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## Solutions to Selected Exercises

0.2.101: Compute and . Then .

0.2.102: Yes.

0.2.103: is a solution for and .

0.2.104: ,

0.2.105:

0.3.101: a) PDE, equation, second order, linear, nonhomogeneous, constant coeﬃcient.
b) ODE, equation, ﬁrst order, linear, nonhomogeneous, not constant coeﬃcient, not autonomous.
c) ODE, equation, seventh order, linear, homogeneous, constant coeﬃcient, autonomous.
d) ODE, equation, second order, linear, nonhomogeneous, constant coeﬃcient, autonomous.
e) ODE, system, second order, nonlinear.
f) PDE, equation, second order, nonlinear.

0.3.102: equation: , solution: .

1.1.101:

1.1.102:

1.1.103:

1.1.104: 170

1.1.105: If , then . If , then .

1.1.106: The equation is for some constant . The snowball will be completely melted in 25 minutes from time .

1.1.107: , so 4 constants.

1.2.101:

is a solution such that .

1.2.102: Yes a solution exists. where . The function is continuous and , which is also continuous near . So a solution exists and is unique. (In fact is the solution).

1.2.103: No, the equation is not deﬁned at .

1.2.104: a) , b) , c) . Justiﬁcation left to reader.

1.2.105: Picard does not apply as is not continuous at . The equation does not have a continuously solution. If it did notice that , by ﬁrst derivative test, for small positive , but then for those we would have , so clearly the derivative cannot be continuous.

1.3.101:

1.3.102:

1.3.103:

1.3.104:

1.3.105:

1.3.106: The range is approximately 7.45 to 12.15 minutes.

1.3.107: a) . b) 102 rabbits after one month, 861 after 5 months, 999 after 10 months, 1000 after 15 months.

1.4.101:

1.4.102:

1.4.103: grams

1.4.104:

1.4.105: , where is a constant with units .

1.5.101:

1.5.102:

1.5.103:

1.5.104:

1.6.101:
a) 0, 1, 2 are critical points.
b) is unstable (semistable), is stable, and is unstable.
c) 1

1.6.102: a) There are no critical points. b)

1.6.103: a) b)

1.6.104: a) is a stable critical point, is an unstable one. b) , c) , d) or DNE.

1.7.101: Approximately: 1.0000, 1.2397, 1.3829

1.7.102:
a) 0, 8, 12
b) , so errors are: 16, 8, 4.
c) Factors are 0.5, 0.5, 0.5.

1.7.103: a) 0, 0, 0 b) is a solution so errors are: 0, 0, 0.

1.7.104: a) Improved Euler: for , for , b) Standard Euler: for , for , c) , so is approximately . d) Approximate errors for improved Euler: for , and for . For standard Euler: for , and for . Factor is approximately for improved Euler, and for standard Euler.

1.8.101: a) , b) , c) , d) .

1.8.102: a) Integrating factor is , the equation becomes . b) Integrating factor is , the equation becomes . c) Integrating factor is , the equation becomes . d) Integrating factor is , the equation becomes .

1.8.103: a) The equation is , and this is exact because , , so . b) , leads to potential function , solving leads to the same solution as the example.

2.1.101: Yes. To justify try to ﬁnd a constant such that for all .

2.1.102: No. .

2.1.103:

2.1.104:

2.1.105:

2.2.101:

2.2.102:

2.2.103:

2.2.104:

2.2.105:

2.2.106:

2.2.107:

2.3.101:

2.3.102: a) , b) , c)

2.3.103:

2.3.104: No. .

2.3.105: Yes. (Hint: First note that is bounded. Then note that and cannot be multiples of each other.)

2.3.106:

2.4.101: (and larger)

2.4.102:
a)
b)
c)

2.4.103: a) , b) , c) 45000 kg, d) 11250 kg

2.4.104: . If , then the system is overdamped and will not oscillate.

2.5.101:

2.5.102: a) , b)

2.5.103:

2.5.104:

2.5.105:

2.6.101:

2.6.102: , where and .

2.6.103: a) , b)

3.1.101: , ,

3.1.102: ,

3.1.103: , ,

3.1.104: , , ,

3.1.105: . Explanation of the intuition is left to reader.

3.2.101:

3.2.102:

3.2.103:

3.2.104: a) b)

3.3.101: Yes.

3.3.102: No.

3.3.103:

3.3.104: a) b)

3.4.101:
a) Eigenvalues: Eigenvectors: , ,
b)

3.4.102:
a) Eigenvalues: , Eigenvectors: ,
b)

3.4.103:

3.4.104:

3.5.101: a) Two eigenvalues: so the behavior is a saddle. b) Two eigenvalues: and , so the behavior is a source. c) Two eigenvalues: , so the behavior is a center (ellipses). d) Two eigenvalues: and , so the behavior is a sink. e) Two eigenvalues: and , so the behavior is a saddle.

3.5.102: Spiral source.

3.5.103:

The solution will not move anywhere if . When is positive, then the solution moves (with constant speed) in the positive direction. When is negative, then the solution moves (with constant speed) in the negative direction. It is not one of the behaviors we have seen.
Note that the matrix has a double eigenvalue 0 and the general solution is and , which agrees with the above description.

3.6.101:

3.6.102: . Solution:

3.6.103:

3.7.101: a) b) No defects. c)

3.7.102:
a)
b) Eigenvalue 1 has a defect of 1
c)

3.7.103:
a)
b) Eigenvalue 2 has a defect of 2
c)

3.7.104:

3.8.101:

3.8.102:

3.8.103: a) b)

3.8.104:

3.9.101: The general solution is (particular solutions should agree with one of these):

3.9.102: The general solution is (particular solutions should agree with one of these):

3.9.103:

3.9.104:

4.1.101:

4.1.102: for (for any )

4.1.103:

4.1.104: General solution is . Since then , and so . Therefore, the solution is always identically zero. One condition is always enough to guarantee a unique solution for a ﬁrst order equation.

4.1.105:

4.2.101:

4.2.102:

4.2.103:

4.2.104:

4.3.101: a) b)

4.3.102: a) b)

4.3.103:

4.3.104: a) b) no.

4.3.105: a) b) is continuous at so the Fourier series converges to . Obtain . c) Using the ﬁrst 4 terms get (quite a bad approximation, you would have to take about 50 terms to start to get to within of ).

4.4.101: a) b)

4.4.102: a) b)

4.4.103: a) b)

4.4.104:

4.4.105:

4.5.101:

4.5.102:

4.5.103:

4.5.104:

4.6.101:

4.6.102:

4.6.103: for some

4.6.104:

4.7.101:

4.7.102:

4.7.103:

4.7.104:

4.8.101:

4.8.102: a)
b) c)

4.8.103: a) b) c)

4.9.101:

4.9.102:

4.10.101:

4.10.102:

4.10.103: a) b)

4.10.104:

5.1.101: , ,

5.1.102: a) , , , , , , . The problem is not regular. b) , , , , , , . The problem is regular.

5.2.101:

5.2.102: , , , .

5.3.101:

5.3.102: Approximately 1991 centimeters

6.1.101:

6.1.102:

6.1.103:

6.1.104:

6.2.101:

6.2.102:

6.2.103:

6.3.101:

6.3.102:

6.3.103:

6.3.104:

6.4.101:

6.4.102:

6.4.103:

6.4.104:

6.4.105:

7.1.101: Yes. Radius of convergence is .

7.1.102: Yes. Radius of convergence is .

7.1.103: so , which converges for .

7.1.104:

7.1.105: is a polynomial. Hint: Use Taylor series.

7.2.101: , , , recurrence relation (for ): , so:

7.2.102: a) , and for we have , so

b)

7.2.103: Applying the method of this section directly we obtain for all and so is the only solution we ﬁnd.

7.3.101: a) ordinary, b) singular but not regular singular, c) regular singular, d) regular singular, e) ordinary.

7.3.102:

7.3.103: (Note that for convenience we did not pick )

7.3.104:

8.1.101: a) Critical points and . At using , the linearization is , . At using , the linearization is , .
b) Critical point . Using , the linearization is , .
c) Critical point . Using , the linearization is , .

8.1.102: 1) is c), 2) is a), 3) is b)

8.1.103: Critical points are , and . The linearization at the origin using variables , , is , , . The linearization at the point using variables , , is , , .

8.1.104: , , .

8.2.101: a) : saddle (unstable), : source (unstable), b) : spiral sink (asymptotically stable), : saddle (unstable), c) : saddle (unstable), : saddle (unstable)

8.2.102: a) , critical points : an unstable saddle, and : a stable center. b) , no critical points. c) , critical point at is a stable center.

8.2.103: Critical point at . Trajectories are , for , these give closed curves around the origin, so the critical point is a stable center.

8.2.104: A critical point is stable if and unstable when .

8.3.101: a) Critical points are , for any integer . When is odd, we have a saddle point. When is even we get a sink. b) The ﬁndings mean the pendulum will simply go to one of the sinks, for example and it will not swing back and forth. The friction is too high for it to oscillate, just like an overdamped mass-spring system.

8.3.102: a) Solving for the critical points we get and . The Jacobian matrix at is whose eigenvalues are and . So the eigenvalues are always real of opposite signs and we get a saddle (In the application however we are only looking at the positive quadrant so this critical point is not relevant). At we get Jacobian matrix . b) For the speciﬁc numbers given, the second critical point is the matrix is , which has eigenvalues . Therefore there is a spiral source. This means the solution will spiral outwards. The solution will eventually hit one of the axis or so something will die out in the forest.

8.3.103: The critical points are on the line . In the positive quadrant the is always positive and so the fox population always grows. The constant of motion is , for any this curve must hit the axis (why?), so the trajectory will simply approach a point on the axis somewhere and the number of hares will go to zero.

8.4.101: Use Bendixson-Dulac Theorem. a) , so no closed trajectories. b) for all except the lines given by (where we get zero), so no closed trajectories. c) for all except the line given by (where we get zero), so no closed trajectories.

8.4.102: Using Poincarè-Bendixson Theorem, the system has a limit cycle, which is the unit circle centered at the origin as , gets closer and closer to the unit circle. Thus we also have that , is the periodic solution.

8.4.103: , . So . The Bendixson-Dulac Theorem says there is no closed trajectory lying entirely in the set .

8.4.104: The closed trajectories are those where , therefore, all the circles with radius a multiple of are closed trajectories.

8.5.101: Critical points: , , . Linearization at using , , is , , . Linearization at using , , is , , . Linearization at using , , is , , .