# Nonlinear first-order differential equation with a simple parametric solution.

I have to solve the nonlinear first-order differential equation $$\frac{a-y'}{\sqrt{1+y'^2}}e^{-a \arctan y'}=bx+c,$$ where $a,b,c$ are constants, and $y$ is a function of $x$.

Obviously, there is no way to put it in the form $y'=f(x)$ and integrate. However, I know that a simple parametric solution exists: \begin{align} x(t) &= c_1 + c_2 e^{-a t}(a\cos t-\sin t) \\ y(t) &= c_3 + c_2 e^{-a t}(a\sin t+\cos t), \end{align} for some constants $c_1,c_2,c_3$ (that depends on $a,b,c$). How can we arrive at this solution? What is the method?

Here is a start, let

$$\arctan(y')=t \implies y' = \tan(t).$$

Subs back in the ode, we have

$$\frac{a-y'}{\sqrt{1+y'^2}}e^{-a \arctan y'}=bx+c \implies \frac{a-\tan(t)}{\sec(t)}e^{-a t}=bx+c$$

$$\implies x(t) = -\frac{c}{b} + \frac{1}{b}(a\cos(t)-\sin(t))e^{-at}$$

$$\implies x(t) = c_1 + c_2e^{-at}(a\cos(t)-\sin(t)).$$

Now, you need to find $y(t)$. I think you can finish the task.