staff: Mentor

The only mistake I see is that you forgot to include the homogeneous solution in your answer. You found C = 1/4, which is correct, but your answer doesn't have the e^(2t) term.

BTW, it's simpler to write e^(2t) than 1/e^(-2t).

4. Aug 29, 2011

### ƒ(x)

What do you mean by the e^(2t) term?

5. Aug 29, 2011

### ƒ(x)

Ah, found it. Thanks. Forgot to multiply my C by e^2t

6. Aug 29, 2011

### Staff: Mentor

The answer you show in the OP is -t/2 + 1/3 *e^(5t) -1/4

The actual answer (the general solution to the initial value problem) is y = -t/2 + 1/3 *e^(5t) -1/4 + 1/4 *e^(2t)

7. Aug 29, 2011

Solved.