# How can I solve $\frac{wy'(x)}{y'^2(x)+1}=kx$?

I stumbled upon this differential equation while thinking of a physics problem.

$$\frac{wy'(x)}{y'^2(x)+1}=kx$$

I found two solutions using wolframalpha.

And when I plot it, the result is magically accurate (according to my intuition). So, I know it's right.

My question is, how can I learn to solve this particular differential equation? Is there a procedural approach to arrive at this solution or does solving it mean plugging in the guess and fiddling with the free parameters to arrive at one particular solution?

• Need a step by step solution for this problem ? >> Commented Dec 27, 2017 at 10:19

Note that \begin{align} \frac{w y'}{(y')^2 + 1} &= kx \\ kx (y')^2 - wy' +kx &= 0 \\ y' &= \frac{w\pm \sqrt{w^2 - 4(kx)^2}}{2kx} \end{align} Therefore we just need to integrate both sides to get the answer.

• And the integral is complicated... I get it. Thanks. Commented Dec 27, 2017 at 2:30
• Basically the point is that the ODE does not contain $y$ on its own, only $y'$.
– yo'
Commented Dec 27, 2017 at 2:34

WLOG, $w=1$ (absorb it in $k$).

If $y'=\tan t$, then

$$\frac{y'}{y'^2+1}=\sin t\cos t=\frac{\sin 2t}2=\frac{\sin(2\arctan y')}2.$$

From this we draw

$$y'=\tan\frac{\arcsin2kx}2,$$ which you can integrate.

The trick is to turn into a form $y'=f(x)$ (as user113988 did, in an easier way).