Find the general solution of the differential equation ` (x + 2y^(3)) (dy)/(dx) =y, y ne 0`

Explanation

The given differential equation may be written as
`y (dx )/(dy) = x + 2y^(3)`
`rArr (dx)/(dy)- (1)/(y) *x = 2y^(2)" " `... (i)
This is of the form ` (dx)/(dy) + Px = Q`, where `P = - (1)/(y) and Q = 2y^(2)` .
Thus, the given equation is linear.
`IF = e ^(int Pdy)= e ^(int - (1)/(y) dy ) = e ^(-log (y^(-1))) = y ^(-1) = (1)/(y)`
So, the required solution is
` x xx IF = int {Q xx IF } dy + C `,
i.e, ` x xx (1)/(y) = int ( 2y ^(2 ) xx (1)/(y) ) dy +C `
`" " = int 2y dy +C =y ^(2) +C`.
Hence, ` x = y^(3) +Cy` is the required solution.