MA2051 - Ordinary Differential Equations
Old Exam II - Solutions


The roots for the characteristic equation are -3 and -2 so two solutions are and . To verify independence, compute the Wronskian and show that it is not zero.
Use the method of undetermined coefficients to find .
The general solution is

The initial data gives and so .

For large values of t, the solution approaches the constant solution (steady state) . (Everything else is transient.)

The model reduces to

(Note that .) The characteristic equation has roots (approximately) and so the general solution is . The initial data determines and .

The period is and the amplitude is . The mass passes through the equilibrium position (x=0) twice during each period and there are 133.7 periods in 60 seconds, so it passes through equilibrium 267 times in a minute.
If damping is present (and the mass still oscillates), then the solution is of the form

where and you don't care about . Let denote the amplitude at t=0. The amplitude (or pseudo-amplitude) at time t=60 is them and the data gives you the equation


The characteristic equation reduces to so the roots are and . The first solution pair is . The second solution pair is . (You can test linear independence with the Wronskian.) The general solution is .
The initial data gives and so , , A sketch of the solution curve starts at the point in the plane and goes off to infinity approaching the line y=z asymptotically. (Notice that is close to 1 for large t.)
If you choose initial data so that the constants in the general solution become and , then you get convergence to the origin; the solution becomes . (You must ``remove'' that nasty growing exponential). Any initial data of the form works. The solution curve in this case follows a straight line to the origin.


© 1996 by Will Brother. All rights Reserved. File last modified on April 25, 1996.