Download Final exam review questions and answers

Final Exam MATH 3321,
Sample Questions.
Fall 2016
1. y2 = Cx3 − 3 is the general solution of a differential equation. Find the equation.
Answer: y0 =
3y2 + 9
2xy
2. y = C1 x2 + C2 is the general solution of a differential equation. Find the equation.
Answer: xy00 − y0 = 0
3. Identify and find the general solution of each of the following first order differential equations.
(a) xy0 = 5x3 y1/2 − 4y
y1/2 =
Answer: Bernoulli,
(b) x y0 + 3 y =
cos 2x
.
x2
Answer: Linear,
(c) x2
1 3
C
x + 2
2
x
y=
sin 2x
C
+ 3
2x3
x
dy
= x2 + xy + y2
dx
Answer: Homogeneous,
(d) 2y0 =
y = x tan(ln x + C)
y2 + 3
.
4y + xy
y2 = C(4 + x) − 3
Answer: Separable,
(e) x2 y0 = 4x3 y3 + xy
Answer: Bernoulli,
y2 =
x2
C − 2x4
(f) x3 y0 = x2 y + 2x3 ey/x
Answer: Homogeneous,
y = x ln
1
C − ln x2
4. Given the one-parameter family y3 = Cx2 + 4.
(a) Find the differential equation for the family.
(b) Find the differential equation for the family of orthogonal trajectories.
(c) Find the family of orthogonal trajectories.
Answer: (a) y0 =
2y3 − 8
.
3xy2
(b) y0 = −
3xy2
.
2y3 − 8
(c) 3x2 y + 2y3 + Cy + 16 = 0.
5. A certain radioactive material is decaying at a rate proportional to the amount present. If a
sample of 100 grams of the material was present initially and after 3 hours the sample lost
30% of its mass, find:
1
(a) An expression for the mass A(t) of the material remaining at any time t.
(b) The mass of the material after 8 hours.
(c) The half-life of the material.
7
Answer: (a) A(t) = 100
10
t/3
(b) A(8) = 100
7
10
8/3
(c) t =
−3 ln 2
ln 7/10
6. Scientists observed that a colony of penguins on a remote Antarctic island obeys the population
growth law. There were 1000 penguins in the initial population and there were 3000 penguins
4 years later.
(a) Give an expression for the number P (t) of penguins at any time t.
(b) How many penguins will there be after 6 years?
(c) How long will it take for the number of penguins to quadruple?
t/4
t
Answer: (a) P (t) = 1000e 4 ln(3) = 1000 (3)
(c) t =
3/2
(b) P (6) = 1000 (3)
.
4 ln(4)
years.
ln(3)
7. A disease is infecting a herd of 1000 cows. Let P (t) be the number of sick cows t days after
the outbreak. Suppose that 50 cows had the disease initially, and suppose that the disease is
spreading at a rate proportional to the product of the time elapsed and the number of cows
who do not have the disease.
(a) Give the mathematical model (initial-value problem) for P .
(b) Find the general solution of the differential equation in (a).
(c) Find the particular solution that satisfies the initial condition.
Answer: (a)
dP
= kt(1000 − P ), P (0) = 50
dt
(c) P (t) = 1000 − 950e−kt
2
(b) P (t) = 1000 − Ce−kt
2
/2
/2
8. Determine a fundamental set of solutions of y00 − 2 y0 − 15 y = 0.
Answer:
e5x, e−3x
9. Find the general solution of y00 + 6y0 + 9y = 0.
Answer: y = C1 e−3x + C2 xe−3x.
10. Find the solution of the initial-value problem y00 − 7y0 + 12y = 0, y(0) = 3, y0 (0) = 0.
Answer: y = 12e3x − 9e4x.
11. The function y = −2 e−3x sin 2x is a solution of a second order, linear, homogeneous differential equation with constant coefficients. What is the equation?
Answer: y00 + 6y0 + 13y = 0
2
12. The function
y = −2 e3x + 4 xe3x
is a solution of a second order, linear, homogeneous differential equation with constant coefficients. What is the equation?
Answer: y00 − 6y0 + 9y = 0
13. Find a particular solution of y00 − 6y0 + 8y = 4e4x .
Answer: z = 2xe4x.
14. Give the form of a particular solution of the nonhomogeneous differential equation
y00 − 8 y0 + 16 y = 2 e4x + 3 cos 4x − 2x + 1.
Answer: z = Ax2 e4x + B cos 4x + C sin 4x + Dx + E.
15. Given the differential equation
y00 − 4 y0 + 4 y =
e2x
x
(a) Give the general solution of the reduced equation.
(b) Find a particular solution of the nonhomogeneous equation.
Answer: (a) y = C1 e2x + C2 xe2x .
(b) z = −xe2x + x e2x ln x or z = x e2x ln x.
4
6
4
16. Find the general solution of y00 − y0 + 2 y = 2 . HINT: The reduced equation has solutions
x
x
x
of the form y = xr .
Answer: y = C1 x2 + C2 x3 +
2
3
17. Find a particular solution of y00 + 4y = 2 tan 2x.
Answer: y = C1 cos 2x + C2 sin 2x −
1
2
cos 2x ln [sec 2x + tan 2x]
18. The general solution of
y(4) − 6 y000 + 17 y00 − 28 y0 + 20 y = 0
is: (HINT: 2 is a root of the characteristic polynomial)
Answer: y = C1 e2x + C2 xe2x + C3 ex cos 2x + C4 ex sin 2x.
19. Give the form of a particular solution of
y(4) + 5 y00 − 36 y = 2x + 3e2x − 2 sin 3x
Answer: z = Ax + B + Cxe2x + Dx cos 3x + Ex sin 3x.
3
20. Find the Laplace transform of the solution of the initial-value problem
y00 − y0 − 6 y = 3; y(0) = 5, y0 (0) = 0
Answer: Y =
3
5s − 5
+
.
s(s2 − s − 6) s2 − s − 6
3
4s + 3
+ 2
.
2
s
s +4
21. Find f(x) = L−1 [F (s)] if F (s) =
Answer: f(x) = 3x + 4 cos 2x +
3
2
sin 2x
22. Find L[f(x)] if
f(x) =
Answer: F (s) =
(
x2 + 2x
0≤x<4
x
x≥4
2
2
2
1
1
+ 2 − e−4s 3 − 9e−4s 2 − 20e−4s .
s3
s
s
s
s
5
3
1
1
s+1
− − 2 e−3s 2 + 3 e−3s + 2e−3s 2
. Find L−1 [F (s)] = f(x).
s2
s
s
s
s + π2


5x,
0≤x<3
Answer: f(x) =
.
2
 3x + 6 − 2 cos πx − sin πx,
x≥3
π
23. F (s) =
24. Given the initial-value problem
y0 − 4 y = 2 e−2x, y(0) = 3.
(a) Find the Laplace transform of the solution.
(b) Find the solution by finding the inverse Laplace transform of your answer to (a).
Answer: (a) Y =
2
3
+
.
(s + 2)(s − 4) s − 4
(b) y =
10 4x
3 e
− 13 e−2x .
25. Given the system of equations
x +2y
2x +5y
−2x −2y
−z
−4z
−2z
=1
=3
=0
(a) Write the augmented matrix for the system.
(b) Reduce the augmented matrix to row-echelon form.
(c) Give the solution set of the system.

1

Answer: (a)  2
−2
(c) x1 = −1 − 3a,
2 −1
5 −4
−2 −2
x2 = 1 + 2a,

1

3 .
0

1 2 −1

(b)  0 1 −2
0 0 0
x3 = a,
a arbitrary.
4

1

1 .
0
26. Given the system of equations
x −2y
x −y
y
+kz
−2z
=1
=3
=k
The value(s) of k, if any, such that the system has infinitely many solutions is (are):
Answer: No values of k.

λ
0

27. Find the values of λ, (if any) such that A =  0
1
−2 −1

3

λ  is nonsingular.
−5
Answer: λ 6= 2, 3

1
2

28. The matrix A =  2
5
−2 −4
Answer: A−1

−7 −2

= 4
1
2
0

0

−3  is nonsingular. Find A−1 .
1

−6

3 .
1
29. The system of equations
2x − y + 3z = 4
y + 2z = −2
x+z = 1
has a unique solution. Find y.
Answer: y = −2
30. Determine whether the vectors
v1 = ( 1, −3, 2 ), v2 = ( 0, −2, −2 ), v3 = ( 1, −5, 0 ), v4 = ( 0, 4, 4 )
are linearly dependent or linearly independent. If they are linearly dependent, find the maximal
number of independent vectors.
Answer: Linearly dependent; the maximum number of independent vectors is 2.

2
2

31. Find the eigenvalues and eigenvectors of  2 −1
−2 −1


2


Answer: λ1 = 6,  1  ;
−1

1


λ2 = −2,  −2  ;
0

5

−6

−3 . Hint: 6 is an eigenvalue.
1


0
 
λ3 = −2,  3 
1

−2
1

32. Find the eigenvalues and eigenvectors of  3 −3
3 −1


0
 
Answer: λ1 = 1,  1  ;
1

−1

4 . Hint: 1 is an eigenvalue.
2


1


λ2 = λ3 = −2,  −1 
−1




1 1 2
2




33. Find the solution of the initial-value problem x0 =  0 2 2 x, x(0) =  0 . HINT:
−1 1 3
1
1 is an eigenvalue.




1
0




Answer: x(t) = 2e2t  1  − et  2 .
0
−1
2 1
−5 4
0
34. Find a fundamental set of solutions of x =
Answer:
(
"
1
1
e3t cos 2t
!
− sin 2t
2 −1
1
4
0
35. Find the general solution of x =
1
−1
Answer: x(t) = C1 e3t
!
0
2
"
!#
!

x.
"
0
2
, e3t cos 2t
!
+ sin 2t
1
1
!#)
.
x.
−2
1
+ C2 e3t
!
!
+ te3t
1
−1
!#

1

−1 x. HINT: 2 is a root of the characteristic
1
1
1

36. Find the general solution of x0 =  2
1
0 −1
polynomial.




 
−3
0




 
Answer: x(t) = C1 e−t  4  + C2 e2t  1  + C3 e2t 
2
−1


5 4 2


37. Find a fundamental set of solutions of x0 =  4 5 2 x.
2 2 2
characteristic polynomial.
 

 



2 
1
0


 10t  
 t
t
Answer:
e  1 , e  0 , e  2  .



1 
−2
−2
6



1
0



0  + te2t  1 .
1
−1
HINT: 10 is a root of the