Ordinary Differential Equation

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Homework Help Overview

The discussion revolves around solving the ordinary differential equation \(\frac{dz}{dt} + 3 t e^{t+z} = 0\). Participants are exploring the correctness of various approaches to rearranging and integrating the equation.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to rearrange the equation and integrate both sides. There are questions about the correctness of the integration steps and the final expression for \(z\). Some participants suggest alternative forms of the logarithmic expression to check for system acceptance.

Discussion Status

The discussion is ongoing, with some participants expressing uncertainty about the integration process and the format of the answer required by the system. There is a mix of agreement on the original poster's approach, but also suggestions for alternative representations of the solution.

Contextual Notes

Participants are working under the constraints of a system that evaluates their answers, which may have specific formatting requirements. There is also a mention of a reference solution provided by a software tool, which adds to the complexity of the discussion.

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.
 

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