Dr. Gabardo (C01), Dr. Yapalparvi (C02), Dr. Kovarik (C03)
SOLUTIONS
1. The general solution of Cauchy-Euler equationx2y00−6y= 0 on the interval 0< x <∞
is given by
(A) y=C1x√6+C2x−√6
→ (B) y=C1x3+C2 1 x2
(C) y=C1e3x+C2e−2x
(D) y=C1x3+C2x3lnx
(E) y=C1x2+C2 1 x3
Solution. The Cauchy-Euler auxiliary equation ism(m−1)−6 = 0 or (m−3) (m+2) = 0 with rootsm= 3 andm=−2. The general solution is thus
y=C1x3+C2x−2=C1x3+C2 1 x2
2. A particular solution of the Cauchy-Euler equationx2y00+x y0−y=x2 on the interval 0< x <∞is given by
(A) yp= 1
4x 4 ex
(B) yp=x2+x
(C) yp=x2+ 1
x
(D) yp =x3
→(E) yp =1
3x 2
Solution. The Cauchy-Euler auxiliary equation for the associated homogenerous equa-tion is m(m−1) +m−1 = 0 or (m−1) (m+ 1) = 0 with rootsm= 1 andm=−1. Choosingy1(x) =xandy2(x) =x−1, the associated Wronskian is
W(y1, y2)(x) =
¯ ¯ ¯ ¯y1y0 y2
1 y02
¯ ¯ ¯ ¯=
¯ ¯ ¯ ¯x x
−1
1 −x−2
¯ ¯ ¯
¯=−2x−1.
When written in standard form, the right-hand side of the DE is f(x) = 1 and the variation of parameters method yields the particular solution
yp(x) =−
µZ
f(x)y2(x) W(y1, y2)(x)dx
¶
y1(x) +
µZ
f(x)y1(x) W(y1, y2)(x)dx
¶
y2(x)
=−
µZ
x−1
−2x−1dx
¶
x+
µZ
x
−2x−1dx
¶
x−1
=
µZ
1 2dx
¶
x−1
2
µZ
x2dx
¶
x−1=³x 2
´
x−
µ
x3 6
¶
x−1
= 1 3x
3. The displacement from the equilibrium position of an undamped spring-mass system subject to an external force f(t) = 2 sin(3t) is given by a function x(t) satisfying the differential equation
x00(t) +k x(t) = 2 sin(3t).
For which value ofkwill “pure resonance” occur?
→(A) k= 9
(B) k= 2
(C) k= 3
(D) k= 16
(E) k= 4
Solution. The auxiliary equation for the associated homogenerous equation is given by m2+k= 0. Fork >0, the complementary solution will have the form
xc(t) =C1 cos(ωt) +C2 sin(ωt)
whereω=√kand resonance occurs when the oscillations of the complementary solution have the same period as the oscillations of the driving force. This will happen whenω= 3 ork=ω2= 9.
4. The chargeq(t) on the capacitor in an LRC circuit satisfies the differential equation
q00(t) + 2q0(t) + 1601q(t) = 0.
If the initial charge is q(0) = 20 and the initial current is i(0) = q0(0) =−20, find the
first timet0>0 at whichq(t0) = 0.
(A) t0= π 20
(B) t0=π 4
(C) t0= π 15
→(D) t0= π 80
(E) t0= π 160
Solution. The auxiliary equation for the associated homogenerous equation is given by m2+ 2m+ 1602 = 0 or (m+ 1)2+ (40)2= 0. The roots are thusm=−1±40iand the general solution is given by
q(t) =C1e−tcos(40t) +C2e−tsin(40t)
The initial conditionq(0) = 20 yieldsC1= 20 and thus
q(t) = 20e−tcos(40t) +C2e−t sin(40t)
and
q0(t) =−20e−t cos(40t)−800e−tsin(40t)−C2e−tsin(40t) +C2e−t40 cos(40t)
The initial conditionq0(0) =−20 yieldsC2= 0. Thus
q(t) = 20e−t cos(40t)
and the charge on the capacitor first become zero when 40t0=π
5. A beam of lengthL=πis embedded at both ends. Find the deflectiony(x) of the beam if a non-uniform load given byw(x) = sinxis distributed along its length andE I= 1. The functiony(x) is the solution of the boundary-value problem
d4y
dx4 = sinx, y(0) =y
0(0) =y(π) =y0(π) = 0.
→(A) y(x) =x(x−π) π + sinx
(B) y(x) =x2(x−π)
π + cosx
(C) y(x) =x(x−π)2
π −sinx
(D) y(x) = x2(x−π)2
π + sinx
(E) y(x) = x(x−π)
π −cosx
Solution. The auxiliary equation for the associated homogenerous equation ism4 = 0 and the complementary solution is
yc(x) =C0+C1x+C2x2+C3x3.
The general form of a particular solution ifyp(x) =A cosx+B sinx. we have
y00
p(x) =Acosx+B sinx= sinx
which shows thatA= 0 and B= 1. The general solution is thus
y(x) =C0+C1x+C2x2+C3x3+ sinx.
and
y0(x) =C1+ 2C2x+ 3C3x2+ cosx.
The boundary conditions yield the equations
C0= 0, C1+ 1 = 0, C0+C1π+C2π2+C3π3= 0, C1+ 2C2π+ 3C3π2−1 = 0
which have solutions
C0= 0, C1=−1, C2= 1
π, C3= 0. Hence,
y(x) =−x+1 πx
2+ sinx=x(x−π)
π + sinx.
A= a 1 are given by:
(A) −1 +√2−ab, −1−√2−ab
(B) −1 +√ab,−1−√ab
→(C) 1 +√ab,1−√ab
(D) −1 +√2−ab,−1−√2−ab
(E) −1,1
Solution. We have
det(A−λ I) =
¯ ¯ ¯
¯1−aλ 1−b λ ¯ ¯ ¯
¯= (1−λ)2−a b
The solutions of det(A−λ I) = 0 are thus given by
(1−λ) =±√ab or λ= 1±√ab.
7. The eigenvectors of the matrix
A=
·
−1 2 72
−1 2 −12
¸
are given by:
(A)
·
1 2
¸
,
·
2
−1
¸
(B)
· √
2 1
¸
,
·
−√2 2
¸
(C)
· 1
2 1 4
¸
,
·
−1 1 4
¸
(D)
· √
2 0
¸
,
·
−√2 0
¸
→(E) None of the above.
Solution. We have
det(A−λ I) =
¯ ¯ ¯ ¯−
1
2−λ 72
−1
2 −12−λ
¯ ¯ ¯
¯= (12 −λ)2+74
and the roots of the characteristic polynomials areλ=−1 2±
√
7
2 iIfλ=−12+
√
7 2 i, the corresponding eigenvector is obtained by solving
"
−√7 2 i 72
−1
2 −
√
7 2 i
# ·
x y
¸
=
·
0 0
¸
which yields the eigenvector
·√
7 i
¸
. Similarly, If λ = −1 2 −
√
7
2 i , the corresponding eigenvector is obtained by solving
"√
7 2 i 72
−1 2
√
7 2 i
# ·
x y
¸
=
·
0 0
8. ComputeA4if
A=
·
a 1 1 0
¸
.
(A) A4=
·
4a 4 4 4a
¸
(B) A4=
·
a4 1 1 a4
¸
(C) A4=
·
a4 4a3 4a3 a4
¸
→(D) A4=
·
a4+ 3a2+ 1 a3+ 2a a3+ 2a a2+ 1
¸
(E) A4=
·
a4+ 2a2 a3+ 2a a3+ 2a a2+ 1
¸
Solution. We have
A2=
·
a 1 1 0
¸ ·
a 1 1 0
¸
=
·
a2+ 1 a a 1
¸
and
A4=A2A2=
·
a2+ 1 a a 1
¸ ·
a2+ 1 a a 1
¸
=
·
(a2+ 1)2+a2 a3+ 2a a3+ 2a a2+ 1
¸
9. A realn×nmatrixAis orthogonal if
(A) ATA=AAT
→(B) ATA=I
(C) The columns ofAare all unit vectors
(D) The columns ofAare orthogonal to the rows of A
(E) The columns ofAare orthogonal to each other
10. Suppose thatAis a real,n×nsymmetric matrix. Which of the following statements is
incorrect?
(A) All the eigenvalues ofAare real
(B) Ahasnlinearly independent eigenvectors
→(C) Ais always non-singular
(D) Acould have repeated eigenvalues
(E) Ais diagonalizable
11. Consider the matrixA=
·
0 9 1 0
¸
.Then a matrixP which makesP−1AP diagonal is
→(A) P=
·
3 3 1 −1
¸
(B) non-existent
(C) P=
·
3 0 0 −3
¸
(D) P =
·1
6 12 1 6 −12
¸
(E) P =
·
0 1 1 9 0
¸
Solution. We have
det(A−λ I) =
¯ ¯ ¯
¯−1λ −9λ ¯ ¯ ¯
¯=λ2−9 = (λ−3) (λ+ 3)
and the roots of the characteristic polynomial areλ=±3. Ifλ= 3 , the corresponding eigenvector is obtained by solving
·
−3 9 1 −3
¸ ·
x y
¸
=
·
0 0
¸
orx−3y= 0 which yields the eigenvector
·
3 1
¸
. Ifλ=−3 , the corresponding eigenvector
is obtained by solving · 3 9 1 3
¸ ·
x y
¸
=
·
0 0
¸
orx+ 3y= 0 which yields the eigenvector
·
3
−1
¸
. Thus we can take
P=
·
3 3 1 −1
¸
.
With this choice, we will have
P−1AP =
·
3 0 0 −3
¸
.
12. Consider the symmetric matrix A =
·
0 −2
−2 3
¸
. Then an orthogonal matrix P which diagonalizesAis
(A) P=
·
2 1
−1 2
¸
(B) P=
·
−1 0 0 4
¸
(C) P=
·
−3 4 −12
−1 2 0
¸
(D) P =
"
2
√
10 1
√
10
−√1 10
2
√
10
#
→(E) P =
"
2
√
5 − 1
√
5 1
√
5 2
√
5
det(A−λ I) =¯¯
−2 3−λ¯¯=λ −3λ−4 = (λ−4) (λ+ 1)
and the roots of the characteristic polynomial are λ = 4 and λ = −1. If λ = 4 , the corresponding eigenvector is obtained by solving
·
−4 −2
−2 −1
¸ ·
x y
¸
=
·
0 0
¸
or 2x+y = 0 which yields the unit eigenvector
"
−√1 5 2
√
5
#
. If λ=−1 , the corresponding
eigenvector is obtained by solving
·
1 −2
−2 4
¸ ·
x y
¸
=
·
0 0
¸
orx−2y= 0 which yields the unit eigenvector
" 2
√
5 1
√
5
#
. We can thus take
P =
"
2
√
5 − 1
√
5 1
√
5 2
√
5
#
.
as an orthogonal matrix which dagonalizesA.
13. The matrix form of the system of differential equations
(
dx
dt = 2y+e−t dy
dt =−x+ 3y,
with the notationX=
·
x y
¸
, is
(A) X0=
·
2 e−t
−1 3
¸
X
(B) X0=
·
2 0
−1 3
¸
X+e−t
→(C) X0=
·
0 2
−1 3
¸
X+
·
e−t
0
¸
(D) X0 =
·
e−t 2
−1 3
¸
X
(E) X0 =
·
2 1 e−t
−1 3 0
¸
14. Consider the system of differential equations
(
dx
dt =−7x+ 4y dy
dt =−4x+ 3y.
Then, one of its solutionsX=
·
x y
¸
is
(A) X=
·
2 1
¸
et
→(B) X=
·
1 2
¸
et
(C) X=
·
−7
−4
¸
et
(D) X=
·
−3
−1
¸
et
(E) X=
·
1 0
¸
et
Solution. The system can be written asX0 =AX, where
A=
·
−7 4
−4 3
¸
We have
det(A−λ I) =
¯ ¯ ¯
¯−7−−4λ 3−4 λ ¯ ¯ ¯
¯=λ2+ 4λ−5 = (λ+ 5) (λ−1)
and the roots of the characteristic polynomial are λ = 1 and λ = −5. If λ = 1, the corresponding eigenvector is obtained by solving
·
−8 4
−4 2
¸ ·
x y
¸
=
·
0 0
¸
or−2x+y= 0 which yields the eigenvector
·
1 2
¸
. Ifλ=−5, the corresponding eigenvector is obtained by solving ·
−2 4
−4 8
¸ ·
x y
¸
=
·
0 0
¸
orx−2y= 0 which yields the eigenvector
·
2 1
¸
. The general solution of the system is
·
x y
¸
=C1et
·
1 2
¸
+C2 e−5t
·
2 1
¸
TakingC1= 1 andC2= 0, we obtain thus the solutionet
·
1 2
¸
−1 0 1
→(A) X=c1et
01
0
+c2
10
1
+c3e2t
−01
1
(B) X=c1et
10
−1
+c2
01
0
+c3e2t
−01
1
(C) X=et
01
0
+
10
1
+e2t
−01
1
(D) non-existent becauseA is singular
(E) X=
01
0
+ cost
10
1
+ cos 2t −01
1
Solution.
We have
det(A−λ I) =
¯ ¯ ¯ ¯ ¯ ¯
1−λ 0 −1 0 1−λ 0
−1 0 1−λ
¯ ¯ ¯ ¯ ¯
¯= (1−λ)
3−(1−λ) = (1−λ)¡(λ−1)2−1¢= (1−λ) (λ−2)λ.
and the roots of the characteristic polynomial areλ= 0, λ= 1 andλ= 2. Ifλ= 0, the corresponding eigenvector is obtained by solving
10 01 −01
−1 0 1
xy
z
=
00
0
or y = 0 and x = z which yields the eigenvector
10
1
. If λ = 1, the corresponding
eigenvector is obtained by solving
00 00 −01
−1 0 0
xy
z
=
00
0
or x = 0 and z = 0 which yields the eigenvector
01
0
. If λ = 2, the corresponding
eigenvector is obtained by solving
−01 −01 −01
−1 0 −1
xy
z
=
00
0
ory= 0 andz=−xwhich yields the eigenvector
−01
1
. The general solution is thus
X=c1et
01
0
+c2
10
1
+c3e2t
−01
1
16. The general solution
·
x y
¸
of the system
(
x0=y
y0=−2x−2y
is (A) c1 · cost sint ¸ +c2 ·
−sint cost
¸
(B) c1e−t
·
cost sint
¸
+c2e−t
·
−sint cost
¸
(C) c1et
·
cost sint
¸
+c2et
·
−sint cost
¸
(D) c1et
·
cost+ sint
−2 sint
¸
+c2et
·
sint cost−sint
¸
→(E) c1e−t
·
cost+ sint
−2 sint
¸
+c2e−t
·
sint cost−sint
¸
Solution. The system can be written asX0 =AX, whereA=
·
0 1
−2 −2
¸
.
We have
det(A−λ I) =
¯ ¯ ¯
¯−−λ2 −21−λ ¯ ¯ ¯
¯=λ2+ 2λ+ 2 = (λ+ 1)2+ 1
and the roots of the characteristic polynomial are λ = −1±i. If λ = −1 +i, the corresponding eigenvector is obtained by solving
¯ ¯ ¯
¯1−−2i −11−i ¯ ¯ ¯ ¯ · x y ¸ = · 0 0 ¸
ory= (−1 +i)xwhich yields the eigenvector
·
1
−1 +i
¸
. This yields the complex solution
Z(t) =e(−1+i)t
·
1
−1 +i
¸
=e−t(cost+isint)
½· 1 −1 ¸ +i · 0 1 ¸¾ = ½
e−t cost
·
1
−1
¸
−e−tsint
· 0 1 ¸¾ +i ½
e−tcost
·
0 1
¸
+e−tsint
·
1
−1
¸¾
=X(t) +iY(t),
where
X(t) =e−t
·
cost
−sint−cost
¸
and Y(t) =e−t
·
sint cost−sint
¸
are two linearly independent real-valued solutions. Note that
U(t) =X(t) +Y(t) =e−t
·
cost+ sint
−2 sint
¸
can replaceX(t), soU(t) andY(t) are also two linearly independent real-valued solutions and the general solutions is given by
c1e−t
·
cost+ sint
−2 sint
¸
+c2e−t
·
sint cost−sint