Calculus of a Single Variable, Chapter 6, 6.4, Section 6.4, Problem 59

y'+3x^2y=x^2y^3
Multiply the above equation by y^(-3)
y^(-3)dy/dx+3x^2y(y^(-3))=x^2
y^-3dy/dx+3x^2y^(-2)=x^2
Taking the transformation v=y^(-2)
(dv)/dx=d/dy(y^(-2))*dy/dx
(dv)/dx=-2y^(-3)dy/dx
-1/2(dv)/dx=y^(-3)dy/dx
Now the Bernoulli equation is transformed as ,
-1/2(dv)/dx+3x^2v=x^2
(dv)/dx-6x^2v=-2x^2
Now the above is a linear equation in the dependent variable v and independent variable y.
The integrating factor is n(x)=e^(int(-6x^2dx))
=e^(-6x^3/3)
=e^(-2x^3)
Then,
e^(-2x^3)*(dv)/dx-6e^(-2x^3)*x^2v=-2e^(-2x^3)*x^2
d/dx(e^(-2x^3)*v)=e^(-2x^3)(dv)/dx+ve^(-2x^3)(-6x^2)
=e^(-2x^3)(dv)/dx-6e^(-2x^3)*x^2v
=-2e^(-2x^3)*x^2
intd/dx(e^(-2x^3)*v)dx=int-2e^(-2x^3)x^2dx
e^(-2x^3)*v=-2inte^(-2x^3)*x^2dx
Let t=x^3
dt=3x^2dx
e^(-2x^3)*v=-2inte^(-2t)*dt/3
=-2/3(e^(-2t)/(-2))+C
=e^(-2t)/3+C
Substitute back t=x^3
e^(-2x^3)*v=1/3e^(-2x^3)+C
Substitute back v=y^(-2)
e^(-2x^3)*y^(-2)=1/3e^(-2x^3)+C
y^-2=1/3+C/e^(-2x^3)
1/y^2=1/3+Ce^(2x^3)
y^2 = 1 / (1/3+Ce^(2x^3))
y = +-sqrt(3)/(Ce^(2x^3) + 1)