Solution for a nonlinear ODE During my PhD thesis im facing the following ODE:
$$\frac{C}{\gamma}\frac{f'(y)}{\sqrt{1+(f'(y))^2}}=y-K_1$$
Where $y$ is a positive variable$(y\geq0)$, $C$ and $\gamma$ are parameters, $K_1$ is a constant (due to previous integration and is necessarily $\leq0$) and $f'(y)=df/dy$  is the function I want to obtain. Squaring both sides, it can be rewritten: 
$$(\frac{df}{dy})^2=\frac{\gamma^2(y-K_1)^2}{C^2-\gamma^2(y-K1)^2}$$
Now, if I guarantee that $C^2-\gamma^2(y-K1)^2 \geq 0$ the solution is easily obtained through:
$$df=\pm\frac{\gamma(y-K_1)}{\sqrt{C^2-\gamma^2(y-K1)^2}}dy$$
But my problem is that in many times this term in the square root will be less than 0, then i don't know how to proceed since the function will not belong to $\mathbb{R}$ anymore, which is phisically impossible in the problem. Could anyone help me in this matter?
Thank you very much!
P.S.: The ODE was obtained after integration of $\frac{Cf''(y)}{(1+f'(y)^2)^{3/2}}=\gamma$

Let us re-write the ODE as
$$\frac{A y'(x)}{\sqrt{1+y'^2(x)}}=x-B \implies A^2 y'^2(x)=(x-B)^2(1+y'^2(x)).$$
$$\implies y'^2(x)(A^2-(x-B)^2)=(x-B)^2.$$
$$\implies y'(x)=\pm \frac{x-B}{\sqrt{A^2-(x-B)^2}}.$$
$$\int dy= \pm \int \frac{x-B}{\sqrt{A^2-(x-B)^2}}dx.$$
$$\implies y(x)=\pm \sqrt{A^2-(x-B)^2}+D$$
$$\implies (y-D)^2+(x-B)^2=A^2.$$
$D$ is constant of integration, $A=C/\gamma$, $B=K_1.$
