Question

For the differential equation, find the general solution:

`(x + 3y^2) dy/dx = y(y > 0)`

Answer 1

`(x + 3y^2) dy/dx = y`

or `y dx/dy = x + 3y^2`

`dx/dy - x/y = 3y`

A vertical differential equation of the form `dx/dy + Px = Q.`

Here `P = - 1/y, Q = 3y`

∴ `I.F. = e^(int P dx) = e^(- int 1/y dy) = e^(- log y) = 1/y`

Hence, the general solution of the differential equation

⇒ `x × I.F. = int Q xx I.F. dx + C`

⇒ `x xx 1/y = int 1/y (3y) dy + C`

⇒ `x/y = 3 int 1 dy + C`

⇒ `x/y = 3y + C`

⇒ x = 3y² + Cy

Which is the required solution.