I'm dealing with this exercise.
"Given the general solution \begin{equation} (1) \;\;\;\;\;\;\;\; y(x)=\pm \sqrt{K|x|-1} \;\;\;,\;\; K>0\end{equation} of an exact ODE. Find a potential function $\Psi (x,y)$ for the exact ODE that is solved by $(1)$.
So what does it actually mean? I have $y(x)$ so I can't really figure out why it ask for a potential function. I know an exact ODE is someting where you have $M(x,y)dx+N(x,y)dy=0$ but I can't figure out what I am supposed to do.
First, you can assume $\,K\left|x\right|-1\,$ is non-negative, cause if it isn't your function $\,y(x)\,$ can't be a (real) solution.
Now, $$\begin{align*}y=\pm\sqrt{K\left|x\right|-1}&\Longrightarrow \;y^2=K\left|x\right|-1\quad (1)\\ 2yy'=K\frac{d}{dx}\left|x\right|=\pm\,K & \Longrightarrow\;2y\frac{dy}{dx}=\pm\, K\end{align*}$$ $$\mp\,Kdx\,+\,2y\,dy=0 \quad(2)\\$$
You can define $\,M(x,y)=\mp\,K,\;N(x,y)=2y.$
Your potential function verifies: $\,\mathbf{F}=(M,N)=\nabla\Psi,\;$ so: $$\Psi=\int\,M\,dx=\int\,N\,dy$$
Solving the equation $\,(2)\,$ for $\,\left|x\right|=x\,$ or $\,\left|x\right|=-x,\,$ you obtain two different potentials (apparently):
$$\begin{align*}\Psi(x,y)&=y^2-Kx\quad \textrm{if}\;x\geq0\\\ \Psi(x,y)&=y^2+Kx\quad\textrm{if}\;x<0\end{align*}$$
So, $\;y^2-K\left|x\right|=-1\;$ is your potential function, which can be obtained just isolating the variables in $\,(1).$