T E ST 2( VE RSI ON 1 )

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_+C2x3_lnx

(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) =

¯ ¯ ¯ ¯y1_y0 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

x2_dx

¶

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

d4_y

dx4 = sinx, y(0) =y

0_{(0) =y(π) =y}0_{(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. ComputeA4_if

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_{+ 3a}2_{+ 1 a}3_{+ 2a} a3_{+ 2a a}2_{+ 1}

¸

(E) A4₌

·

a4_{+ 2a}2 _a3_{+ 2a} a3_{+ 2a a}2_{+ 1}

¸

Solution. We have

A2₌

·

a 1 1 0

¸ ·

a 1 1 0

¸

=

·

a2_{+ 1 a} a 1

¸

and

A4_=A2_A2₌

·

a2_{+ 1 a} a 1

¸ ·

a2_{+ 1 a} a 1

¸

=

·

(a2_{+ 1)}2_+a2 _a3_{+ 2a} a3_{+ 2a a}2_{+ 1}

¸

9. A realn×nmatrixAis orthogonal if

(A) AT_A=AAT

→(B) AT_A=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−1_{AP 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 ₋9_λ ¯ ¯ ¯

¯=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



 

x_y

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



 

x_y

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



 

x_y

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) =

¯ ¯ ¯

¯−₋λ_{2 −2}1_−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−t_sint

· 0 1 ¸¾ +i ½

e−t_cost

·

0 1

¸

+e−t_sint

·

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