Download Math 133A, Fall 2015, Term Test II Study Sheet

Math 133A, Fall 2015, Term Test II Study Sheet Answers
Ordinary Differential Equations
Date:
Time:
Lecture Section:
Instructor:
Thursday, November 19
1:30-2:45 p.m.
002
Matthew Johnston
Surname (Family Name):
Given Name:
SJSU Student ID Number:
Instructions
1. Fill out this cover page completely.
FOR EXAMINERS’ USE ONLY
2. Answer questions in the space provided, using scratch paper for rough
work.
3. Show all the work required to obtain
your answers.
4. No calculators are permitted but you
may consult a one page hand-written
cheat sheet.
Page
Mark
2
/∞
3
/∞
Total
/∞
Math 133A Term Test II Study Sheet
Page 2 of 7
Name:
1. Short Answer:
(a) Convert the following higher-order differential equations in a system of first-order
differential equations (all derivatives with respect to x):
(i) y 00 + 5y 0 + 8y
 = sin(x)
 y1 (x) = y(x), y2 (x) = y 0 (x)
y 0 = y2
Solution:
 10
y2 = −8y1 − 5y2 + sin(x)
(ii) y 00 − 4xy = 
ex
 y1 (x) = y(x), y2 (x) = y 0 (x)
y 0 = y2
Solution:
 10
y2 = 4xy1 + ex
(iii) (x + 1)y 00 +
xy 0 + x2 y = 1
0
 y1 (x) = y(x), y2 (x) = y (x)
y 0 = y2
Solution:
 10
x2
x
y2 = − x+1
y1 − x+1
y2 + 1
(iv) y (4) + y 00 = 
0





Solution:





y1 (x) = y(x), y2 (x) = y 0 (x), y3 (x) = y 00 (x), y4 (x) = y 000 (x)
y10 = y2
y20 = y3
y30 = y4
y40 = −y3
(b) Determine the value of c for which the following differential equations, corresponding to a mechanical pendulum/spring system obeying Newton’s second law, are
critically damped :
(i) 4x00 + cx0 + x = 0
Solution: c2 = 4mk =⇒ c2 = 4(4)(1) =⇒ c = 4
(ii) 9x00 + cx0 + 4x = 0
Solution: c = 12
(iii) x00 + cx0 + 3x = 0√
Solution: c = 2 3
(iv) 6x00 + cx0 + 83 x = 0
Solution: c = 8
Math 133A Term Test II Study Sheet
Page 3 of 7
Name:
(c) Supposing that a system of differential equations has the given general solution,
determine the fundamental matrix Φ(t) with the property Φ(0) = I.
3
x(t)
1
−2t
(i)
= C1
e + C2
e2t
y(t)
1 −2
1 2e−2t + 3e2t 3e−2t − 3e2t
Solution: Φ(t) =
5 2e−2t − 2e2t 3e−2t + 2e2t
x(t)
1
1
1
t
(ii)
= C1
e + C2
t+
et
y(t)
2
2
0
(t + 1)et − 12 tet
Solution: Φ(t) =
2tet
(1 − t)et
3
0
x(t)
−2t
cos(3t) −
sin(3t)
(iii)
= C1 e
1 1 y(t)
3
0
−2t
sin(3t) +
cos(3t)
.
+C2 e
1
1
−2t
e (cos(3t) − sin(3t))
3e−2t sin(3t)
Solution: Φ(t) =
− 23 e−2t sin(3t)
e−2t (cos(3t) + sin(3t))
2. Linear Systems and Vector Fields:
For the initial value problems, sketch the vector field and then find the solution. Clearly
label the nullclines and the identify the direction of flow (%, &, ., -) in each region.

 x0 = −x + 2y
y 0 = −x − 4y
(a)

x(0) = 3, y(0) = −3
x(t) = 3e−3t
Solution:
y(t) = −3e−3t

 x0 = x − 2y
y 0 = −x + 2y
(b)

x(0) = 0, y(0) = 3
x(t) = −2e3t + 2
Solution:
y(t) = 2e3t + 1

 x0 = −3x − 9y
y 0 = x + 3y
(c)

x(0) = 1, y(0) = 0
x(t) = −3t + 1
Solution:
y(t) = t
Math 133A Term Test II Study Sheet
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
 x0 = −x + 2y
y 0 = −x + y
(d)

x(0) = 2, y(0) = −1
x(t) = 2 cos(t) − 4 sin(t)
Solution:
y(t) = − cos(t) − 3 sin(t)
√

3y
 x0 = 2x + √
0
(e)
y = −3x
√ + 2y √

x(0) = ( 2, y(0) = 2
√ √
x(t) = 2e√ 2t (cos(3t) + sin(3t))
√
Solution:
y(t) = 2e 2t (cos(3t) − sin(3t))

 x0 = −x + y
y 0 = −4x − 5y
(f)

x(0) = 0, y(0) = −2
x(t) = −2te−3t
Solution:
y(t) = (4t − 2)e−3t
3. Second-Order Differential Equations:
(a) Determine the trial function yp (x) for the following second-order differential equations. Do not attempt to solve for the constants.
(i) y 00 + 2y 0 − 3y = e−x + ex
Solution: yc (x) = C1 ex + C2 e−3x =⇒ yp (x) = Ae−x + Bxex
(ii) y 00 + 2y 0 − 3y = xe−x
Solution: yp (x) = Axe−x + Be−x
(iii) y 00 + 2y 0 − 3y = 2xe−3x
Solution: yp (x) = Ax2 e−3x + Bxe−3x
(iv) y 00 + 4y 0 + 5y = e−2x
Solution: yc (x) = C1 e−2x cos(x) + C2 e−2x sin(x) =⇒ yp (x) = Ae−2x
(v) y 00 + 4y 0 + 5y = sin(x) + cos(x)
Solution: yp (x) = A sin(x) + B cos(x)
(vi) y 00 + 4y 0 + 5y = xe−2x sin(x)
Solution: yp (x) = Ax2 e−2x cos(x)+Bx2 e−2x sin(x)+Cxe−2x cos(x)+Dxe−2x sin(x)
(b) Solve the following initial value problems:
 00
 2y + 5y 0 − 3y = 10 sin(x),
y(0) = 1,
(i)
 0
y (0) = 0
1
Solution: y(x) = 2e 2 t − cos(x) − sin(x)
 00
 y − 2y 0 + 5y = 5x − 2,
y(0) = 1,
(ii)
 0
y (0) = −2
Solution: y(x) = ex cos(2x) − 2ex sin(2x) + x
Math 133A Term Test II Study Sheet
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 00
 y + 3y 0 + 2y = 6ex ,
y(0) = 1,
(iii)
 0
y (0) = 0
Solution: y(x) = e−2x − e−x + ex
 00
 4y + 9y = 12 sin 23 x ,
y(0) = 0,
(iv)
 0
y (0) = 0
Solution: y(x) = 23 sin 23 x − x cos
Name:
3
x
2
4. Applications: (Mechanical Systems)
(a) A 1 kg mass is attached to the end of a pendulum which has a restoring constant
of 5 N/m and damping constant of 4 N/(m/s). Suppose the system is driven by an
external force of f (t) = 8 sin(t) N. Formulate a second-order differential equation
to describe the system, and solve it. Rewrite the forcing function and periodic
portion of the solution x(t) in the form R cos(ω0 t + δ). Comment on the long-term
behavior of the solution, whether it depends upon the initial conditions, and the
relationshipbetween the forcing and long-term response.
x00 + 4x0 + 5x = 8 sin(t)


 x(t) = C e−2t cos(t) + C e−2t sin(t) − cos(t) + sin(t)
1
2
√
Solution:
π
),
−
cos(t)
+ sin(t) = 2 cos(t − 2.3562) √
f
(t)
=
8
cos(t
−

2


Magnitude of forcing (8) is ∼ 5 times large than response ( 2)
(b) An 8 kg mass is attached to the end of a spring which has a restoring constant of
1 N/m and damping constant of 6 N/(m/s). Suppose the system is driven by an
external force of f (t) = (5 sin( 12 t) + 5 cos( 12 t)) N. Formulate a second-order differential equation to describe the system, and solve it. Rewrite the forcing function and
periodic portion of the solution x(t) in the form R cos(ω0 t + δ). Comment on the
long-term behavior of the solution, whether it depends upon the initial conditions,
and the relationship
between the forcing and long-term
response.

1
1
00
0
8x
+
6x
+
x
=
5
cos(
t)
+
sin(
t)

2
2


1
1
−1t
−1t
x(t) = C√
1 e 2 + C2 e 4 − 2 cos( 2 t) + sin( 2 t)
√
Solution:
1
1
1

f
(t)
=
2
5
cos(t
−
0.7854),
−2
cos(
t)
+
sin(
t)
=
5 cos(√
t − 2.6779)

2
2
2
√

Magnitude of forcing ( 5) is twice large than response (2 5)
Math 133A Term Test II Study Sheet
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5. Applications: (Chemical Reactions)
Consider the reversible chemical reaction X Y occuring in a tank subject to continuous inflow of X and outflow of both X and Y . Suppose the process can be modeled by
the following system of first-order differential equations:
(
x0 = −5x + 2y + 7
x(0) = 1
(1)
y 0 = 4x − 3y
y(0) = 0.
(a) Solve the non-homogeneous
system of differential equations (1) for x(t) and y(t).
x(t) = −2e−t + 3
Solution:
y(t) = −4e−t + 4
(b) Describe the long-term behavior of the chemical species X and Y .
Solution: As t → 0, x(t) → 3 and y(t) → 4. That is, the concentrations equilibriate.
Math 133A Term Test II Study Sheet
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Name:
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