Thread: Seperable Differential Equation

1. 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.

2. 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.

3. 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.