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May 23

Differential Equation

Does the differential equation x² * d²x/dt² = k have a name? (All I've figured out about this so far is that I don't remember enough about differential equations. I'm not getting anything on solving it except errors.) Thank you. RJFJR (talk) 02:45, 23 May 2024 (UTC)

As to a solution, you could guess that one might be some power of $t$ and accordingly substitute $x=A{t}^p$ (where $A$ and $p$ are constants), then solve for $p$ and then solve for $A$. catslash (talk) 09:02, 23 May 2024 (UTC)

Then, since nothing in the equation depends on the absolute value of $t$, you could apply an arbitrary time-shift to get a slightly more general solution $x=A(t-t_0{)}^p$. catslash (talk) 09:20, 23 May 2024 (UTC)

You're right. Thank you. p=2/3, I was not expecting that. I appreciate it. RJFJR (talk) 14:14, 23 May 2024 (UTC)

This is an autonomous second order equation. If you write $\ddot{x}-\frac{k}{x^2}=0$ and multiply with $\dot{x}$, you find $\frac{d}{dt}(\frac{1}{2}{\dot{x}}^2+\frac{k}{x})=0$ so $\frac{1}{2}{\dot{x}}^2+\frac{k}{x}=C$ for some constant $C$. Solve for $\dot{x}$ and you get a separable first order equation. —Kusma (talk) 14:35, 23 May 2024 (UTC)

Thank you. I need to dig out the old textbook and start reading. RJFJR (talk) 19:25, 23 May 2024 (UTC)

catslash's solution is a one-parameter family (indexed by $t_0$) of very nice solutions, but in general you should be able to solve the initial value problem for any initial values of $x$ and $\dot{x}$, so you'll get a two parameter family. It is easy to show that the solution exists; you can get an implicit formula from Mathematica or other symbolic computation software. —Kusma (talk) 12:17, 24 May 2024 (UTC)

Autonomous system (mathematics)#Special case: x″ = f(x) gives $t$ as a two-parameter function of $x$, but this function looks uninvertable except for the choice of the parameter $C_1$ which makes it correspond to my guessed solution. catslash (talk) 22:43, 25 May 2024 (UTC)