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?


1 Answer 1


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.


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