Question

Solution of the differential equation $y^{\prime}=\frac{3 x^2 y}{x^3+2 y^4}$ is
(where $C$ is constant of integration)

• A. $ x^3 y^{-1}=\frac{2}{3} y^3+C $
• B. $ x^2 y^{-1}=\frac{2}{3} y^3+C $
• C. $x y^{-1}=\frac{2}{3} y^3+C $
• D. $ x^{-1} y^3=\frac{2}{3} y^3+C$

Answer 1

Answer: (A)

$\left(x^3+2 y^4\right) \frac{d y}{d x}-3 x^2 y=0$
$x^3 d y-3 x^2 y d x+2 y^4 d y=0$
$\frac{x^3 d y-3 x^2 y d x}{y^2}+2 y^2 d y=0$
$d\left(-\frac{x^3}{y}\right)+2 y^2 d y=0 $
$C-\frac{x^3}{y}+\frac{2 y^3}{3}=0$
$x^3 y^{-1}=\frac{2}{3} y^3+C$