Question

Solve the differential equation: $y^{2}dx+(xy+x^2)dy=0$

• A. $x{y}^2=c(y+2x)$
• B. $x{y}^2=c(2y+x)$
• C. $y{x}^2=c(y+2x)$
• D. $y{x}^2=c(2y+x)$

Answer 1

Answer: (B)

$x{y}^2=c(2y+x)$

$y^{2}dx+(xy+x^2)dy=0$
$\frac{dy}{dx}=-\frac{y^2}{xy+x^2}.$

Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow \frac{(v+1)}{v(2v+1)}dv+\frac{dx}{x}=0\Rightarrow (\frac{1}{v}-\frac{1}{2v+1})dv+\frac{dx}{x}=0$.

Integrating we get, $\log v-\frac{1}{2}\log (2v+1)+logx=\log k$

$\frac{vx}{k}=\sqrt{(2v+1)}\Rightarrow v^2{x}^2=k^2(2\cdot \frac{y}{x}+1)\Rightarrow x{y}^2=c(2y+x).$