Solve the following first order differential equation 2dy/dx +y=y cube (x-1)?

To solve the differential equation 2dy/dx + y = y^3(x-1), we first need to rewrite it in standard form.

Dividing through by 2, we get:

dy/dx + y/2 = (y^3/2)(x-1)

Now, we can use the integrating factor method to solve this differential equation. The integrating factor is given by:

μ(x) = e^∫(1/2)dx
μ(x) = e^(x/2)

Multiplying both sides of the differential equation by the integrating factor, we get:

e^(x/2)dy/dx + (e^(x/2)y)/2 = (e^(x/2)y^3)(x-1)

Now, we can rewrite the left side of the equation as the derivative of the product of the integrating factor and y:

d/dx(e^(x/2)y) = (e^(x/2)y^3)(x-1)

Integrating both sides with respect to x, we get:

e^(x/2)y = ∫(e^(x/2)y^3)(x-1)dx + C

Solving the integral on the right side, we get:

e^(x/2)y = ∫(e^(x/2)y^3)(x-1)dx + C
e^(x/2)y = (1/4)e^(x/2)y^3(x^2 - 3x + 2) + C

Now, we can solve for y by dividing through by e^(x/2):

y = (1/4)y^3(x^2 - 3x + 2) + Ce^(-x/2)

This is the general solution to the differential equation 2dy/dx + y = y^3(x-1).