find the general solution of du/dt=e^(u+2t)
Refreshing my memory of differential equations, I think we have a separable equation here
We can write
du/dt = e^(u) * e^(2t)......and separating the variables, we have
e^(-u) du = e^(2t) dt....and integrating both sides gives us
∫ e^(-u) du = ∫ e^(2t) dt.....now, perform the integration, and we get either
-e^(-u) = e^(2t) / 2 + C or -e^(-u) = e^(2t) / 2 - C
Note that we can't use the first one, because we would have to take a negative LN in the next step (which isn't possible)........i.e., we have to "fudge" the problem a bit and take the negative sign in front of the constant, C..........multiplying the second one through by - 1, we have
e^(-u) = C - e^(2t) / 2 Now take the LN of both sides
LN (e^(-u)) = LN (C - e^(2t) / 2) ........ =
-u = LN (C - e^(2t) / 2)........... =
u(t) = - LN (C - e^(2t) / 2)
(Notice that we're assuming that C > e^(2t) / 2)
I believe that's the answer........I checked this with an online solver and it arrived at the same solution, but if any of the other forum members spot a flaw in my reasoning....I'll stand corrected!!
+128731 
