4—Differential Equations 85For whatever reason, I would like to get a non-zero solution forxandy. Can I? The condition depends
on the determinant, so take the determinant and set it equal to zero.
k(k−1)^2 −(1−k)(k−1) = 0, or (k+ 1)(k−1)^2 = 0
There are two roots,k=− 1 andk= +1. In thek=− 1 case the two equations become
−x− 2 y= 0, and 2 x+ 4y= 0
The second is just− 2 times the first, so it isn’t a separate equation. The family of solutions is all those
xandysatisfyingx=− 2 y, a straight line.
In thek= +1case you have
x+ 0y= 0, and 0 = 0
The solution to this isx= 0andy=anything and it is again a straight line (they-axis).
4.10 Simultaneous ODE’s
Single point masses are an idealization that has some application to the real world, but there are many
more cases for which you need to consider the interactions among many masses. To approach this,
take the first step, from one mass to two masses.
k 1 k 3 k 2
x 1 x 2
Two masses are connected to a set of springs and fastened be-
tween two rigid walls as shown. The coordinates for the two masses
(moving along a straight line) arex 1 andx 2 , and I’ll pick the zero
point for these coordinates to be the positions at which everything is at
equilibrium — no total force on either. When a mass moves away from
its equilibrium position there is a force on it. Onm 1 , the two forces are
proportional to the distance by which the two springsk 1 andk 3 are stretched. These two distances are
x 1 andx 1 −x 2 respectively, soFx=maxapplied to each mass gives the equations
m 1
d^2 x 1
dt^2
=−k 1 x 1 −k 3 (x 1 −x 2 ), and m 2
d^2 x 2
dt^2
=−k 2 x 2 −k 3 (x 2 −x 1 ) (4.45)
I’m neglecting friction simply to keep the algebra down. These are linear, constant coefficient, homo-
geneous equations, just the same sort as Eq. (4.4) except that there are two of them. What made
the solution of (4.4) easy is that the derivative of an exponential is an exponential, so that when you
substitutedx(t) =Aeαtall that you were left with was an algebraic factor — a quadratic equation in
α. Exactly the same method works here.
The only way to find out if this is true is to try it. The big difference is that there are two
unknowns instead of one, and the amplitude of the two motions will probably not be the same. If one
mass is a lot bigger than the other, you expect it to move less.
Try the solution
x 1 (t) =Aeαt, x 2 (t) =Beαt (4.46)
When you plug this into the differential equations for the masses, all the factors ofeαtcancel, just the
way it happens in the one variable case.
m 1 α^2 A=−k 1 A−k 3 (A−B), and m 2 α^2 B=−k 2 B−k 3 (B−A) (4.47)
Rearrange these to put them into a neater form.
(