Separation of variables problem, please check.

**tttcomrader** · September 9th 2007, 10:32 AM

Question: $\frac{du}{dt} = e^{4u+3t}$ , $u(0) = 10$

My answer: $u = \frac{ln(-\frac{4e^{3t}}{3}+e^{-40}+\frac{4}{3})}{-4}$

I want to believe this is right, but when I submit the answer it turns out wrong...

Did I do something wrong here?

Thanks.

**topsquark** · September 9th 2007, 10:58 AM

Originally Posted by tttcomrader

Question: $\frac{du}{dt} = e^{4u+3t}$ , $u(0) = 10$

My answer: $u = \frac{ln(-\frac{4e^{3t}}{3}+e^{-40}+\frac{4}{3})}{-4}$

I want to believe this is right, but when I submit the answer it turns out wrong...

Did I do something wrong here?

Thanks.

$\frac{du}{dt} = e^{4u+3t} = e^{4u}e^{3t}$ , $u(0) = 10$

$e^{-4u}du = e^{3t}dt$

$\int e^{-4u}du = \int e^{3t}dt$

$-\frac{1}{4}e^{-4u} = \frac{1}{3}e^{3t} + C$

We have $u(0) = 10$ , so
$-\frac{1}{4}e^{-4 \cdot 10} = \frac{1}{3}e^{3 \cdot 0} + C$

$C = -\frac{1}{4}e^{-40} - \frac{1}{3}$ .

Thus
$-\frac{1}{4}e^{-4u} = \frac{1}{3}e^{3t} - \frac{1}{4}e^{-40} - \frac{1}{3}$

$e^{-4u} = -\frac{4}{3}e^{3t} + e^{-40} + \frac{4}{3}$

$-4u = ln \left ( -\frac{4}{3}e^{3t} + e^{-40} + \frac{4}{3} \right )$

$u = -\frac{1}{4} \cdot ln \left ( -\frac{4}{3}e^{3t} + e^{-40} + \frac{4}{3} \right )$

It looks good to me.

Perhaps it's looking for a factored version of the argument of the logarithm?
$u = -\frac{1}{4} \cdot ln \left ( \frac{4}{3} \left [1 - e^{3t} \right ] + e^{-40} \right )$

or possibly

$u = ln \left ( \frac{1}{\sqrt[4]{-\frac{4}{3}e^{3t} + e^{-40} + \frac{4}{3}}} \right )$

or something.

-Dan

Thread: Separation of variables problem, please check.

LinkBack

Thread Tools

Display

Separation of variables problem, please check.

Similar Math Help Forum Discussions

Separation of variables problem then satisfying initial condition

Separation of Variables Problem

rate problem with separation of variables

Separation of Variables

Separation of variables...

Search Tags