# Separable Differential Equations

• Mar 25th 2010, 11:41 AM
littlesohi
Separable Differential Equations
I need help in this problem!

$\displaystyle dy/dx=3x^2 (y-2)^2$

I moved the dx to the other side and divided by (y-2)^2 but I got stuck in that part. I know that I have to integrate but I don't know how to continue:

$\displaystyle \int dy/(y-2)^2 = \int 3x^2dx$
• Mar 25th 2010, 11:48 AM
General
Quote:

Originally Posted by littlesohi
I need help in this problem!

$\displaystyle dy/dx=3x^2 (y-2)^2$

I moved the dx to the other side and divided by (y-2)^2 but I got stuck in that part. I know that I have to integrate but I don't know how to continue:

$\displaystyle \int dy/(y-2)^2 = \int 3x^2dx$

To integrate the left side, use the partial fractions method or substitute $\displaystyle u=y-2$ ..
the right side should be easy ..
• Mar 25th 2010, 12:03 PM
littlesohi
So the left side would be -1/u?
• Mar 25th 2010, 12:26 PM
General
Quote:

Originally Posted by littlesohi
So the left side would be -1/u?

Yes .. which is $\displaystyle \frac{-1}{y-2}$ ..