Find the general solution of the differential equation $$\begin{equation*} \frac{dy}{dx}=4y+2e^{x}y^{1/2} \end{equation*}$$

Solution We begin by writing the equation in the form of (3): $$\begin{equation*} \frac{dy}{dx}-4y=2e^{x}y^{1/2} \end{equation*}$$

This is a Bernoulli differential equation with $P(x)=-4$, $Q(x)=2e^{x}$, and $n=\dfrac{1}{2}$. Then we follow the steps for solving a Bernoulli equation.

Step 1 Multiply the differential equation by $y^{-1/2}$ to obtain $$\begin{equation*} y^{-1/2}\frac{dy}{dx}-4y^{1/2}=2e^{x} \end{equation*}$$

Step 2 Let $v=v( x) =y^{1/2}$. Then $\dfrac{dv}{dx}= \dfrac{1}{2}y^{-1/2}\dfrac{dy}{dx}$.

Step 3 Substitute $v$ and $\dfrac{dv}{dx}$ into $y^{-1/2}\dfrac{dy }{dx}-4y^{1/2}=2e^{x}$. $$\begin{eqnarray*} 2\dfrac{dv}{dx}-4v& =&2e^{x} \\ \dfrac{dv}{dx}-2v& =&e^{x} \end{eqnarray*}$$

This is a first-order linear differential equation in $x$ and $v$.

Step 4 Multiply the differential equation by the integrating factor $e^{\int -2\,dx}=e^{-2x}$. $$\begin{eqnarray*} e^{-2x}\dfrac{dv}{dx}-2e^{-2x}v& =&e^{-x} \\ \dfrac{d}{dx}(e^{-2x}v)& =&e^{-x} \\ e^{-2x}v& =&\int e^{-x}dx=-e^{-x}+C\quad C > 0 \\ v& =&Ce^{2x}-e^{x} \end{eqnarray*}$$

Step 5 Using $v=y^{1/2}$, we write the solution in terms of $x$ and $y$. $$\begin{eqnarray*} y^{1/2}& =&Ce^{2x}-e^{x} \\ y& =&(Ce^{2x}-e^{x})^{2} \end{eqnarray*}$$