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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.