4—Differential Equations 103The second is just− 2 times the first, so it isn’t a separate equation. The family of solutions is all thosexandy
satisfyingx=− 2 y, a straight line.
In thek= +1case you have
x+ 0y= 0, and 0 = 0The solution to this isx= 0andy=anything and it is again a straight line (they-axis).
4.8 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 2x 1 x 2Two masses are connected to a set of springs and fastened between two rigid
walls as shown. The coordinates for the two masses (moving along a straight line
only) 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 arex 1 andx 1 −x 2 respectively, soFx=maxapplied to each mass gives the
equations
m 1d^2 x 1
dt^2=−k 1 x 1 −k 3 (x 1 −x 2 ), and m 2d^2 x 2
dt^2=−k 2 x 2 −k 3 (x 2 −x 1 ) (27)I’m neglecting friction simply to keep the algebra down. These are linear, constant coefficient, homogeneous
equations, just the same sort as Eq. ( 4 ) except that there are two of them. What made the solution of ( 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 inα. 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