# Differential equation

• Oct 24th 2008, 11:22 PM
roshanhero
Differential equation
How to solve the equation-
(x^2+y^2)dx+2xydy=0
• Oct 25th 2008, 01:06 AM
mr fantastic
Quote:

Originally Posted by roshanhero
How to solve the equation-
(x^2+y^2)dx+2xydy=0

I'd write it as $\displaystyle \frac{dy}{dx} = -\frac{1}{2} \left(\frac{x^2 + y^2}{xy} \right) = -\frac{1}{2} \left(\frac{x}{y} + \frac{y}{x}\right)$.

Now make the usual substitution $\displaystyle \frac{y}{x} = v \Rightarrow y = xv$.
• Oct 25th 2008, 01:36 AM
Peritus
you can easily see that this is an exact ODE

$\displaystyle \frac{\partial } {{\partial y}}\left( {x^2 + y^2 } \right) = \frac{\partial } {{\partial x}}2xy$

thus we can find the solution as follows:

$\displaystyle \psi (x,y) = \int {2xy} dy = xy^2 + f(x)$

we know that:

$\displaystyle \begin{gathered} \frac{{\partial \psi (x,y)}} {{\partial x}} = y^2 + f'(x) = x^2 + y^2 \hfill \\ \Leftrightarrow f(x) = \frac{{x^3 }} {3} + k \hfill \\ \end{gathered}$

thus the implicit solution is:

$\displaystyle \frac{{x^3 }} {3} + xy^2 = c$
• Oct 25th 2008, 04:30 AM
roshanhero
Quote:

Originally Posted by mr fantastic
I'd write it as $\displaystyle \frac{dy}{dx} = -\frac{1}{2} \left(\frac{x^2 + y^2}{xy} \right) = -\frac{1}{2} \left(\frac{x}{y} + \frac{y}{x}\right)$.

Now make the usual substitution $\displaystyle \frac{y}{x} = v \Rightarrow y = xv$.

Thanks-
I subsituted v=y/x and got the expression-
2v/1+3v^2 dv=-dx/x.
I tried to integrate the left side with respect to v, but I am just completely lost in this case.I wasnot able to integrate it
So please suggest me the stage after this.
• Oct 25th 2008, 05:16 AM
mr fantastic
Quote:

Originally Posted by roshanhero
Thanks-
I subsituted v=y/x and got the expression-
2v/1+3v^2 dv=-dx/x.
I tried to integrate the left side with respect to v, but I am just completely lost in this case.I wasnot able to integrate it
So please suggest me the stage after this.

$\displaystyle \frac{dy}{dx} = -\frac{1}{2} \left(\frac{x}{y} + \frac{y}{x}\right) \Rightarrow x \frac{dv}{dx} = \frac{-3v^2 - 1}{2v}$

(the algebra leading to this should be routine at this level. You do realise that if y = vx then dy/dx is v + x dv/dx, right?).

This DE is seperable: $\displaystyle \frac{dx}{x} = \frac{-2v \, dv}{3v^2 + 1}$.