# Seemingly impossible differential equation $y' = \sqrt{|y|(1-y)}$ with $y < 0$

I've been staring at the following seemingly easy differential equation problem for a couple of hours now and I cannot seem to figure out what I'm doing wrong. I'd really appreciate if someone could provide a hint (rather than present an all worked out answer) to me.

The differential equation is

$$y' = \sqrt{|y|(1-y)}, \hspace{1cm} y < 0$$

I proceeded in the following manner: rewrite to account for the $y < 0$ condition to $y' = \sqrt{y^2-y}$. Separate variables and then integrate to get rid of the derivative

$$\int \frac{y'}{\sqrt{y^2 - y}}dx = \int dx \Rightarrow y = \frac{\cosh(x + C) + 1}{2} > 0$$

So it seems to me that there is no solution, yet there is asked for one. I integrated by first completing squares, substituting $u = y - \frac{1}{2}$, multiplying by 2 to remove fraction in denominator, substituting $s = 2u$ to obtain the standard integral $\int \frac{s'}{\sqrt{s^2 - 1}} dx = \cosh^{-1}(s)$ and finally substituting everything back (and then solve for $y$ of course).

EDIT:

I decided to add all my integration steps explicitly to remove any ambiguity of what I did. So

$$\int \frac{y'}{\sqrt{y^2 - y}}dx = \int dx \Rightarrow \int \frac{y'}{\sqrt{(y-\frac{1}{2})^2-\frac{1}{4}}}dx = \int dx \underbrace{\Rightarrow}_{u = y - \frac{1}{2}} \int \frac{u'}{\sqrt{u^2-\frac{1}{4}}}dx = \int dx$$

$$\Rightarrow 2 \int \frac{u'}{\sqrt{4u^2-1}}dx = \int dx \underbrace{\Rightarrow}_{s = 2u} \int \frac{s'}{\sqrt{s^2 - 1}}dx = \int dx \underbrace{\Rightarrow}_{\text{integral for arccosh}} \cosh^{-1}(s) = x + C$$

$$\Rightarrow s = \cosh(x + C) \Rightarrow y = \frac{\cosh(x + C) + 1}{2} > 0$$

• wolframalpha.com/input/?i=integral+1%2F%28sqrt%28x^2-x%29%29+dx seems to disagree with one of your steps (they get an inverse sinh instead of inverse cosh). – Ian Sep 14 '15 at 21:51
• @Ian They are the same function (the function your link leads to and arccosh). Plot them in GeoGebra to see. Don't know why Wolfram is making it look so complex... – Jori Sep 14 '15 at 22:14
• Since $s \lt -1$ and domain of $acosh(s)$ is $s \ge 1$, the integral should be $acosh(-s)$ – xidgel Sep 14 '15 at 22:35
• @Did, I don't know why everyone is providing complete answers (some obviously wrong though), I only asked for a hint or more importantly the place I went wrong. – Jori Sep 15 '15 at 8:09
• @Narasimham Huh? The first sentence of my last comment makes for a decent hint, I believe. – Did Sep 15 '15 at 8:44

$$y' = \sqrt{|y|(1-y)}, \hspace{1cm} y < 0$$ Let $z=-y$ hence $|y]=|z|=z$ in $z>0$ $$-z' = \sqrt{z(1+z)}, \hspace{1cm} z > 0$$ $$z=\sinh^2\left(\tfrac12(x-c)\right)$$ $$y=-\sinh^2\left(\tfrac12(x-c)\right)$$
$$y'=-\sinh\left(\tfrac12(x-c)\right)\cosh\left(\tfrac12(x-c)\right)$$ $y'>0$ implies $\sinh\left(\tfrac12(x-c)\right)<0$ hence $x-c<0$
This limits the range of $x$ and the solution is :
$$y=-\sinh^2\left(\tfrac12(x-c)\right) \qquad x <c$$