Ordinary Differential Equation

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stunner5000pt
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Homework Statement


Homework Statement


Solve [tex]\frac{dz}{dt} + 3 t e^{t+z} = 0[/tex]


Homework Equations


None that I can think of...


The Attempt at a Solution


"Rearranging" the given question, we get:

[tex]\int \frac{dz}{e^z} = -3\int t e^t dt[/tex]

[tex]-e^{-z} = -3 \left( t e^t - e^t \right) + C[/tex]
[tex]e^{-z} = 3 \left( t e^t - e^t \right) + C[/tex]
[tex]z = - ln \left( 3 t e^t - 3 e^t + C \right)[/tex]

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 [tex]\frac{dz}{dt} + 3 t e^{t+z} = 0[/tex]


Homework Equations


None that I can think of...


The Attempt at a Solution


"Rearranging" the given question, we get:

[tex]\int \frac{dz}{e^z} = -3\int t e^t dt[/tex]

[tex]-e^{-z} = -3 \left( t e^t - e^t \right) + C[/tex]
[tex]e^{-z} = 3 \left( t e^t - e^t \right) + C[/tex]
[tex]z = - ln \left( 3 t e^t - 3 e^t + C \right)[/tex]

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

Let's hope its that... I only get one more shot :(
 
I think you have to take the natural log of each part individually. [tex]lne^{-z}=ln3te^t-ln3e^t+lnc[/tex] that would give [tex]z=-ln3te^t+ln3e^t+lnc[/tex] Which simplifies to [tex]z=ln({\frac{c3e^t}{3te^t})}[/tex]
 
Jesse H. said:
I think you have to take the natural log of each part individually. [tex]lne^{-z}=ln3te^t-ln3e^t+lnc[/tex] that would give [tex]z=-ln3te^t+ln3e^t+lnc[/tex] Which simplifies to [tex]z=ln({\frac{c3e^t}{3te^t})}[/tex]

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.