final exam - penn mathfinal exam december 20, 2011 math 420 - ordinary di erential equations no...

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Final Exam December 20, 2011 Math 420 - Ordinary Differential Equations No credit will be given for answers without mathematical or logical justification. Simplify answers as much as possible. Leave solutions in implicit form when appropriate. Scientific calculators only. No notes or books Part I (35 points) Problem 1. (5 pts) Find the real part of e ln(7)- π 4 i . Problem 2. (5 pts) Compute, directly from the definitions, the Laplace transform of u 2 (t). 1

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Page 1: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

Final Exam December 20, 2011

Math 420 - Ordinary Differential Equations

• No credit will be given for answers without mathematical or logical justification.

• Simplify answers as much as possible.

• Leave solutions in implicit form when appropriate.

• Scientific calculators only. No notes or books

Part I (35 points)

Problem 1. (5 pts) Find the real part of eln(7)−π4 i.

Problem 2. (5 pts) Compute, directly from the definitions, the Laplace transform of u2(t).

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Page 2: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

Problem 3. (10 pts) Find the particular solution.

dx

dt=− 2 t x

t2 + 1+ 1, x(0) = 0. (1)

Problem 4. (10 pts) Find the general solution.

dx

dt=

x2 + 2

2 t x(2)

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Page 3: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

Problem 5. (5 pts) Write the following nonlinear second order equation as a system of firstorder equations.

d2y

dt2+

(dy

dt

)2

− y = 0 (3)

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Page 4: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

Part II (65 points)

Problem 6. (65 points total) Consider the damped harmonic oscillator

d2y

dt2+ 2p

dy

dt+ ω2

0y = cos(ωt)u(t) (4)

If ω20 − p2 > 0, meaning the system is underdamped, define ω1 =

√ω20 − p2.

Let z be the fundamental solution of (4). That is, z(t) satisfies d2zdt2 + 2pdz

dt + ω20z = δ(t)

and z(t) = 0 for all t < 0.

6a) (5 pts) Let Z(s) = L(z(t))(s) be the Laplace transform of z(t). Assume the system isunderdamped, and prove that Z(s) = 1

(s+ p)2 +ω21.

6b) (10 pts) Find z(t).

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Page 5: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

6c) (10 pts) Let y(t) be the solution of the forced equation (4) with initial conditionsy(0) = 0, y′(0) = 0, and let Y (s) be its Laplace transform. Prove that

Y (s) =s

(s2 + ω2) [(s+ p)2 + ω21 ]

(5)

Do not attempt to perform partial fraction separation.

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Page 6: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

6d) (10 pts) Using the partial fraction expansion technique, we have

Y (s) =1

(ω20 − ω2)2 + (2pω)2

[(ω2

0 − ω2) · s + 2pω2

s2 + ω2− (ω2

0 − ω2) · (s+ p) + p(ω2 + ω20)

(s+ p)2 + ω21

].

Do not attempt to prove this. For this question, simply indicate which part of thisexpression corresponds to the steady-state solution and which corresponds to the transientsolution. State why.

6e) (10 pts) Use the table of Laplace transforms to prove that the steady-state solution is

y(t) =

[ω20 − ω2

(ω20 − ω2)2 + (2pω)2

cos(ωt) +2 pω

(ω20 − ω2)2 + (2pω)2

sin(ωt)

]u(t). (6)

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Page 7: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

6f) (20 pts) Using the result from question (6e), show that the total amplitude of thesteady-state solution as a function of ω is

A(ω) =1√

(ω20 − ω2)2 + (2pω)2

. (7)

Assuming ω0 = 2 and p = 12 , sketch the frequency response plot. Indicate clearly any critical

points.

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Page 8: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

Part III (100 points)

Problem 7. (25 points total) Consider the following non-linear system.

dx

dt= −

(1 − x

6

)x +

1

2xy

dy

dt=(

1 − y

4

)y − 1

3xy

(8)

7a) (5 pts) If interpreted as a predator-prey system, which is the prey and which is thepredator?

7b) (5pts) In the absence of any prey species, what type of equation does the predatorspecies follow?

7c) (5pts) In the absence of any predator species, what type of equation does the preyspecies follow?

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Page 9: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

7b) (10 pts) Determine all critical points of the system (8).

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Page 10: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

Problem 8. (40 points total) Consider again the system

dx

dt= −

(1 − x

6

)x +

1

2xy

dy

dt=(

1 − y

4

)y − 1

3xy

(9)

8a) (20 pts) Linearize the system at the critical point (x, y) = (2, 43 ). Determine all eigen-values. Sketch the graph of the linearized system, including any straight-line solutions.

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Page 11: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

8b) (20 pts) Linearize the system at the critical point (x, y) = (6, 0). Determine all eigen-values. Sketch the graph of the linearized system, including any straight-line solutions.

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Page 12: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

Problem 9. (35 points total) Consider again the system

dx

dt= −

(1 − x

6

)x +

1

2xy

dy

dt=(

1 − y

4

)y − 1

3xy

(10)

Most predator-prey systems are not Hamiltonian. However, set

H(x, y) = x · y ·(x

6+y

4− 1

). (11)

9a) (5 pts) Verify that the system (10) is a Hamiltonian system.

9b) (5 pts) Prove that H is a Hamiltonian for this system.

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Page 13: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

9c) (10 pts) Sketch the level-sets of H in the first quadrant. To do this, make a precisesketch of the level-set H = 0, then make reasonable approximations for the level-sets H = αfor values of α both larger and smaller than 0.

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Page 14: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

9d) (10 pts) Sketch the trajectories of the system (10) in the phase plane. Restrict your dia-gram to the first quadrant. Clearly label all critical points by type, and label any separatricesyou find.

9e) (5 pts) What is the ultimate fate of the system with initial condition (x0, y0) = (1, 1)?What if initial conditions are (x0, y0) = (10, 10)?

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Page 15: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

Extra Credit Problem. (20 points total) Consider the following slightly modified versionof (8):

dx

dt= −7

8

(1 − x

7

)x +

15

32xy

dy

dt=(

1 − y

4

)y − 1

3xy

(12)

and again set

H(x, y) = x · y ·(x

6+y

4− 1

). (13)

a) (10 points) Show that the function H is a Lyapunov function for (12), when the systemis restricted to x ≥ 0, y ≥ 0. It may help to rewrite the first equation as

dx

dt=

[−(

1− x

6

)x +

1

2xy

]− x

8

[y4

+x

3− 1

](14)

b) (10 points) On a scratch page, sketch the trajectories in the phase plane for this system.What is the ultimate fate of a system with initial conditions (x0, y0) = (1, 1)?

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Page 16: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

Scratch

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Page 17: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

Scratch

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Page 18: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

Scratch

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Page 19: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

Trigonometric Formulas

Basic formulas

cos(α± β) = cos(α) cos(β) ∓ sin(α) sin(β)

sin(α± β) = sin(α) cos(β) ± cos(α) sin(β)

From these follow the sum-to-product formulas:

cos(α− β) + cos(α+ β) = 2 cos(α) cos(β)

cos(α− β) − cos(α+ β) = 2 sin(α) sin(β)

sin(α+ β) + sin(α− β) = 2 sin(α) cos(β)

sin(α+ β) − sin(α− β) = 2 cos(α) sin(β)

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Page 20: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

Laplace Transforms

Definition

F (s) =

∫ ∞0

e−stf(t) dt

Table of Common Transforms

t-domain function s-domain function

δ(t), δa(t) 1, e−as

u(t), ua(t)1s ,

e−as

s

t u(t) 1s2

tn u(t) n!sn+1

eat u(t) 1s− a

sin(ωt)u(t) ωs2 +ω2

cos(ωt)u(t) ss2 +ω2

eat sin(ωt)u(t) ω(s−a)2 +ω2

eat cos(ωt)u(t) (s−a)(s−a)2 +ω2

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Page 21: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

Properties of the Laplace Transform

Propertyt – domain function s – domain function

Name

f(t) F (s) =∫∞

0 e−stf(t) dt

Linearity α f(t) + β g(t) αF (s) + β G(s)

dfdt

sF (s) − f(0)∫f 1

s F (s)

t f(t) −dFds

1t f(t) −

∫F

Time Shiftf(t − a) eas F (s)

Theorem

Phase Shifte−at f(t) F (s+ a)

Theorem

Time Scalef(at) 1

|a| F ( sa)Theorem

Convolution(f ∗ g)(t) F (s)G(s)

Theorem

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Page 22: Final Exam - Penn MathFinal Exam December 20, 2011 Math 420 - Ordinary Di erential Equations No credit will be given for answers without mathematical or logical justi cation. Simplify

Convolutions

Definition

(f ∗ g)(t) =

∫ ∞−∞

f(t − u) g(u) du

Property Expression

Symmetry f ∗ g = g ∗ f

Associativity f ∗ (g ∗ h) = (f ∗ g) ∗ h

Linearity f ∗ (αg + βh) = α(f ∗ g) + β(f ∗ h)

Differentiation ddt(f ∗ g) = df

dt ∗ g = f ∗ dgdt

Antidifferentiation∫

(f ∗ g) =∫f ∗ g = f ∗

∫g

The Identity f ∗ δ = f

Integration f ∗ u =∫ t−∞ f(v) dv

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