Find the general solution of the differential equation `y dx-(x+2y^2)dy=0`.

Explanation

`ydx-(x+2y^(2))dy=0`
`implies y(dx)/(dy)-x-2y^(2)=0`
`implies (dx)/(dy)-(1)/(y)x=2y`
which is in the form of `(dx)/(dy)+Px=Q`
Here `P=-(1)/(y)`, `Q=2y`
`:. I.F.=e^(intPdy)=e^(int-(1)/(y)dy)`
`=e^(-logy)`
`=e^(log(y^(-1))=y^(-1)=(1)/(y)`
and general solution is
`x*(1)/(y)=int2y(1)/(y)dy+c`
`=int2dy+c`
`=2y+c`
`impliesx=2y^(2)+cy`.