Find the general solution of `(x+2y^(3))(dy)/(dx)=y`

Options

A. `x^2=y^(3)+Cy`
B. `x=y^(2)+Cy`
C. `2x=y^(3)+Cy`
D. `x=y^(3)+Cy`

Answer

D

Explanation

Given that, `(x+2y^(3))(dy)/(dx)=y`
`y(dy)/(dx)=x+2y^(3)`
`Rightarrow (dy)/(dx)=(x)/(y)+2y^(2)["dividing throughout by y"]`
Which linear different equation.
On comparing it with `(dx)/(dy)+Px=0` , we get
`P=-(1)/(y),Q=2y^(2)`
`IF=e^(int-(1)/(y)dy)=e^(-1(1)/(y)dy)`
`=e^(-logy)=(1)/(y)`
The genergal solution is `x.(1)/(y)=int2y^(2).(1)/(y)dy+C`
`Rightarrow (x)/(y)=(2y^(2))/(2)+C`
`Rightarrow (x)/(y)=y^(2)+C`
`Rightarrow x=y^(3)+Cy`