Find the general and singular solutions (if any) of…

Find the general and singular solutions (if any) of $(y')^2-2yy'=y^2(e^x-1)$.

Step 1: Quadratic formula in $p$

With $p=y'$: $p^2-2yp-y^2(e^x-1)=0$. Sympy's quadratic solve gives \[p=y\left(1\pm\sqrt{e^{x}}\right)=y\left(1\pm e^{x/2}\right),\] and substituting back confirms the original relation $p^2-2yp-y^2(e^x-1)=0$ is satisfied identically (residual $0$) for both roots.

Step 2: Solve each linear branch by separation

Branch 1: $y'=y(1+e^{x/2})\Rightarrow \dfrac{dy}{y}=(1+e^{x/2})dx\Rightarrow \ln y = x+2e^{x/2}+\text{const}$, so $y=C_1e^{x+2e^{x/2}}$.
Branch 2: $y'=y(1-e^{x/2})\Rightarrow y=C_2e^{x-2e^{x/2}}$.
Sympy's dsolve reproduces both results exactly, and direct differentiation confirms each satisfies its respective linear ODE (residual $0$).

Step 3: Verify both families in the ORIGINAL equation

Substituting $y=C_1e^{x+2e^{x/2}}$ and $y=C_2e^{x-2e^{x/2}}$ (with $y'$ computed by differentiation) into $(y')^2-2yy'-y^2(e^x-1)$, sympy simplifies both residuals exactly to $0$. (The source's own "Step 9" got confused mid-calculation — this direct symbolic check removes any doubt: both families are correct.)

Step 4: $p$-discriminant

$F=p^2-2yp-y^2(e^x-1)$, $F_p=2p-2y=0\Rightarrow p=y$; substituting into $F=0$ gives $y^2-2y^2-y^2(e^x-1)=0\Rightarrow -y^2e^x=0\Rightarrow y=0$.

Step 5: Check whether $y=0$ is genuinely singular

$y=0$ satisfies the original equation trivially, but it is already the $C_1=0$ (equivalently $C_2=0$) member of both general families above, so it is not a singular solution — matching the source's conclusion.