Question

The solution of the differential equation $x^{2}dy+y(x+y)dx=0$ is:

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

Answer 1

Answer: (A)

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

$x^2\frac{dy}{dx}+xy+y^2=0$

Dividing $x^2{y}^2$, we get

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

Substitute $\frac{1}{y}=v$ so that $-\frac{1}{y^2}\frac{dy}{dx}=\frac{dv}{dx}$

$\therefore -\frac{dy}{dx}+\frac{1}{x}v=-\frac{1}{x^2}\Rightarrow \frac{dv}{dx}-\frac{1}{x}v=\frac{1}{x^2}$ $($ linear in $v)$

Here $P=-\frac{1}{x},Q=\frac{1}{x^2}.\int Pdx=-\log x=\log x^{-1}$

$\therefore I.F.=e^{\log x^{-1}}=x^{-1}=\frac{1}{x}$

The solution is $v\frac{1}{x}=\int \frac{1}{x}.\frac{1}{x^2}dx+c=\frac{x^{-2}}{-2}+c$

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

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