1. ## separable differential equation

I was working on my hw and i'm stuck on this question.

Solve the separable differential equation for ,
Use the following initial condition:

[

2. $e^{4u+7t} = e^{4u} \cdot e^{7t}$

separate variables ...

$e^{-4u} \, du = e^{7t} \, dt$

take it from here?

3. $-[ln(-4/7)*7*t-48]/4$

i ended up with this but it's wrong

4. $e^{-4u} \, du = e^{7t} \, dt$

$-\frac{1}{4}e^{-4u} = \frac{1}{7}e^{7t} + C$

$e^{-4u} = -\frac{4}{7}e^{7t} + C_2$

$-4u = \ln\left(C_2 - \frac{4}{7}e^{7t}\right)$

$u = -\frac{1}{4}\ln\left(C_2 - \frac{4}{7}e^{7t}\right)$

$u(0) = 12$

$12 = -\frac{1}{4}\ln\left(C_2 - \frac{4}{7}\right)$

$C_2 = e^{-48} + \frac{4}{7}$

$u = -\frac{1}{4}\ln\left(e^{-48} + \frac{4}{7} - \frac{4}{7}e^{7t}\right)$

not very pretty.