Question

Solve: $(x^2-y{x}^2)dy+(y^2+x{y}^2)dx=0$

Answer 1

Solution

Given , Differential Equation is

$(x^2-y{x}^2)dy+(y^2+x{y}^2)dx=0$

This can be Simplified as

$(y{x}^2-x^2)dy=(y^2+x{y}^2)dx$

$x^2(y-1)dy=y^2(1+x)dx$

$\frac{dy(y-1)}{y^2}=\frac{dx(1+x)}{x^2}$

Now On Integrating both side , we get

$\int \frac{dy(y-1)}{y^2}=\int \frac{dx(1+x)}{x^2}$

$\int [\frac{1}{y}-\frac{1}{y^2}]dy=\int [\frac{1}{x}+\frac{1}{x^2}]dx$

$\int [\frac{dy}{y}-\frac{dy}{y^2}]=\int [\frac{dx}{x}+\frac{dx}{x^2}]$

$ln\left|y\right|-\frac{1}{y}=ln\left|x\right|-\frac{1}{x}+lnC$

$ln\left|y\right|-ln\left|x\right|-lnC=\frac{1}{y}-\frac{1}{x}$

$ln\left[\frac{\left|y\right|}{c\left|x\right|}\right]=\frac{x-y}{xy}$