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

*Post-script version*

- 1.
- Solve the initial value problems
- (a)
- .The characteristic equation,
*r*^{2} - 5*r* + 6=0 has roots 2,3.
The general solution is *x*_{g} = *C*_{1} *e*^{2t} + *C*_{2} *e*^{3t}.

The initial conditions require *C*_{1} = 1 and *C*_{2} = -1. Therefore,

*x*(*t*) = *e*^{2t} - *e*^{3t}

- (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 *C*_{1} = 0 and *C*_{2} = 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*'' + 5*x*' + 4*x* = 0

The characteristic equation, *r*^{2} + 5*r* + 4 = 0, has real roots -1,-5
so the system is overdamped.
The characteristic equation for general mass *m* is

*mr*^{2} + 5*r* + 4 = 0,

has discriminant 25-16*m*. The system is critically damped when
- 4.
- For the first-order equation
*y*' = *y*^{2} - 3*y*, where are the equilibria?
For each equilibrium, determine if it is stable or if it is unstable.

The equilibria are the roots of *y*^{2} - 3*y*, or *y*=0 and *y*=3.

*y*^{2}-3*y*>0 if *y*>3 or *y*<0, and *y*^{2} - 3*y*<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*' = -3*y*. 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 *y*^{2} - 3*y* = 3(*y*-3)+(*y*-3)^{2}
so *p*' = 3*p* + *p*^{2}. The linearization around *y*=3, or *p*=0, is
*p*' = 3*p*, which has an unstable equilibrium.

Therefore the original equation is unstable at *y*=3.

*Nathan Gibson*

*12/11/1997*