Solving separable differential equation Seems straight-forward but I've been unable to get it right. Here are my steps:
$$y'(x) = \sqrt{-2y(x) + 28},\hspace{20 pt} y(-4)=-4$$
$$\int {1 \over \sqrt{28-2y} }\hspace{2 pt}\text{d}y = \int \text{d}x$$
$$-\sqrt{28-2y} = x + c$$
$$(28-2y) = (x+c)^2$$
$$y = -1/2x^2 - cx - c^2/2 + 14$$
$$c = {-2, 10}$$
$$\Rightarrow y = -1/2x^2 - 10x - 36$$
I've checked it over many countless times but for the life of me I can't figure out why it won't work. I've tried plugging the result back into the original equation and it seems to me like it checks out if you take the negative of the square root..
 A: Your second line should be
$$\int {1 \over \sqrt{28-2y} }\hspace{2 pt}\text{d}y = \int \text{d}x.$$
A: As Robin Chapman has pointed out, the correct choice is $c=-2$, not $c=10$. This gives
$$y(x)=\frac{28-(x-2)^2}{2} = -\frac{x^2}{2} + 2x + 12.$$
However, this formula describes a parabola $y(x)$ which hits its highest point $y=14$ at $x=-2$, and this is a decreasing function for $x>-2$, which disagrees with the ODE. (Since the square root in the right-hand side of the ODE can't be negative, the solution $y(x)$ can't be decreasing.) Moreover, the right-hand side of the ODE is undefined for $y>14$, so the solution can't continue to increase beyond 14. The conclusion is that the only possible continuation is $y(x)=14$ for all $x>-2$. Hence, the solution is:
$$y(x)=\frac{28-(x-2)^2}{2} = -\frac{x^2}{2} + 2x + 12, \quad -4 \le x \le -2,$$
$$y(x)=14, x>-2.$$
A: So what we want here is a particular solution to our ODE given our condition: $y(-4) = -4$
$$y'(x) = \sqrt{-2y(x) + 28},\hspace{20 pt} y(-4)=-4$$
$$\Rightarrow \dfrac{dy}{dx} = \sqrt{-2y(x) + 28}$$
$$\Rightarrow \int {1 \over \sqrt{28-2y} }\hspace{2 pt}\text{d}y = \int {1}~\text{d}x$$
$u = 28-2y$
$du = -2dx$
$dx = -\dfrac{1}{2}du$
$$\Rightarrow -\dfrac{1}{2}\int {1 \over \sqrt{u} }\hspace{2 pt}\text{d}y = \int {1}~\text{d}x$$
$$\Rightarrow -\dfrac{1}{2}\int {u^{-\dfrac{1}{2}}} \hspace{2 pt}\text{d}y = \int {1}~\text{d}x$$
$$\Rightarrow -\dfrac{1}{2}2u^{\dfrac{1}{2}} = x$$
$$\Rightarrow -\sqrt{28-2y} = x + c$$
$$\Rightarrow \sqrt{28-2y} = -c - x,~~y(-4) = -4$$
$$\Rightarrow \sqrt{28-2(-4)} = -c - (-4)$$
$$\Rightarrow \sqrt{36} = -c + 4$$
$$\Rightarrow 6 = -c + 4$$
$$\Rightarrow c = -2$$
$$\Rightarrow \sqrt{28-2y}^{2} = (-c-x)^2$$
$$\Rightarrow 28-2y = (-(-2)-x)$$
$$\Rightarrow 28-2y = (2-x)^{2}$$
$$\Rightarrow 28-2y = 4-4x+x^{2}$$
$$\Rightarrow 2y = 28-4+4x-x^{2}$$
$$\Rightarrow y(x) = \dfrac{28-4+4x-x^{2}}{2}$$
$$\Rightarrow y(x) = \dfrac{28}{2}-\dfrac{4}{2}+\dfrac{4x}{2}-\dfrac{x^{2}}{2}$$
$$\Rightarrow y(x) = 14-2+2x-\dfrac{1}{2}x^{2}$$
$$\Rightarrow y(x) = -\dfrac{1}{2}x^{2}+2x+12.$$
Hence, 
$~~~~~~~~~~~~~~~~~~~~~~~~y(x) = -\dfrac{1}{2}x^{2}+2x+12$
is our particular solution found to our original first-order seperable linear ordinary differential equation. $\blacksquare$
I hope this helped out, and hopefully I did not make any mistakes to cause any type of confusion
here.
