Question

The solution of the differential equation $\frac{d y}{d x}=\frac{2 x y}{x^2-1-2 y}$ is

• A. $ \frac{x^2}{y}=\frac{1}{y}-2 \ell n y+c$
• B. $\frac{x^2}{y^2}=\frac{1}{y}-2 \ell n y+c$
• C. $x^2 y=\ln y+c$
• D. $\frac{x^2}{y}=2 \ln y+c$

Answer 1

Answer: (A)

$\frac{d y}{d x}=\frac{2 x y}{x^2-1-2 y} $
$\Rightarrow x^2 d y-(1+2 y) d y=2 x y d x$
or $2 x y d x-x^2 d y=-(1+2 y) d y$
or $\frac{y d\left(x^2\right)-x^2 d y}{y^2}=-\left(\frac{1}{y^2}+\frac{2}{y}\right) d y$
$d\left(\frac{x^2}{y}\right)=-\left(\frac{1}{y^2}+\frac{2}{y}\right) d y$
Integrating $\frac{x^2}{y}=\frac{1}{y}-2 \ell$ ny $+c$