91Ó°ÊÓ

The given differential equation is$xydx+\left(y^2-x^2\right)dy=0$

A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself to its derivatives of various orders.

Here the given differential equation is

$xydx+\left(y^2-x^2\right)dy=0$

Simplify the equation then,

$\begin{array}{rcl}\left(0^2-x^2\right)dy& =& -xydx\\ \frac{dy}{dx}& =& \frac{-xy}{\left(y^2-x^2\right)}\\ \frac{dy}{dx}& =& \frac{xy}{\left(y^2-x^2\right)}n\\ \frac{dy}{dx}& =& \frac{\left(x^2-y^2\right)}{xy}\\ \frac{dx}{dy}& =& \frac{x}{y}-\frac{y}{x}\end{array}$

Here the differential equation is a homogeneous differential equation,

Therefore to solve the equation, let us assume,

x = vy

Differentiate both sides with respect to x then,

$\begin{array}{rcl}\frac{dx}{dy}& =& v+y\frac{dy}{dy}...\end{array}$

Here put the value of equation 2 in equation 1 then,

$\begin{array}{rcl}v+y\frac{dy}{dy}& =& v-\frac{1}{v}y\frac{dv}{dy}\\ & =& -\frac{1}{v}\end{array}$

Now simplify the equation then,

$\begin{array}{rcl}& & vyv\frac{dy}{r}\end{array}$

Now integrate both sides and the solution of the differential equation is

$\begin{array}{rcl}∫vyv& =& -∫\frac{dy}{y}\frac{v^2}{2}\\ & =& -Iny+c\\ v& =& -2Iny+2c\end{array}$

Now put the value of $\begin{array}{rcl}v& =& \frac{x}{y}\end{array}$, then solution of the differential equation is,

$\begin{array}{rcl}y& =& C_2{e}^{-x^2/y^2}\end{array}$