Exam # 2 - Solutions Math 308.200 Spring 2002 Name: Student ID:
Signature:
The following exam consists of 8 problems, worth a total of 100 points. There is a total of 10 extra credit points. Partial credit will be awarded according to the completeness of
Solutions to Homework from Section 2.3
Problem 6 The equation can be re-written as
dr
= 3r + 3
d
The equation is linear (in r) but not seperable.
Problem 9 The equation is already in standard form
dr
+ r tan = sec
d
The integrating factor is
1
sin()
d) =
Solutions to Homework from Section 3.4
Problem 18 Assuming the force due to (sliding) friction is proportional to the normal force, the equations of motion of an object sliding down an inclined plane are given by: dv = mg sin() - mg cos() dt
m
If = 30 deg
Final Exam - Solutions Math 308.200 Spring 2002 Name: Signature: The following exam consists of 11 problems, worth a total of 200 points. There is a total of 10 extra credit points. Partial credit will be awarded according to the completeness of work show
Exam # 3 - Solutions Math 308.200 Spring 2002 Name: Student ID:
Signature: The following exam consists of 6 problems, worth a total of 100 points. There is a total of 15 extra credit points. Partial credit will be awarded according to the completeness of
Solutions to Homework from Section 1.5
Problem 5 Applying Eulers method (euler.m) gives the following results xn 1.0000 1.2000 1.4000 1.6000 1.8000 yn 1.0000 1.4000 1.9600 2.7888 4.1096
Problem 12 Given the initial value problem y = y, Eulers method yield
Solutions to Homework from Section 1.1
Problem 15 From the statement of the problem dT (M - T (t) dt which means that dT = k(M - T (t) dt
for some constant k. Problem 16 In a similar fashion to 15, dA A(t)2 dt or dA = kA2 (t) dt
for some constant k. Probl
Exam # 1 Math 308.200 Spring 2002 Name: Student ID:
Signature:
The following exam consists of 6 problems, worth a total of 100 points. Partial credit will be awarded according to the completeness of work shown. Write the answers to each problem on the bla
Solutions to Homework from Section 4.7
Problem 13 a) The difference of any two particular solutions is a homogeneous solution, therefore y1 (x) = x3 - x and y2 (x) = x2 - x are homogeneous solutions. These are linearly independent. b) Using x3 as a partic
Solutions to Homework from Section 9.6
Problem 3 The matrix has eigenvector/eigenvalue pairs equal to 1 0 1 = 1, u1 = 0 -1 - 2i 1 2 = 1 + i, u2 = i -1 + 2i 1 3 = 1 - i, u3 = -i The general solution is therefore 1 -1 - 2i -1 + 2i + c3 e(1-i)t 1 1 c1 et 0 +
Solutions to Homework from Section 5.5
Problem 10
Using Khirchoffs law on the left loop we get
20 = 10I3 (t) + 10I1 (t)
Using Khirchoffs law on the right loop we get
0 = 40I2 (t) + 30I2 (t) 10I3 (t)
The remaining Khirchoffs law gives
I1 (t) = I2 (t) + I3
Solutions to Homework from Section 9.1
Problem 7 The damped mass-spring oscillator equation
my (x) + by (x) + ky(x) = 0
can be written as a matrix ode in normal form by using the change of variables
y1 = y and y2 = y . This leads to the set of equations
y
Solutions to Homework from Section 4.4
Problem 6 Given one solution, y1 = ex , find a second solution y2 (x), to the differential equation: xy + (1 - 2x)y + (x - 1)y = 0, x > 0 Substituting the form y2 = v(x)ex into the equation, we get xv (x) + v (x) = 0
Solutions to Homework from Section 5.2
Problem 18 The critical points are given by solving
x(7 x 2y) = 0
y(5 x y) = 0
which gives cfw_(0,0),(0,5),(7,0),(3,2).
The phase plane, with direction elds in the vicinity of the critical points, is given
below
x =