# Separation of variables problem, please check.

• September 9th 2007, 10:32 AM
Separation of variables problem, please check.
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.
• September 9th 2007, 10:58 AM
topsquark
Quote:

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