For each of the differential equations 1-5,
give the general solution.
1. y'' + 6 y' + 13 y = 0
2. y'' - 4 y' + 4 y = 0
3. y'' - 2 y' + 26 y = 0
4. y'' - 4 y' + 13 y = 0
5. y'' - 5.5 y' + 7.5 y = 0
For each of the differential equations 6-10,
give the solution satisfying the initial conditions.
6. y'' + 2 y' + 17 y = 0, y(0) = 1, y'(0) = 0.
7. y'' + 5 y' + 7.25 y = 0, y(0) = 0, y'(0) = 1.
8. y'' + 6 y' + 9 y = 0, y(0) = -1, y'(0) = 0.
9. y'' + 5 y' + 5.25 y = 0, y(0) = 0 y'(0) = -1.
10. y'' - 2 y' + 2 y = 0, y(0) = 1, y'(0) = -1.
A mass of 1 g is attached to the free end of a horizontal
spring with spring constant 13 dynes/cm and to a damper with
damping constant p dynes/(cm/sec). This leads to the
differential equation
y'' + py' + 13y =0,
where y = y(t) is the elongation of the spring in cm at time
t seconds.
The spring is stretched to elongation
5 cm and then released at time t=0 with no initial velocity.
For the given values of p find the equation giving the
position of the mass for times t > 0, and graph this
equation for t < 2 secs. (Copy the graph from your
graphing calculator, but label axes and units carefully.)
11. p = 6
12. p = 4
11. p = 2
11. p = 0
Anthony Phillips Math Dept, Stony Brook University
tony at math dot sunysb dot edu
November 4 2000