The solution of the differential equation (dy/dx)=(3x2y4+2xy/x2-2x3 y3) is

Question

The solution of the differential equation (dy/dx)=(3x2y4+2xy/x2-2x3 y3) is (A) (y2/x)-x3y2=c (B) (x2/y2)+x3y3=c (C) (x2/y)+x3y2=c (D) (x2/3y)-2 x3y2=c

Tardigrade Admin · Accepted Answer

Rewrite the differential equation as (2xy dx-x2 dy)+y2(3x2 y2 dx+2x3y dy)=0 Dividing by y2, we get (y 2x dx-x2 dy/y2) +y2 3x2 dx+x3 2y dy=0 or d=((x2/y))+d (x3 y2)=0 Integrating, we get the solution (x2/y)+x3y2=c

Q. The solution of the differential equation $\frac{d y}{d x} = \frac{3 x ^{2} y ^{4} + 2 x y}{x ^{2} - 2 x ^{3} y ^{3}}$ is

$\frac{y ^{2}}{x} - x^{3} y^{2} = c$

$\frac{x ^{2}}{y ^{2}} + x^{3} y^{3} = c$

$\frac{x ^{2}}{y} + x^{3} y^{2} = c$

$\frac{x ^{2}}{3 y} - 2 x^{3} y^{2} = c$

Q. The solution of the differential equation dxdy​=x2−2x3y33x2y4+2xy​ is

xy2​−x3y2=c

y2x2​+x3y3=c

yx2​+x3y2=c

3yx2​−2x3y2=c

Rewrite the differential equation as (2xydx−x2dy)+y2(3x2y2dx+2x3ydy)=0 Dividing by y2, we get y2y2xdx−x2dy​+y23x2dx+x32ydy=0 or d=(yx2​)+d(x3y2)=0 Integrating, we get the solution yx2​+x3y2=c

Q. The solution of the differential equation $\frac{d y}{d x} = \frac{3 x ^{2} y ^{4} + 2 x y}{x ^{2} - 2 x ^{3} y ^{3}}$ is

$\frac{y ^{2}}{x} - x^{3} y^{2} = c$

$\frac{x ^{2}}{y ^{2}} + x^{3} y^{3} = c$

$\frac{x ^{2}}{y} + x^{3} y^{2} = c$

$\frac{x ^{2}}{3 y} - 2 x^{3} y^{2} = c$