ISI2018-MMA-25

Question

The solution of the differential equation

$(1 + x^2y^2)ydx + (x^2y^2 − 1)xdy = 0$ is

• A. $xy = \log\ x − \log\ y + C$
• B. $xy = \log\ y − \log\ x + C$
• C. $x^2y^2 = 2(\log\ x − \log\ y) + C$
• D. $x^2y^2 = 2(\log\ y − \log\ x) + C$

Answer 1

This differential equation can be solved by using the method of inspection.

Rearranging the terms of the D.E. , we get   $xdy-ydx=(x^{2}y^{2})(ydx+xdy)$

$\Rightarrow$    $\frac{xdy-ydx}{xy}=xy(xdy+ydx)$

$\Rightarrow$    $d\left ( log\left ( \frac{y}{x} \right ) \right )=xyd(xy)$

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

Rearranging the above equation, we get   $x^{2}y^{2}=2(logy-logx)+C$

Option D is correct answer.