Math 211 Fall 2007 Exam 2
Instructor: S. Cautis
Thursday, November 8, 2007
Instructions: This is a closed book, closed notes exam. Use of calculators is
not permitted. You have one hour and fifteen minutes. Do all 7 problems.
Please do all your work on the paper provided. You must show your work
to receive full credit on a problem. An answer with no supporting work or
explanation will receive little to no credit.
Please print you name clearly here.
Print name:
Upon finishing please sign the pledge below:
On my honor I have neither given nor received any aid on this exam.
Grader’s use only:
1.
/10
2.
/10
3.
/20
4.
/17
5.
/20
6.
/15
7.
/8
1. Find the general solutions of the following differential equations:
(a) [5 points] y ′′ + 2y ′ + 5y = 0
Characteristic polynomial is λ2 + 2λ + 5 = 0 which has roots λ = −1 ± 2i.
So the general solution is e−t (A cos(2t) + B sin(2t)).
(b) [5 points] y ′′ + 11y ′ = 0
Characteristic polynomial is λ2 + 11λ = 0 which has roots λ = 0 and
λ = −11. So the general solution is A + Be−11t .
2. [10 points]
Find a particular solution to y ′′ + 3y ′ + 2y = 1 + et .
Look for a particular solution of the form yp = A + Bet . Then yp′ = Bet
and yp′′ = Bet . Substituting back gives Bet + 3Bet + 2A + 2Bet = 1 + et
so A = 1/2 and B = 1/6. So a solution is 1/2 + 1/6et .
3. [20 points]
′
Find a particular solution to y ′′ − yt − 4t2 y = t2 using the “variation
of parameters” method. (you may use that the associated homogeneous
′
2
2
equation y ′′ − yt − 4t2 y = 0 has solutions et and e−t )
2
2
We look for a solution of the form y = uet + ve−t . Then
2
2
2
2
y ′ = u′ et + 2tuet + v ′ e−t − 2tve−t
2
2
which we simplify to y ′ = 2tuet − 2tve−t by requiring that
2
2
u′ et + v ′ e−t = 0
2
2
2
2
2
2
Next y ′′ = 2u′ tet + u(2et + 4t2 et ) − 2v ′ (te−t ) − v(2e−t − 4t2 e−t ).
Substituting back and simplifying we get
2
2
2u′ tet − 2v ′ te−t = t2
or
2
2
u′ et − v ′ e−t = t/2
Combining the two equations involving u′ and v ′ we can solve to get
2
2
u′ = te−t /4 and v ′ = −tet /4. We can integrate by substituting s = t2
2
2
to get u = −e−t /8 and v = −et /8. So a solution is
2
2
2
2
(−e−t /8) · et + (−et /8) · e−t = −1/8 − 1/8 = −1/4
4. Recall L(eat ) =
1
s−a ,
s
s2 +a2
L(cos(at)) =
and L(sin(at)) =
a
s2 +a2 .
(a)[5 points] State the definition of the Laplace transform L(f ) of f (t).
L(f ) =
∞
Z
f (t)e−ts dt
0
(b)[7 points] Find L−1
2
s2 +2s+2
.
Notice that s2 + 2s + 2 = (s + 1)2 + 1. Since L(sin(t)) =
1
that L(e−t sin(t)) = (s+1)
2 +1 and hence
L−1
2
s2 + 2s + 2
1
s2 +1
it follows
= 2e−t sin(t)
(c)[3 points] Give the definition of the convolution f (t) ∗ g(t) of two functions.
f (t) ∗ g(t) =
Z
t
f (u)g(t − u)du
0
(d)[2 points] Compute L(4 cos(t) ∗ et ∗ t−1 ).
By the property of the Laplace transform of convolutions we have
L(4 cos(t) ∗ et ∗ t−1 ) = L(4 cos(t))L(et )L(t−1 ) =
4s
1
·
· L(t−1 )
+1 s−1
s2
5. [20 points]
Solve for the Laplace L(y) of y given that y is a solution of y ′′ +ty ′ +2y = 0
with y(0) = y ′ (0) = 0.
We apply L to the left side. Notice that L(y ′′ ) = s2 L(y) since y(0) =
y ′ (0) = 0. Also
L(ty ′ ) = −
d
d
(L(y ′ )) = − (sL(y)) = −L(y) − sL(y)′
ds
ds
Thus, if we let L(y) = Y (s) we get
s2 Y − Y − sY ′ + 2Y = 0
so (s2 + 1)Y = sY ′ which means
1
Y′
=s+
Y
s
Integrating the left side gives
ln(Y ) =
or
2
Y = es
s2
+ ln(s)
2
/2+ln(s)
2
= ses
/2
6. (a)[12 points] Solve the following system of linear equations and write the
answer in parametric form:
x1 − 3x2 + 2x3 + 4x4 + x5 = 0
−2x1 + 6x2 − 3x3 − 5x4 = 5
x1 − 3x2 + x3 + 2x4 = −3
Adding twice the first row to the second and subtracting the first row from
the third gives
x1 − 3x2 + 2x3 + 4x4 + x5 = 0
x3 + 3x4 + 2x5 = 5
−x3 − 2x4 − x5 = −3
Adding the third row to the second and multiplying the third row by −1
gives
x1 − 3x2 + 2x3 + 4x4 + x5 = 0
x4 + x5 = 2
x3 + 2x4 + x5 = 3
So if we let x5 = t, then x4 = 2 − t and x3 = 3 − 2x4 − x5 = 3 − 4 + 2t − t =
t − 1. Then we let x2 = s and
x1 = 3x2 − 2x3 − 4x4 − x5 = 3s − 2t + 2 − 8 + 4t − t = 3s + t − 6
So the solution is parametric form is

  
 
 
3s + t − 6
−6
1
3

 0
0
1
s

  
 
 
 t − 1  = −1 + t  1  + s 0

  
 
 
 2−t   2 
−1
0
t
0
1
0
(b) [3 points] What does the solution set look like?
The solution set is a plane in R5 which passes through the point (−6, 0, −1, 2, 0)
and is parallel to the vectors (1, 0, 1, −1, 1) and (3, 1, 0, 0, 0).
L
L
2
1
2
3
2
7. Given the linear
 maps
 shown R −−→ R −−→ R where L1 is given by the
1 2
−1 2 4
matrix A = 0 1 and L2 given by B =
.
0 3 1
1 0
(a)[5 points] Calculate the matrix product AB. We



1 2 −1
−1 2 4
AB = 0 1
= 0
0 3 1
1 0
−1
have

8 6
3 1
2 4
(b)[3 points] The composition L2 ◦ L1 : R2 → R2 of the two linear maps
above is given by which matrix, AB or BA? (choose one and briefly
explain why)
L2 ◦ L1 is represented by BA. To see this note that if v is a vector in
R2 then L2 ◦ L1 (v) = L2 (L1 (v)) = B(A(v)) = (BA)v so the matrix
representing L2 ◦ L1 must be BA.
Moreover, AB is a 3 × 3 matrix so it cannot represent the map L2 ◦ L1 :
R2 → R2 .