Question

If c is an arbitrary constant then solution of differential equation $x^{2}dy-y^{2}dx-x{y}^2(x-y)dy=0$ can be

• A. $ln\left|\frac{xy}{x-y}\right|+\frac{y^2}{2}=c$
• B. $ln\left|\frac{x-y}{xy}\right|+\frac{y^2}{2}=c$
• C. $(x-y)e^{\frac{y^2}{2}}=cxy$
• D. $(x-y)e^{-\frac{y^2}{2}}=cxy$

Answer 1

Answer: (B), (C)

$(x-y)e^{\frac{y^2}{2}}=cxy$
$ln\left|\frac{x-y}{xy}\right|+\frac{y^2}{2}=c$

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

$\frac{dy}{y^2}-\frac{dx}{x^2}-y[\frac{1}{y}-\frac{1}{x}]dy=0$

$\Rightarrow \frac{\frac{dy}{y^2}-\frac{dx}{x^2}}{\frac{1}{y}-\frac{1}{x}}-ydy=0$

$\Rightarrow -ln|\frac{1}{y}-\frac{1}{x}|-\frac{y^2}{2}=c$

$\Rightarrow ln\left|\frac{x-y}{xy}\right|+\frac{y^2}{2}=c$

$\left|\frac{x-y}{xy}\right|=e^{c-\frac{y^2}{2}}$

$\left|\frac{x-y}{xy}\right|=C.e^{\frac{-y^2}{2}}$

$(x-y)e^{\frac{y^2}{2}}=cxy$