# Second order Autonomous Differential Equations

1. Aug 20, 2006

### roamer

Hi All,
I have a differntial equation that I came up with on a little engineering problem posted here https://www.physicsforums.com/showthread.php?t=129247 that I can't solve. It is d^2R/dt^2=W^2*R where R is radial position and W is angular velocity and t is time. I think it is an autonomous diff eq. but don't know, its been a while since math class. Any ideas?? Thanks

2. Aug 21, 2006

### qbert

W is a constant?

if so we can solve it relatively easy.

Introduce v = dR/dt. Then the differential equation is
v dv/dR = W^2 R.

Integrating once gives
v^2 - v0^2 = W^2 R^2 - W^2 R0^2
Where i have assumed v(t=0) = v0 and R(t=0) = R0.

a quick arrangement
v = +/- sqrt( v0^2 - W^2 R0^2 + W^2 R^2 )

and thus dR/dt = +/- sqrt( v0^2 - W^2 R0^2 + W^2 R^2 )

This is a seperable first order ODE

define a such that W^2 a^2 = v0^2 - W^2 R0^2

Then
dR/dt = +/- sqrt(W^2 a^2 + W^2 R^2)

dR/sqrt(a^2 + R^2) = +/- W dt

The right hand side is +/- Wt.
To integrate the left hand side Put R = a*sinh(x) so that
dR = a cosh(x) dx then
sqrt( a^2 + R^2 ) = sqrt(a^2 + a^2 sinh^2(x)) = a sqrt(1 + sinh(x)^2)
= a sqrt(cosh^2(x)) = a*cosh(x).

This dR/sqrt(a^2 + R^2) -> dx
The integral is thus

arcsinh(R/a) - arcsinh(R0 / a) = +/- Wt
and therefore

R = a sinh( +/- Wt + arcsinh(R0 / a))
and a = sqrt(V0^2/W^2 - R0^2).

Plugging this in and checking shows us that the
- sign gives the v(0) = - v0

Therefore

R = a sinh( Wt + arcsinh(R0 / a))

Last edited: Aug 21, 2006
3. Aug 21, 2006

### qbert

just shows to go you, when your tool is a hammer....
i was thinking about the trick v =dR/dt so that was the
first way i did this problem.

ergh! the easier (more standard way) assume
R = e^kt then the differential equation gives
k^2 = W^2 so that

R = A exp(Wt) + B exp(-Wt)

introducing the initial values you get the same

4. Aug 21, 2006

### roamer

qbert,
Thanks for the answers. I know exactly what you mean about going about problems with the wrong tool, still it is at least good to know you can grind out answers in more than one way even if your first choice isn't the easiest.

5. Aug 23, 2006

### HallsofIvy

"autonomous" equations are those in which the independent variable does not appear explicitly. Second order autonomous equations are those of the form $\frac{d^2y}{dx^2}= f(y,y')$ and, as qbert said, letting v= y' is a standard method (its' called "quadrature"). If v= y', then
$$\frac{d^2y}{dx^2}= \frac{dv}{dt}= \frac{dv}{dy}\frac{dy}{dt}= v\frac{dv}{dy}$$
so that the second order equation reduces to a first order equation
[tex]v\frac{dv}{dy}= f(y,v)[/itex]. After you have solved for v, integrate to find y.