Question

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

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

Answer 1

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

Given $(1+x^2{y}^2)ydx+(x^2{y}^2-1)xdy=0$
$\Rightarrow (1+x^2{y}^2)ydx=-(x^2{y}^2-1)xdy$
$\Rightarrow \frac{(1+x^2{y}^2)}{x}dx=\frac{(1-x^2{y}^2)}{y}dy$
$\Rightarrow (\frac{1}{x}+x{y}^2)dx=(\frac{1}{y}-x^{2}y)dy$
Integrating both sides

$\Rightarrow \int (\frac{dx}{x}-\frac{dy}{y})+\int (x{y}^{2}dx+x^{2}ydy)=0$
$\Rightarrow \left(\ln \frac{x}{y}\right)+\int \frac{1}{2}d\left(x^2{y}^2\right)+c_1=0$
$\Rightarrow \ln \frac{x}{y}+\frac{1}{2}{x}^2{y}^2+c_2=0$
$\Rightarrow x^2{y}^2+2\log \frac{x}{y}+c_3=0$
$\therefore x^2{y}^2=2\log \frac{y}{x}+c$