Beginner Differential Equations

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October 20th 2012, 02:31 PM
astuart

Beginner Differential Equations

Hello,

I've just started on separating differential equations and am just having a little trouble on how to proceed with this sort of question.

$\frac {dy}{dx} = y^2 \,e^(2x)$

I understand that the y needs to be bough to the left, and then the formal separation of $\frac {dy}{dx}$ occurs - as my tutor puts it.

$=\frac{1}{y^2}\,dy = e^{2x} dx$

$= \int \frac{1}{y^2}dy = \int e^{2x}$

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

This is where it breaks down a little bit for me...I'd assume that we need to get y to its simplest form, but I don't know how to do that...

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

That isn't right (I'm assuming), so can someone point me in the direction of the correct step.

Thanks :)
October 20th 2012, 02:55 PM
TheEmptySet

Re: Beginner Differential Equations

Your seperation is good but

$\int \frac{1}{y^2}dy = -\frac{1}{y}$

So you should get

$-\frac{1}{y}=\frac{1}{2}e^{2x}+C$
October 20th 2012, 02:59 PM
MarkFL

Re: Beginner Differential Equations

First, when you separate the variables to get:

$\frac{1}{y^2}\cdot\frac{dy}{dx}=e^{2x}$

you should note that you are losing the trivial solution $y\equiv0$ .

Then integrating with respect to $x$ , we have:

$\int y^{-2}\,dy=\int e^{2x}\,dx$

As noted above, you have integrated incorrectly.
October 20th 2012, 03:00 PM
astuart

Re: Beginner Differential Equations

Quote:

Originally Posted by TheEmptySet

Your seperation is good but

$\int \frac{1}{y^2}dy = -\frac{1}{y}$

So you should get

$-\frac{1}{y}=\frac{1}{2}e^{2x}+C$

Ahh, of course...It's basically $y^{-2}$ .

I was treating it as $y^{-1}$ or something like that.

Thank you!! :)