While studying a particular physical system, I arrived at the following difference differential equation:
$$\frac{dx_n(t)}{dt} = -g \left\{\sqrt{(n + 1)(n + 2)}x_{n+1}(t) - (2n +1)x_n(t)\right\},$$
where $g$ is a constant and the initial conditions are is $x_n(0) = 0$ for $n \geq 0.$
How do I solve this?
Thank you!
My work:
Writing out the first equations, I got:
$$x_0'(t) = -g \{\sqrt{2}x_1(t) - x_0(t)\},$$
$$x_1'(t) = -g \{\sqrt{6}x_2(t) - 3x_1(t)\},$$
$$\vdots$$
Since these first order differential equations are interdependent, my solution to a given one of them will have to be restricted by the next one. That's where I need your help.
Given $x_n(0)=0$. Plug in $t=0$ to get $x_n'(0)=0$. Differentiate the equation, plug in $t=0$ to get $x_n''(0)=0$, and so on: all derivatives of $x_n$ at $t=0$ are zero.
But start with any function $x_0(t)$ with all derivatives $0$ at $0$., then recursively plug in to get solutions for all the other $n$ in terms of that. For example $x_0(t) = \exp(-1/t^2)$ with of course $x_0(0)=0$.
$$ x_{{1}} \left( t \right) ={\frac { \left( x_{{0}} \left( t \right) g-x'_{{0}} \left( t \right) \right) \sqrt {2}} {2g}} $$
$$ x_{{2}} \left( t \right) ={\frac { 3\,x_{{0}} \left( t \right) {g}^{2}-4\, x'_{{0}} \left( t \right) g+x''_{{0}} \left( t \right) }{{g}^{2}}} $$
$$ x_{{3}} \left( t \right) ={\frac {15\,x_{{0}} \left( t \right) { g}^{3}-23\, x'_{{0}} \left( t \right) {g }^{2}+9\, x''_{{0}} \left( t \right) g-x'''_{{0}} \left( t \right) }{12{g}^{3}}} $$