<s>[INST] 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?
 [/INST] 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.
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