<s>[INST] Question about initial value problem This seems like it should be something simple. 
While solving some HW problems I ran across this: 
$dy/dx = ({1-2x})/y$
which I separated and integrated:
$\int{Ydy} = \int{1-2x}dx$
$y^2/2= x-x^2 + c$
$y^2=2(x-x^2+c)$
$y=\sqrt{2(x-x^2+c)}$
$y=\sqrt{2x-2x^2+2c}$
I was given $y(1)=-2$ and asked to solve. So i did this: 
$-2=\sqrt{2(1)-2(1^2)+2c}$
$-2=\sqrt{2-2+2c}$
$-2=\sqrt{2c}$
So, I assumed c must be some complex number? However, the final solution given by the book pops a negative sign into the equation and just has $c=2$, so the solution looks like this:
$y=-\sqrt{2x-2x^2+2c}$
where $c=2$ 
$y=-\sqrt{2x-2x^2+4}$
which is obviously a true statement. I'm just confused where the negative sign on the right side of the equation came from.  I'm sure its some silly algebra thing I keep overlooking. 
 [/INST] $$\frac {y^2}{2} = -x^2+x+c \iff y=-\sqrt2 \sqrt{-x^2+x+c}$$ or $$y=\sqrt2 \sqrt{-x^2+x+c}$$
The first one when solved for the initial condition gives $c=2$ while the second one has no solution, thus can't satisfy the initial condition, ergo no solution exists. So you're left with only $$y=-\sqrt2 \sqrt{-x^2+x+2}$$
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