### Exam given in term E97, July 24, 1997: Solutions

Post-script version

1.
Solve the initial value problems
(a)
.The characteristic equation, r2 - 5r + 6=0 has roots 2,3.

The general solution is xg = C1 e2t + C2 e3t.

The initial conditions require C1 = 1 and C2 = -1. Therefore,

x(t) = e2t - e3t

(b)
.

The general solution to the homogeneous solution is

The particular solution has the form

That gives

so A=-2 and B=0.

Therefore, the general solution to the nonhomogeneous equation is

The initial conditions require C1 = 0 and C2 = 1.

Therefore, the solution to the initial-value problem is

2.
For the differential equation
(a)
Find the general solution to this equation. The roots of the characteristic equation are .

The general solution to the homogeneous equation is

The particular solution has the form

The method of undetermined coefficients gives

The general solution is therefore

(b)
What part of the solution will become unimportant after a long time (the transient)? What part of the solution will be important after a long time (the limiting behavior)?

The transient is

The limiting solution is

(c)
Find the amplitude and period of the limiting solution

The amplitude is

3.
Write a differential equation that models a spring-mass system with mass 1 kg, damping coefficient 5 kg/s, and a spring constant 4 N/m. Is the system underdamped or overdamped? Using the same damping coefficient and spring constant, what mass will make the system critically damped?

The differential equation is

x'' + 5x' + 4x = 0

The characteristic equation, r2 + 5r + 4 = 0, has real roots -1,-5 so the system is overdamped.

The characteristic equation for general mass m is

mr2 + 5r + 4 = 0,

has discriminant 25-16m. The system is critically damped when

4.
For the first-order equation y' = y2 - 3y, where are the equilibria? For each equilibrium, determine if it is stable or if it is unstable.

The equilibria are the roots of y2 - 3y, or y=0 and y=3.

y2-3y>0 if y>3 or y<0, and y2 - 3y<0 if 0<y<3.

Therefore solutions are

• increasing () if y>3
• decreasing () if 0<y<3
• increasing () if y<0
So the equilibrium at y=3 is unstable and the equilibrium at y=0 is stable.

Alternate solution to the question of stability:

The linearization about y=0 is y' = -3y. The linearized equation is stable at y=0, so the original equation is stable at y=0.

Linearize about y=3: Put y=3 + p; then y2 - 3y = 3(y-3)+(y-3)2 so p' = 3p + p2. The linearization around y=3, or p=0, is p' = 3p, which has an unstable equilibrium.

Therefore the original equation is unstable at y=3.

Nathan Gibson
12/11/1997