Ordinary Differential Equation

stunner5000pt
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Homework Statement


Homework Statement


Solve \frac{dz}{dt} + 3 t e^{t+z} = 0


Homework Equations


None that I can think of...


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 I am wrong :(
 
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stunner5000pt said:

Homework Statement


Homework Statement


Solve \frac{dz}{dt} + 3 t e^{t+z} = 0


Homework Equations


None that I can think of...


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 I am wrong :(

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

Let's hope its that... I only get one more shot :(
 
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})}
 
Jesse H. said:
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})}

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.
 
Ray Vickson said:
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.

Ah, apologies.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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