Differential equation

roshanhero · #1 $Old$ October 25th, 2008, 12:22 AM

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

mr fantastic · #2 $Old$ October 25th, 2008, 02:06 AM

Quote:

Originally Posted by roshanhero $View Post$

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

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

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

Peritus · #3 $Old$ October 25th, 2008, 02:36 AM

you can easily see that this is an exact ODE

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

thus we can find the solution as follows:

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

we know that:

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

thus the implicit solution is:

$\frac{{x^3 }}{3} + xy^2 = c$

roshanhero · #4 $Old$ October 25th, 2008, 05:30 AM

Quote:

Originally Posted by mr fantastic $View Post$

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

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

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.

mr fantastic · #5 $Old$ October 25th, 2008, 06:16 AM

Quote:

Originally Posted by roshanhero $View Post$

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.

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

(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: $\frac{dx}{x} = \frac{-2v \, dv}{3v^2 + 1}$ .