seperable differential equation...?

Question

du/dt = e^(3u - 14t)

initial condition: u(0) = 0

HINT:
To determine the constant of integration after you integrated both sides, DO NOT take natural logs, but rather just set u = 0 and t = 0 with u and t both still exponents. After determining the constant, then you need to take logs on both sides to solve for u.


The hint was given in a practice problem. Let me see if y'all can show me something of it.

Your answer is off amigo. You were obviously on the right track, but you didn't solve for u. You also ended up putting 1/13 instead of 1/3.

The answer i came up with is. I think this is right

u(t) = ln((3/14)(e^(-14t)+11/3))/-3

Answer 1

du /e^(3u )= e^(- 14t) dt
e^(-3u ) du = e^(- 14t) dt

-1/3* e^(-3u) = -1/14* e^(- 14t) +C
if u=0 and t=0
-1/3=-1/14+C
C=1/14-1/13
C=-1/182

-1/3* e^(-3u) = -1/14* e^(- 14t) -1/182

e^(-3u) = 3/14* e^(- 14t) -3/182