# Ordinary Differential Equation

1. Mar 5, 2013

### stunner5000pt

1. The problem statement, all variables and given/known data
1. The problem statement, all variables and given/known data
Solve $$\frac{dz}{dt} + 3 t e^{t+z} = 0$$

2. Relevant equations
None that I can think of...

3. The attempt at a solution
"Rearranging" the given question, we get:

$$\int \frac{dz}{e^z} = -3\int t e^t dt$$

$$-e^{-z} = -3 \left( t e^t - e^t \right) + C$$
$$e^{-z} = 3 \left( t e^t - e^t \right) + C$$
$$z = - ln \left( 3 t e^t - 3 e^t + C \right)$$

Is this all correct? The system into which I need to enter this answer is saying im wrong :(

2. Mar 5, 2013

### Ray Vickson

Maybe it does not like the '-' sign; have you tried entering
$$\ln \left( \frac{1}{3 t e^t - 3 e^t + C}\right)?$$

3. Mar 5, 2013

### stunner5000pt

Let's hope its that... I only get one more shot :(

4. Mar 5, 2013

### MostlyHarmless

I think you have to take the natural log of each part individually. $$lne^{-z}=ln3te^t-ln3e^t+lnc$$ that would give $$z=-ln3te^t+ln3e^t+lnc$$ Which simplifies to $$z=ln({\frac{c3e^t}{3te^t})}$$

5. Mar 5, 2013

### Ray Vickson

No, you cannot do that. The answer I gave in my previous post was the one that Maple gave as the solution to the DE. The OP's workings were perfectly correct, as was the answer he gave.

6. Mar 5, 2013

### MostlyHarmless

Ah, apologies.