How to solve the equation-

(x^2+y^2)dx+2xydy=0

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- Oct 24th 2008, 11:22 PMroshanheroDifferential equation
How to solve the equation-

(x^2+y^2)dx+2xydy=0 - Oct 25th 2008, 01:06 AMmr fantastic
- Oct 25th 2008, 01:36 AMPeritus
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 AMroshanhero
- Oct 25th 2008, 05:16 AMmr fantastic
$\displaystyle \frac{dy}{dx} = -\frac{1}{2} \left(\frac{x}{y} + \frac{y}{x}\right) \Rightarrow x \frac{dv}{dx} = \frac{-3v^2 - 1}{2v}$

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