Solve the Bernoulli differential equation $y^{\prime }+2y=y^3$.

Step 1 ：The given differential equation is a Bernoulli equation of the form $y^{\prime }+p(x)y=q(x)y^n$ 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 $u^{\prime }=-(1-n)y^{\prime }{y}^{-n-1}=-2y^{\prime }{y}^{-3}$.

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

Step 4 ：Simplifying this gives $u^{\prime }+4u=0$, which is a first order linear differential equation.

Step 5 ：The solution to this equation is $u=C{e}^{-4x}$ for some constant $C$.

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

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