# Messy Differential Equation!?

• Feb 21st 2010, 05:06 AM
Yehia
Messy Differential Equation!?
I have this differential equation:

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

• Feb 21st 2010, 05:13 AM
Ted
Quote:

Originally Posted by Yehia
I have this differential equation:

$\displaystyle (y-x) {\color{red}dy/dy}=2(y^2-dy/dx)$

The red one is $\displaystyle \,\ \frac{dy}{dx}$ ?
• Feb 21st 2010, 05:20 AM
Yehia
Quote:

Originally Posted by Ted
The red one is $\displaystyle \,\ \frac{dy}{dx}$ ?

YES it is! sorry my bad, it is dy/dx. youre right. can you help?
• Feb 21st 2010, 06:09 AM
shawsend
Isolate the derivative and then put it into differential form:

$\displaystyle (y-x)y'=2y^2-2y'$

$\displaystyle y'(y-x+2)=2y^2$

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

Now, can you find an integrating factor?
• Feb 21st 2010, 11:42 AM
Yehia
Quote:

Originally Posted by shawsend
Isolate the derivative and then put it into differential form:

$\displaystyle (y-x)y'=2y^2-2y'$

$\displaystyle y'(y-x+2)=2y^2$

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

Now, can you find an integrating factor?

I follow you till there. But whats an integrating factor? and how can you move the x's to the dx's and the y's to the dy's???? please help further! thank!!