Question

The solution of the differential equation $dx–dy(1+x{y}^2)xy=0$ is

• A. $\frac{1}{y}=2-x^2+k{e}^{-\frac{1}{2}{x}^2}$
• B. $\frac{1}{x}=2-y^2+k{e}^{-\frac{1}{2}{y}^2}$
• C. $x=2-y^2+k{e}^{-\frac{1}{2}{y}^2}$
• D. $\frac{1}{y}=2+x^2-k{e}^{-\frac{1}{2}{x}^2}$

Answer 1

The correct option is B $\frac{1}{x}=2-y^2+k{e}^{-\frac{1}{2}{y}^2}$

We have,
$\frac{1}{x^2}\cdot \frac{dx}{dy}-\frac{1}{x}\cdot y=y^3|\begin{array}{c}\text{Put}\frac{\text{1}}{\text{x}}=t\\ \therefore \frac{1}{x^2}\frac{dx}{dy}=-\frac{dt}{dy}\end{array}$

$\therefore -\frac{dt}{dy}-t\cdot y=y^3$
$\frac{dt}{dy}+t\cdot y=-y^3$
which is linear in $t$
on solving, we get,
$\frac{1}{x}=2-y^2+k{e}^{-\frac{1}{2}{y}^2}$