How to find singular solutions from a general one

Question

Find singular solutions, if given general solution

\begin{equation*} y= Ce^x + \dfrac{4}{C} \end{equation*}

Usually after solving the equation you will use the parameter for finding the singular solutions

$\left\{\begin{array}{l}F(x, y, p)=0 \\ \frac{\partial F(x, y, p)}{\partial p}=0\end{array}\right.$

I supposed that $\left\{\begin{array}{l}y= Ce^x + \dfrac{4}{C}=0 \\ \frac{\partial F(x, y, C)}{\partial C}=e^x -\frac{4}{C^2}=0\end{array}\right.$

Answer 1

$$\begin{equation*} y= Ce^x + \dfrac{4}{C} \end{equation*}$$ $y'=Ce^x$ then $C=\dfrac {y'}{e^x} \implies y= {y'}+\dfrac {4e^x} {y'}$

Now we have: $$F(x,y,y')= {y'}+\dfrac {4e^x} {y'}-y=0$$ Differentiate with respect to $y'(y'=p)$ and solve: $$\partial_{y'}F(x,y,y')= 1-\dfrac {4e^x} {y'^2}=0$$