# dP/dt =, how do you find the solution?

• Apr 14th 2010, 11:23 AM
AlphaRock
dP/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 PM
TKHunny
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 PM
AlphaRock
Quote:

Originally Posted by TKHunny
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.

Yes. I got c = 0.

Placing that c in the equation:

-e^(-P) = te^t - e^t + 0.

Then when I lned both sides, I got
-P = ln(e^t - te^t)
P = - ln(e^t - te^t)

which I don't think is right. How do we find the answer if this isn't the answer?
• Apr 14th 2010, 03:05 PM
TKHunny
How did you get C = 0? Please try that again. $e^{0}\;=\;1\;\ne\;e$

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

Hmm... "ln" really isn't a verb. Try not to speak pigeon math.
• Apr 14th 2010, 04:59 PM
AlphaRock
Quote:

Originally Posted by TKHunny
How did you get C = 0? Please try that again. $e^{0}\;=\;1\;\ne\;e$

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

Hmm... "ln" really isn't a verb. Try not to speak pigeon math.

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

P(0) = 1
Therefore,
-e^(-1) = 0e^t - e^0 + c.
-1 = -1 + c
-1 + 1 = c
0 = c
c= 0
• Apr 14th 2010, 09:41 PM
TKHunny
You seem to be saying that $e^{-1}\;=\;1$.

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

Be more careful.