1. ## separable differential equation

So I've attempted this problem sooo many times! And I don't know where I'm going wrong. I've also consulted many friends in my class and they're also stuck. So here it goes....

du/dt = e^(2.6t-3.1u), u(0)= 1.9
Find u(t).

So what I did was:
du/dt = e^(2.6t)/e^(3.1u)
e^(3.1u) du = e^(2.6t) dt
INTEGRATE BOTH SIDES
1/3.1 (e^3.1u) = 1/2.6 (e^2.6t) + C

and then I solved for C:
C= (1/3.1)e^(5.89) - (1/2.6)

and then my u(t) equation was:
[(2.6t(ln(1/2.6)) + 5.89(ln(1/3.1)) - ln (1/2.6))/(ln(1/3.1))]/(3.1)

which is wrong.
so if anyone can help out that would be great!!

2. $\begin{gathered}
\frac{{du}}
{{dt}} = e^{2.6t - 3.1u} \hfill \\
e^{3.1u} du = e^{2.6t} dt \hfill \\
\frac{1}
{{3.1}}e^{3.1u} = \frac{1}
{{2.6}}e^{2.6t} + C \hfill \\
\end{gathered}
$

now apply the initial condition...

3. yes, I did that.
And I get C= (1/3.1)*e^(5.89) - (1/2.6)

is that correct???

4. correct