Question:
Solve the separable differential equation for Use the following initial condition: .
?
My work:
(Attached)
Any help would be greatly appreciated!
Thanks in advance!
$\displaystyle \frac{du}{dt} = e^{4u+8t} = e^{4u}\times e^{8t},u(0)=-6$
Therefore
$\displaystyle e^{-4u}~du= e^{8t}~dt$
$\displaystyle \frac{-1}{4}e^{-4u}= \frac{1}{8}e^{8t}+C$
$\displaystyle e^{-4u}= \frac{-1}{2}e^{8t}+C$
$\displaystyle -4u= \ln(\frac{-1}{2}e^{8t}+C)$
$\displaystyle u= \frac{-1}{4}\ln(\frac{-1}{2}e^{8t}+C)$