Question

Solve the differential equation (x + 2y³) dy = y dx.

Answer 1

(x + 2y³) dy = y dx

Divide both sides by y . dy

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

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

Now, bring the term with x to the left side:

`dx/dy - 1/y x = 2y^2`

This is a linear differential equation of the form `dx/dy + P(y)x = Q(y)`, where:

P(y) = `-1/y`

Q(y) = 2y²

The integrating factor is I.F. = `e^(In(y^-1)) = 1/y`

I.F. = `e^(int-1/y dy) = e^(-ln y) = e^(In (y-1)) = 1/y`

Solve the equation:

`x . (I.F.) = int Q(y) . (I.F.) dy`

`x . 1/y = int 2y^2 . 1/y dy`

`x/y = int 2y dy`

`x/y y^2 + C`

Multiply both sides by y:

x = y³ + Cy