Question

The general solution of the differential equation $x{y}^2\frac{dy}{dx}=1-x^2+y^2-x^2{y}^2$ is (where $c$ is a constant of integration and log is considered with base ${}^{\prime }{e}^{\prime }$)

• A. $y-{\sin }^{-1}y=\log |x|+e^x+c$
• B. $y-{\tan }^{-1}y=\log |x|-\frac{x^2}{2}+c$
  
• C. $x-{\sin }^{-1}y=\log |x|+e^x+c$
• D. $y-{\tan }^{-1}x=\log |x|+\frac{x^2}{2}+c$
  

Answer 1

The correct option is B $y-{\tan }^{-1}y=\log |x|-\frac{x^2}{2}+c$


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

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

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

Integrating both sides, we have

$y-{\tan }^{-1}y=\log |x|-\frac{x^2}{2}+c$