Solve the Bernoulli differential equation \(y' + 2y = y^3\).

Step 1 ：The given differential equation is a Bernoulli equation of the form \(y' + p(x)y = q(x)y^n\) where \(p(x) = 2\), \(q(x) = 0\) and \(n = 3\).

Step 2 ：To solve this, we make a substitution to turn it into a linear differential equation. Let \(u = y^{1-n} = y^{-2}\). Then \(u' = -(1-n)y' y^{-n-1} = -2y' y^{-3}\).

Step 3 ：Substituting \(y = u^{-1/2}\) and \(y' = -u' / 2u^{3/2}\) into the original equation, we get \(-u' / 2u^{3/2} + 2/u^{1/2} = u^{-3/2}\).

Step 4 ：Simplifying this gives \(u' + 4u = 0\), which is a first order linear differential equation.

Step 5 ：The solution to this equation is \(u = Ce^{-4x}\) for some constant \(C\).

Step 6 ：Substituting back, we get \(y = u^{-1/2} = (Ce^{-4x})^{-1/2} = \sqrt{\frac{1}{C}} e^{2x}\).

Step 7 ：Thus, the solution to the original differential equation is \(y = \sqrt{\frac{1}{C}} e^{2x}\) for some constant \(C\).