Seperable Differential Equation

**Marconis** · April 10th 2011, 06:14 PM

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?
$<br /> \frac{dy}{dx}=\frac{\sqrt{x}}{e^y}$

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

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

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

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

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

Thank you.

**Chris L T521** · April 10th 2011, 06:22 PM

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?
$<br /> \frac{dy}{dx}=\frac{\sqrt{x}}{e^y}$

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

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

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

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

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

Thank you.

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

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

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.

**Marconis** · April 10th 2011, 06:25 PM

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.

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