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

Find the general and singular solutions (if any) of $y'=2+\sqrt{y-2x}$.

Step 1: Substitute $u=y-2x$

$u'=y'-2$, so the equation becomes $u'+2=2+\sqrt u$, i.e. $u'=\sqrt u$ (valid for $u\ge0$).

Step 2: Solve the separable equation

For $u>0$: $u^{-1/2}\,du=dx\Rightarrow2\sqrt u=x+C\Rightarrow u=\dfrac{(x+C)^2}{4}$.

Step 3: Return to $y$

Since $u=y-2x$: $y=2x+\dfrac{(x+C)^2}{4}$, valid on $x+C\ge0$ (where the original $\sqrt{u}=\sqrt{y-2x}$ branch matches).

Step 4: Identify the envelope

Eliminating $C$ from $G=y-2x-\frac{(x+C)^2}{4}=0$ and $G_C=-\frac{(x+C)}{2}=0$ gives $C=-x$, and substituting back gives $y=2x$. Directly: $y=2x\Rightarrow y'=2$ and $2+\sqrt{y-2x}=2+\sqrt0=2$, so $y=2x$ does solve the ODE, and it cannot be written as $2x+\frac{(x+C)^2}{4}$ for a single fixed $C$ valid over an interval — so it is a genuine singular solution.