dP/dt = te^(t+P) ; where P = 1 when t = 0.

I got to the part where:

dP/(e^P) = te^t dt

Found the integral:

-e^(-P) = te^t - e^t + c

Not sure how to find the solution. (we also can't ln a negative).

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- Apr 14th 2010, 11:23 AMAlphaRockdP/dt =, how do you find the solution?
dP/dt = te^(t+P) ; where P = 1 when t = 0.

I got to the part where:

dP/(e^P) = te^t dt

Found the integral:

-e^(-P) = te^t - e^t + c

Not sure how to find the solution. (we also can't ln a negative). - Apr 14th 2010, 12:11 PMTKHunny
Have you consdiered substituting the given values for p and t?

Note: Is it "p" or "P". Normally, it's bad form to be switching about. - Apr 14th 2010, 12:37 PMAlphaRock
- Apr 14th 2010, 03:05 PMTKHunny
How did you get C = 0? Please try that again. $\displaystyle e^{0}\;=\;1\;\ne\;e$

It may help to write it as $\displaystyle -e^{-P}\;=\;e^{t}\cdot (t-1) + C$

Hmm... "ln" really isn't a verb. Try not to speak pigeon math. - Apr 14th 2010, 04:59 PMAlphaRock
- Apr 14th 2010, 09:41 PMTKHunny
You seem to be saying that $\displaystyle e^{-1}\;=\;1$.

Please reconsider, particularly since later you have $\displaystyle e^{0}\;=\;1$. You can't have it both ways!

Be more careful.