What is the solution of dy/dx = (x^2+y^2)/(xy)

Tuesday, September 15, 2009

What is the solution of dy/dx = (x^2+y^2)/(xy)

The differential equation $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{{{x}}^{{2}}+{{y}}^{{2}}}}{{{x}{y}}}$ has to be solved.

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{{{x}}^{{2}}+{{y}}^{{2}}}}{{{x}{y}}}$

This can be written as

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{{x}}^{{2}}}{{{x}{y}}}+\frac{{{y}}^{{2}}}{{{x}{y}}}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{x}}{{y}}+\frac{{y}}{{x}}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={{\left(\frac{{y}}{{x}}\right)}}^{{-{{1}}}}+\frac{{y}}{{x}}$

Let $\displaystyle{f{=}}\frac{{y}}{{x}}$

$\displaystyle{y}={f{\cdot}}{x}$

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={f{+}}{x}\cdot\frac{{{d}{f}}}{{{\left.{d}{x}\right.}}}$

Now substituting for $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}$ and $\displaystyle\frac{{y}}{{x}}$ in the equation we get

$\displaystyle{f{+}}{x}\cdot\frac{{{d}{f}}}{{{\left.{d}{x}\right.}}}={{f}}^{{-{{1}}}}+{f{}}$

$\displaystyle{x}\cdot\frac{{{d}{f}}}{{{\left.{d}{x}\right.}}}=\frac{{1}}{{f{}}}$

$\displaystyle{f{\cdot}}{d}{f{=}}\frac{{\left.{d}{x}}}{{x}}$

Take the integral of both the sides

$\displaystyle\int{f{\cdot}}{d}{f{=}}\int\frac{{\left.{d}{x}}}{{x}}$

$\displaystyle\frac{{{f}}^{{2}}}{{2}}={\ln{{x}}}+{C}$

The constant C can be included in the logarithm of x as $\displaystyle{\ln{{\left({k}\cdot{x}\right)}}}$

$\displaystyle\frac{{{\left(\frac{{y}}{{x}}\right)}}^{{2}}}{{2}}={\ln{{\left({k}\cdot{x}\right)}}}$

$\displaystyle{{y}}^{{2}}={2}\cdot{{x}}^{{2}}\cdot{\ln{{\left({k}\cdot{x}\right)}}}$

$\displaystyle{y}=\pm\sqrt{{{2}\cdot{{x}}^{{2}}\cdot{\ln{{\left({k}\cdot{x}\right)}}}}}$

The solution of the differential equation $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{{{x}}^{{2}}+{{y}}^{{2}}}}{{{x}{y}}}$ is $\displaystyle{y}=\pm\sqrt{{{2}\cdot{{x}}^{{2}}\cdot{\ln{{\left({k}\cdot{x}\right)}}}}}$

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Tuesday, September 15, 2009