Question

The solution of the differential equation $(1+x^2{y}^2)ydx+(x^2{y}^2-1)xdy=0$ is

• A. $xy=\log \frac{x}{y}+C$
• B. $xy=2\log \frac{y}{x}+C$
• C. $x^2{y}^2=\log \frac{y}{x}+C$
• D. $x^2{y}^2=2\log \frac{y}{x}+C$

Answer 1

The correct option is D $x^2{y}^2=2\log \frac{y}{x}+C$

Given differential equation is $(1+x^2{y}^2)ydx+(x^2{y}^2-1)xdy=0$ .... $\left(i\right)$

$⟹(1+x^2{y}^2)ydx=-(x^2{y}^2-1)xdy$

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

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

Integrating both sides

$⟹\int (\frac{1}{x}+x{y}^2)dx=\int (\frac{1}{y}-x^{2}y)dy$
$\Rightarrow \int \left(\frac{dx}{x}-\frac{dy}{y}\right)+\int \left(x{y}^{2}dx+x^{2}ydy\right)=0$
$\Rightarrow \int d\left(\log \frac{x}{y}\right)+\int \frac{1}{2}d\left(x^2{y}^2\right)+C=0$
$\therefore \log \frac{x}{y}+\frac{1}{2}{x}^2{y}^2+C=0$