Ordinary Differential Equation

Click For Summary
The discussion revolves around solving the ordinary differential equation \(\frac{dz}{dt} + 3 t e^{t+z} = 0\). The original poster attempted to solve it by rearranging and integrating, arriving at the solution \(z = -\ln(3 t e^t - 3 e^t + C)\). However, they encountered an error when submitting their answer to a system, leading to confusion about the correctness of their solution. Other participants suggested alternative forms for the logarithmic expression and debated the validity of the original poster's steps. Ultimately, it was confirmed that the original poster's solution was indeed correct, despite the system's rejection.
stunner5000pt
Messages
1,447
Reaction score
5

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 :(
 
Physics news on Phys.org
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
6
Views
2K
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 12 ·
Replies
12
Views
1K
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K