Find the general and singular solutions (if any) of…
Find the general and singular solutions (if any) of $(y')^2-y^2=0$.
Explanation of Method
Factor into two linear equations $y'=\pm y$ and solve each; then apply the $p$-discriminant test to check for a singular solution, and verify that the candidate it produces is already contained in the general solution.
Step 1: Factor and solve both branches
$(y'-y)(y'+y)=0$ gives $y'=y$ or $y'=-y$, with solutions $y=C_1e^{x}$ and $y=C_2e^{-x}$ respectively.
Step 2: $p$-discriminant test
With $F(x,y,p)=p^2-y^2$: $F_p=2p=0\Rightarrow p=0$; substituting into $F=0$ gives $y^2=0$, i.e. $y=0$.
Step 3: Check whether $y=0$ is genuinely singular
$y=0$ satisfies the ODE ($0-0=0$), but it is already the member $C_1=0$ (or $C_2=0$) of the general solution — it is not a new, non-reproducible curve, so it is not a singular solution.
Result
Final answer:
\[\boxed{\,y=C_1e^{x}\ \ \text{or}\ \ y=C_2e^{-x}\,}\]
No singular solution exists: the only $p$-discriminant candidate, $y=0$, is already part of the general solution (take $C_1=0$ or $C_2=0$). (Verified by direct substitution: both families satisfy $(y')^2-y^2\equiv0$.)