# Can't seem to figure out how to solve for C! Diff EQ question

1. Jan 14, 2006

### mr_coffee

Hello everyone i'm stuck, i got right to the end of this intial value problem and it doens't seem to want to work.

Solve the separable differential equation for u
{du}/{dt} = e^{3 u + 10 t}.
Use the following initial condition: u(0) = 6.
u = ?

I might have screwed up here:
du = e^(3u)*e^(10*t) dt
I took the ln of both sides but can u take the ln of du? thats where i might have screwed up, if it is, what do u suggest?

Here is my work:

I tried this and it was wrong, i also tried the answers of 1, ln(5), also was wrong, any ideas on where i screwed up? also the problem says find U, not C, but i don't think i have my C right, once i find the correct C, then do i plug it into one of the equations above invovling C? Like this:
u = (e^(10*t^2/2)+e^(ln(5)/3))^3; This is also incorrect i just submitted!
Thjanks!

Last edited: Jan 14, 2006
2. Jan 14, 2006

### Pyrrhus

How does

$$du =e^{3u} e^{10t} dt$$

goes to

$$du = 3u10t dt$$

3. Jan 14, 2006

### benorin

Here you go, the easy way

Solve the separable differential equation for u:

$$\frac{du}{dt} = e^{3u + 10t}$$

Use the following initial condition: u(0) = 6.

Rearranging the given differential equation gives

$$e^{-3u}du= e^{10t}dt$$

integrate both sides

$$\int e^{-3u}du = \int e^{10t}dt$$

to get

$$-\frac{1}{3}e^{-3u} = \frac{1}{10}e^{10t} +C$$

solving for u gives

$$u(t)=-\frac{1}{3}\ln \left( -\frac{3}{10}e^{10t}-3C\right)$$

to solve for the value of C that satisfies the initial condition u(0)=6 (the easy way,) substitute t=0 and u(0)=6 into

$$-\frac{1}{3}e^{-3u} = \frac{1}{10}e^{10t} +C$$

to get

$$-\frac{1}{3}e^{-3(6)} = \frac{1}{10}e^{10(0)} +C$$

which simplifies to

$$-\frac{1}{3}e^{-18} = \frac{1}{10} +C$$

yielding

$$C= -\frac{1}{10} -\frac{1}{3}e^{-18}$$

so the particular solution is

$$u(t)=-\frac{1}{3}\ln \left( -\frac{3}{10}e^{10t}-3\left( -\frac{1}{10} -\frac{1}{3}e^{-18} \right) \right) = -\frac{1}{3}\ln \left( -\frac{3}{10}e^{10t}+\frac{3}{10} +e^{-18} \right)$$

4. Jan 14, 2006

### d_leet

Yea that's waht I was wondering as well. Taking the natural log when you did is really not advisable, and doesn't help because you get ln(du) which is just going to make things harder for you. You started out pretty well changin the sum in the exponent into the product, but you should notice then that the equation is seperable and fairly simple to integrate once you seperate it.

5. Jan 14, 2006

### mr_coffee

Thank you everyone for your tips and yes Benion that is exactly what i got too after redoing it! thanks everyone! :)