How do we find envelope ODE of a modulated oscillator obeying
$$y^{''}(t)+y (t)\left(\dfrac{2 \pi}{t-c}\right)^2 = 0 $$
after obtaining solution say for initial conditions $y(0) = 1, y'(0) = 0?$. Does it oscillate indefinitely?
Could not readily apply $\; c-, p-$ discriminant methods. Please help.
Numerical solution with $ c=t _{max} =6$
Envelope sketched by hand.
On
If you set $x=t-c$, then $$x^2y''(x+c)+4\pi^2y(x+c)=0$$ is an Euler-Cauchy equation with characteristic polynomial $$ 0=r(r-1)+4\pi^2=(r-\tfrac12)^2+4\pi^2-\tfrac14 $$ which has for the domain $x<0$ basis solutions $(-x)^{\frac12\pm iω}=(-x)^{\frac12}e^{\pm iω\ln(-x)}$ which combine to the real basis solutions $$ (-x)^{\frac12}\cos(\omega\ln(-x)+\phi_0) ~\text{ and }~ (-x)^{\frac12}\sin(\omega\ln(-x)+\phi_0), $$ where $ω=\frac12\sqrt{16\pi^2-1}$. The initial conditions and setting $\phi_0=-\ln(c)$ then lead to the formula in the other answer.
If I did not make any mistake, the solution is given by $$y=\frac{1}{k }\sqrt{\frac{c-t}{c}} \left(\sin \left(\frac{k}{2} \log \left(\frac{c}{c-t}\right)\right)+k \cos \left(\frac{k}{2} \log \left(\frac{c}{c-t}\right)\right)\right)$$ where $k=\sqrt{16\pi^2-1}$.
Now, close to $c^-$, $y$ oscillates indefinitely.
Edit
Defining $\alpha=\sqrt{\frac{c-t}{c}}$, we have $$y=\alpha \left(\cos [k \log (\alpha )]-\frac{1}{k}\sin [k \log (\alpha )]\right)$$
The first derivative cancels at $$t_n=c \left(1-e^{\frac{2 \pi n}{k}}\right)$$ At these points $$y(t_n)=(-1)^n e^{\frac{ \pi n}{k}}\qquad \text{and}\qquad y''(t_n)=(-1)^{n+1}\frac{1+k^2}{4c^2}e^{-\frac{2 \pi n}{k}}$$
So, if $n$ is odd, these points correspond to maxima and, if $n$ is even, they correspond to minima and then the upper and lower envelopes.