# Seperable Differential Equation

• Apr 10th 2011, 06:14 PM
Marconis
Seperable Differential Equation
We covered this last Tuesday, and had an exam on Thursday, so instead of doing HW I studied for the exam. I am a little confused on DE, and forgot a bit of what we went over in class. Could you check my work, please?
$\displaystyle \frac{dy}{dx}=\frac{\sqrt{x}}{e^y}$

$\displaystyle e^y dy= \sqrt{x}dx$

$\displaystyle \int e^y dy= \int \sqrt{x}dx$

$\displaystyle e^y= \frac{2}{3}x^\frac{3}{2} + C$

$\displaystyle ylne=\frac{2}{3}x^\frac{3}{2} + C$

$\displaystyle y=ln(\frac{2}{3}x^\frac{3}{2} + C)$

Thank you.
• Apr 10th 2011, 06:22 PM
Chris L T521
Quote:

Originally Posted by Marconis
We covered this last Tuesday, and had an exam on Thursday, so instead of doing HW I studied for the exam. I am a little confused on DE, and forgot a bit of what we went over in class. Could you check my work, please?
$\displaystyle \frac{dy}{dx}=\frac{\sqrt{x}}{e^y}$

$\displaystyle e^y dy= \sqrt{x}dx$

$\displaystyle \int e^y dy= \int \sqrt{x}dx$

$\displaystyle e^y= \frac{2}{3}x^\frac{3}{2} + C$

$\displaystyle ylne=\frac{2}{3}x^\frac{3}{2} + C$

$\displaystyle y=ln(\frac{2}{3}x^\frac{3}{2} + C)$

Thank you.

You're pretty much correct, but note the minor thing you missed:

$\displaystyle y\ln e = \frac{2}{3}x^{3/2}+C$ should really be $\displaystyle y\ln e = \ln(\frac{2}{3}x^{3/2}+C)$.

When you take ln of one side, you have to do the same for the other.

Besides that, I don't see anything wrong with your solution.
• Apr 10th 2011, 06:25 PM
Marconis
Those are the kinds of silly mistakes that tend to cost me on exams. Thanks for pointing that out, and thank you for the speedy reply.