1. ## 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

2. ## Re: Beginner Differential Equations

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

So you should get

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

3. ## 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.

4. ## Re: Beginner Differential Equations

Originally Posted by TheEmptySet
$\int \frac{1}{y^2}dy = -\frac{1}{y}$
$-\frac{1}{y}=\frac{1}{2}e^{2x}+C$
Ahh, of course...It's basically $y^{-2}$.
I was treating it as $y^{-1}$ or something like that.