Find the general solution of the differential equation \[ \begin{equation*} \frac{dy}{dx}=4y+2e^{x}y^{1/2} \end{equation*} \]
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 \(y\). \[ \begin{eqnarray*} y^{1/2}& =&Ce^{2x}-e^{x} \\ y& =&(Ce^{2x}-e^{x})^{2} \end{eqnarray*} \]