More elegant way of verifying the solution of $y'=\frac{-(x+2)+\sqrt{x^2+4x+4y}}{2}$? The question asks to verify that for any value of $c$, $y=c^2+cx+2c+1$ satisfies the solution.
$$y'=\frac{-(x+2)+\sqrt{x^2+4x+4y}}{2}\tag{1}$$
I am aware we can just prove this by direct substitution.

However, I think it's no coincidence that the author specifically put the DE in the form of a solution to a quadratic equation. $(1)$ can be interpreted as a solution of the quadractic equation $$(y')^2+(x+2)y'+(1-y)=0$$
Which can be factored to $$(y'-c)(y'+c+x+2)=0\tag{2}$$
The first factor gives $y'=c$, which comes from the solution we were already given $(y=c^2+cx+2c+1)$. The second factor we just added when we constructed the quadratic equation, and it's not necessarily a solution of $(1)$. 
However, I don't think we can conclude that $y=c^2+cx+2c+1$ is a solution of $(1)$ just because $y'=c$ is a solution of $(2)$. Is there any way to complete the question from here, or is another approach neccesary?
 A: Hint.
$$y'=\frac{-(x+2)+\sqrt{x^2+4x+4y}}{2}\tag{1}
\quad\Longleftrightarrow\quad 2y'+x+2=\sqrt{x^2+4x+4y}.
$$
Set $z=x^2+4x+4y$, then our equation becomes
$$
\frac{z'}{2}=\sqrt{z}.
$$
A: Since you're given $y = c^2+cx+2c+1$, we can first verify that $y^{\prime} = c$ by taking the first derivative.
Next we can arrange $y = c^2+cx+2c+1$ into a standard form polynomial of $c$.
$$y = c^2+cx+2c+1 \implies 0 = c^2+cx+2c+1 - y = (1)c^2+(x+2)c+(1 - y)$$
Now using the quadratic formula we find the roots of our polynomial to solve for $c$. Our goal isn't really to solve $c$, but rather the form that results is quite familiar.
$$c = \frac{-(x+2) \pm \sqrt{(x^2+2)-4(1)(1-y)}}{2(1)} = \frac{-(x+2) \pm \sqrt{x^2+4x+4y}}{2}$$
Since we know from our derivative and the given differential equation that $c = y^{\prime} = \frac{-(x+2) + \sqrt{x^2+4x+4y}}{2}$ and solving the candidate solution for $c$ tells us that $c = \frac{-(x+2) \pm \sqrt{x^2+4x+4y}}{2}$, we simply take the solution for $c$ to be the branch where we add the radical to show that our candidate solution satisfies the differential equation.
A: $$
y-1=y'(x+2)+(y')^2\iff y=xy'+(y'+1)^2
$$
is a Clairaut differential equation. As such, the linear functions $y'=c$ are solutions. The derivative of the equation gives $y''=0$, which corresponds to above linear family, and
$$
x+2(y'+1)=0\implies y=-\frac{x(x+2)}2+\frac{x^2}4=-x-\frac{x^2}4
$$
which is the singular solution and envelope of the linear family.

In the given form of the ODE, one has to make sure that the term under the square root remains non-negative. This distinction may get lost in the Clairaut equation, meaning that it could have solutions or segments thereof that are not admissible for the original equation. This appears to be not the case.
Next the sign of the square root is always positive, so that
$$
2c=-(x+2)+|x+2+2c|\iff x+2+2c\ge 0
$$
restricts the domain of the solution.
