1.

Find the general solution of the differential equations:`(x+3y^2)(dx)/(dy)=y(y >0)`

Answer» Here, given equation is,
`(x+3y^2)dy/dx = y`
`=>dx/dy = (x+3y^2)/y`
`=>dx/dy -x/y = 3y`
Comparing it with `dx/dy +Px = Q`
`P = -1/y and Q = 3y`
Integrating factor, `I.F. = e^(intPdy)`
`I.F. = e^(int-1/ydy) = e^-lny = y^-lne = y-1 = 1/y`
Now, general solution will be,
`x(I.F.) = int(I.F.)Qdy`
`=>x/y = int(1/y)(3y)dy`
`=>x/y = 3y+c`
`=>x = 3y^2+cy`


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