Solution of the differential equation y prime=(3 x2 y/x3+2 y4) is (where C is constant of integration)

Question

Solution of the differential equation y prime=(3 x2 y/x3+2 y4) is (where C is constant of integration) (A)  x3 y-1=(2/3) y3+C  (B)  x2 y-1=(2/3) y3+C  (C) x y-1=(2/3) y3+C  (D)  x-1 y3=(2/3) y3+C

Tardigrade Admin · Accepted Answer

(x3+2 y4) (d y/d x)-3 x2 y=0 x3 d y-3 x2 y d x+2 y4 d y=0 (x3 d y-3 x2 y d x/y2)+2 y2 d y=0 d(-(x3/y))+2 y2 d y=0 C-(x3/y)+(2 y3/3)=0 x3 y-1=(2/3) y3+C

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

$x^{3} y^{- 1} = \frac{2}{3} y^{3} + C$

$x^{2} y^{- 1} = \frac{2}{3} y^{3} + C$

$x y^{- 1} = \frac{2}{3} y^{3} + C$

$x^{- 1} y^{3} = \frac{2}{3} y^{3} + C$

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

Q. Solution of the differential equation y′=x3+2y43x2y​ is (where C is constant of integration)

x3y−1=32​y3+C

x2y−1=32​y3+C

xy−1=32​y3+C

x−1y3=32​y3+C