Solving Coupled Differential Equations with Second Derivative, but No First Derivative How can I solve this system of coupled differential equations?
$\frac{d^2\rho}{d\lambda^2}=\frac{5\rho}{(5\rho^2+4t^2)^2}$ $\frac{d^2t}{d\lambda^2}=\frac{4t}{(5\rho^2+4t^2)^2}$
Is it something I could input in the Wolfram calculator?
 A: I hope you'll excuse me if I write $x(t)$ and $y(t)$ instead of $\rho(\lambda)$ and $t(\lambda)$.
The system has the form
$$
\begin{pmatrix} \ddot x \\ \ddot y \end{pmatrix}
=
\frac{1}{(5x^2+4y^2)^2}
\begin{pmatrix} 5x \\ 4y \end{pmatrix}
=
- \begin{pmatrix} \partial V/\partial x \\ \partial V/\partial y \end{pmatrix}
,
$$
where
$$
V(x,y) = -\frac{1/2}{5x^2+4y^2}
.
$$
This means that the total energy (kinetic + potential) $E = \frac12 (\dot x^2 + \dot y^2) + V(x,y)$ is a constant of motion.
Moreover, in polar coordinates the potential $V$ takes the form
$$
V = - \frac{1/2}{r^2 (4+\cos^2 \phi)}
,
$$
which matches the general form for potentials that are separable in polar coordinates,
$$
V = f(r) + \frac{1}{r^2} \cdot g(\phi)
,
$$
with $f(r)=0$ and $g(\phi) = -\dfrac{1/2}{4+\cos^2 \phi}$.
What this means is that you can apply the Hamilton–Jacobi method in polar coordinates, which in principle allows you to integrate the equations explicitly (although in practice it can get a bit messy). Unfortunately this method is quite complicated, so I won't try to describe it here; see some textbook in classical mechanics, such as Landau & Lifshitz (Vol. 1, §48).
A: $$\frac{d^2\rho}{d\lambda^2}=\frac{5\rho}{(5\rho^2+4t^2)^2},\tag{1}$$
$$\frac{d^2t}{d\lambda^2}=\frac{4t}{(5\rho^2+4t^2)^2}.\tag{2}$$
$$2\rho' \frac{d^2\rho}{d\lambda^2}=\frac{10\rho \rho '}{(5\rho^2+4t^2)^2},\tag{1}$$
$$2t'\frac{d^2t}{d\lambda^2}=\frac{8t't}{(5\rho^2+4t^2)^2}.\tag{2}$$
Add both DE's and integrate:
$$2t'\frac{d^2t}{d\lambda^2}+2\rho' \frac{d^2\rho}{d\lambda^2}=\frac{8t't+10 \rho \rho'}{(5\rho^2+4t^2)^2}$$
$$t'^2+\rho'^2 =-\frac{1}{5\rho^2+4t^2}+C_1$$
$$t'^2+\rho'^2 =-\rho\rho''-tt''+C_1$$
$$(t't)'+(\rho \rho ')' =C_1$$
Integrate again:
$$t't+\rho \rho ' =C_1\lambda +C_2$$
Integrate again:
$$t^2+\rho ^2 =C_1\lambda^2 +2C_2\lambda+C_3$$
$$t^2+\rho ^2 =C_1\lambda^2 +C\lambda+C_3$$
You can use this result to eliminate $t$ or $\rho$ in the original DE. Unfortunately the DE won't be linear and therefore it's hard to solve.
