1. Differential Equation

I was wondering if someone could help me with where I am going wrong on this question. I will post the work I have completed, which gave me the wrong solution.
The Question is:
du/dt= e^(1t-3.2u) u(0)=2.6

Find the function u(t) from the differential equation and the initial conditions...

u(t)=__________

This is what I have done:

du/dt = e^t/e^(-3.2u)
e^(-3.2u) du = e^t dt
integrate both sides
-1/3.2 e^(-3.2u) = e^t + c, where c is constant
e^(-3.2u) = -3.2 e^t + c
-3.2 u = ln(-3.2 e^t +c)
u = -1/3.2 ln(-3.2 e^t + c)

Solving for c by using the initial condition
u(0) = -1/3.2 ln(-3.2 + c) = 2.6
Solve for c:
ln(-3.2+c) = -8.32
c = e^-8.32 + 3.2
c = 3.2002

u(t) = -1/3.2 ln (-3.2 e^t + 3.2002)

But this is not correct

2. I was wondering if someone could help me with where I am going wrong on this question. I will post the work I have completed, which gave me the wrong solution.
The Question is:
du/dt= e^(1t-3.2u) u(0)=2.6
$\frac{du}{dt}=e^{1t-3.2u}=e^{1t} \cdot e^{-3.2u}$

Seperating the equation we get

$e^{3.2u}du=e^tdt$ if we integrate both sides

$\int e^{3.2u}du=\int e^tdt \iff \frac{1}{3.2}e^{3.2u}=e^t+C$

You should be able to finish from here

3. Originally Posted by BKennedy
I was wondering if someone could help me with where I am going wrong on this question. I will post the work I have completed, which gave me the wrong solution.
The Question is:
du/dt= e^(1t-3.2u) u(0)=2.6

Find the function u(t) from the differential equation and the initial conditions...

u(t)=__________

This is what I have done:

du/dt = e^t/e^(-3.2u)
e^(-3.2u) du = e^t dt
integrate both sides
-1/3.2 e^(-3.2u) = e^t + c, where c is constant
e^(-3.2u) = -3.2 e^t + c
-3.2 u = ln(-3.2 e^t +c)
u = -1/3.2 ln(-3.2 e^t + c)

Solving for c by using the initial condition
u(0) = -1/3.2 ln(-3.2 + c) = 2.6
Solve for c:
ln(-3.2+c) = -8.32
c = e^-8.32 + 3.2
c = 3.2002

u(t) = -1/3.2 ln (-3.2 e^t + 3.2002)

But this is not correct

$\frac{du}{dt} = e^{t-3.2u}$

$\frac{du}{dt} = \frac{e^t}{e^{3.2u}}$