Worcester Polytechnic Institute

MA2051--C97 --- Solutions for TEST II

1. (a) The roots for the characteristic equation are -1 and 3/2, so two solutions are e-t and e(3/2) t. To verify independence, compute the Wronskian and show that it is not zero.
(b) Use the method of undetermined coefficients to find yp(t) = 1/2 e-3t.
(c) The general solution is

y(t) = c_1 e^{-t} + c_2 e^{(3/2)t} + y_p(t)\; . \end{displaymath}

The initial data give c1 = -6/5 and c2 = +6/5, so y(t) = (6/5) e (3/2)t - (6/5) e- t + (1/2) e-3t.
(d) For large values of t, the solution approaches the exponential term (6/5)e(3/2)t -- everything else is transient. There is no limiting periodic solution. <\br><\br> 2. (a) The data give you m = 6 and k = (6)(9.8)/(0.02) = 2940, so the model reduces to

6x^{\prime\prime}+ 2940x = 0,\quad x(0)=-0.03,\; x^{\prime}(0) = 0\;

(b) The characteristic equation has roots $r = \pm 22.21 i$ (approximately) and so the general solution is $ x(t) = c_1\sin(22.21 t) + c_2\cos(22.21 t)$. The initial data determine c1 = 0 and c2 = -0.03.
(c) The amplitude is 0.03. The period is $T = 2\pi/22.21 \approx 0.283 $ and it takes one half of a period to travel from highest position to the lowest position.
(d) The velocity is $ x^{\prime}(t) = +0.03 \omega_n \sin(\omega_n t)$ and the mass passes through equilibrium at $t^* = \pi / 2\omega_n$. The velocity is therefore $ x^{\prime}(t^*) = 0.03 \omega_n = 0.666$ (meters per second).

3. (a) The characteristic equation reduces to (r+4)(r-2) = 0 so the roots are r1 = -4 and r2 = +2. The first solution pair is (y1,z1) = (e-4t,3e-4t). The second solution pair is (y2,z2) = (e2t,e2t). (You can test linear independence with the Wronskian.) The general solution is (y,z) = c1(y1,z1) + c2(y2,z2).
(b) The initial data gives $c_1 = 0,\; c_2 = 50$and so y= 50e2t, z= 50e2t. (This is a straight line and the solution moves away from the origin as t increases.)
(c) If you choose initial data so that the constants in the general solution become $c_1 \neq 0$ and c2 = 0, then you get convergence to the origin; the solution becomes (y,z) = c1(e-4t, 3 e-4t). (You must ``remove'' that nasty e+2t.) Any initial data of the form y(0) = (1/3) z(0) works. The solution curve in this case follows a straight line to the origin.

4. The critical damping coefficient is p = 4 and the solution in this case is

x(t) = xi e-2t + ( vi + 2xi) t e-2t

Set this equal to zero and solve for exactly one value:

t^* = { -x_i \over v_i + 2 x_i }\end{displaymath}

If xi > 0, this value of t is positive if and only if vi < -2xi. (You have to push the mass toward the origin to be sure it gets there.)

Arthur Heinricher <heinrich@wpi.edu>
Last modified: Wed Oct 8 08:36:08 EDT 1997