# Differential Equation

• Oct 5th 2008, 02:58 PM
Oblivionwarrior
Differential Equation
Hey guys, I am wondering if anyone can help me with this question.

Solve the given differential equation $\displaystyle \frac{dy}{dx} = -\frac{2xy+y^2+2}{x^2+2xy}$

Now the methods of solving I have learned so far are, using an integrating factor, separating the equation and the technique for exact equations using partial derivatives. Thanks!
• Oct 5th 2008, 03:36 PM
Chris L T521
Quote:

Originally Posted by Oblivionwarrior
Hey guys, I am wondering if anyone can help me with this question.

Solve the given differential equation $\displaystyle \frac{dy}{dx} = -\frac{2xy+y^2+2}{x^2+2xy}$

Now the methods of solving I have learned so far are, using an integrating factor, separating the equation and the technique for exact equations using partial derivatives. Thanks!

Note that $\displaystyle \frac{\,dy}{\,dx}=-\frac{2xy+y^2+2}{x^2+2xy}\implies (x^2+2xy)\,dy=-(2xy+y^2+2)\,dx$ $\displaystyle \implies (2xy+y^2+2)\,dx+(x^2+2xy)\,dy=0$

Can you try to take it from here? It should somewhat be obvious now...

--Chris
• Oct 5th 2008, 03:58 PM
Oblivionwarrior
Quote:

Originally Posted by Chris L T521
Note that $\displaystyle \frac{\,dy}{\,dx}=-\frac{2xy+y^2+2}{x^2+2xy}\implies (x^2+2xy)\,dy=-(2xy+y^2+2)\,dx$ $\displaystyle \implies (2xy+y^2+2)\,dx+(x^2+2xy)\,dy=0$

Can you try to take it from here? It should somewhat be obvious now...

--Chris

Yea I got to that point, but the equation isn't exact since the partial derivatives are not equal. So I need to find an integrating factor. (Unless I did the partial derivatives wrong and it is exact).
• Oct 5th 2008, 04:03 PM
Chris L T521
Quote:

Originally Posted by Oblivionwarrior
Yea I got to that point, but the equation isn't exact since the partial derivatives are not equal. So I need to find an integrating factor. (Unless I did the partial derivatives wrong and it is exact).

It is exact:

$\displaystyle M(x,y)=2xy+y^2+2$ and $\displaystyle N(x,y)=x^2+2xy$

Its exact when $\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

$\displaystyle \frac{\partial M}{\partial y}=\color{red}2x+2y$

$\displaystyle \frac{\partial N}{\partial x}=\color{red}2x+2y$

Can you continue now?

--Chris
• Oct 5th 2008, 04:25 PM
Oblivionwarrior
Wow it is exact, no wonder. (Headbang) I thought M(x,y) = $\displaystyle x^2 +2xy$