20d - homework assignment 5

Brian Bowers (TA for Hui Sun)MATH 20D

Homework Assignment 5November 8, 2013

20D - Homework Assignment 5

First, I present the list of all matrix row operations. We use combinations of these steps to row reducematrices. Note that I will use ri to represent the ith row of the matrix.

(1) Swap: We are allowed to swap two entire rows of a matrix. [I will notate swapping rows ri and rj by“ri ↔ rj”.]

(2) Multiply: We are allowed to multiply any one row of a matrix by any nonzero scalar expression (like1, π, et

√t2 + 7, etc.) [I will notate multiplying row ri by k as “ri = kri”.]

(3) Combine: Given two rows ri and rj, we are allowed to replace ri by (ri + krj) for any scalar k. [I willnotate this as “ri = ri + krj”.]

For our purposes, row reduction is the process of trying to get a matrix to “have 1s along the diagonal” and“have 0s above and below the diagonal”. In other words, we’re trying to get our matrix to look like this:

1 ? ? ? · · ·0 1 ? ? · · ·0 0 1 ? · · ·...

......

.... . .

and then like this:

1 0 0 ? · · ·0 1 0 ? · · ·0 0 1 ? · · ·...

......

.... . .

.

Note that we can’t always get our matrix to look exactly like the ones above, but our goal is to always getas close as possible.

7.3 #13,14 In each problem, determine whether the members of the given set of vectors are linearlyindependent for −∞ < t < ∞. If they are linearly dependent, find the linear relation among them. [Thevectors are written as row vectors to save space.]

(13) x(1)(t) = (e−t, 2e−t), x(2)(t) = (e−t, e−t), x(3)(t) = (3e−t, 0).

(14) x(1)(t) = (2 sin t, sin t), x(2)(t) = (sin t, 2 sin t).

A set of (column) vectors x(1),x(2), . . . ,x(n) (each with k entries) is considered to be linearly dependentif and only if there exist scalars c1, c2, . . . , cn (where at least one of these constants is not 0) such that

c1x(1) + c2x

(2) + . . .+ cnx(n) = 0.

We could rewrite the above equation in matrix form as

(x(1) x(2) · · · x(n)

)c1c2...cn

=

00...0

,

where the 0 vector at the end of the equation has k entries.Using the above setup as motivation, we construct the following steps to determine if n vectors (each

with k entries) are linearly independent or linearly dependent:

1

(1) Augment: Write the augmented matrix below:(x(1) x(2) · · · x(n) 0

),

where 0 represents the vector with k entries, all of which are 0.

(2) Row Reduce: Row reduce the above matrix.

(3) Conclude: There are two conclusions to be reached here:

(i) If the row reduction gives us a matrix that looks like1 0 0 · · · 0 00 1 0 · · · 0 00 0 1 · · · 0 0...

......

. . . 0 00 0 0 · · · 1 0

,

then we conclude that the vectors are linearly independent.

(ii) Otherwise, we conclude that the vectors are linearly dependent. If we’re deciding between depen-dent/independent, then we’re done.

However, sometimes we’re asked to find a “linear relation” among the vectors if they are lin-early dependent. In this case, each row of the row-reduced matrix gives us an equation (involvingc1, c2, . . . , cn) that we can use to find such a “linear relation” among the vectors. For example, thematrix (

1 0 2 00 1 −1 0

)gives us the system of equations

1c1 + 0c2 + 2c3 = 0

0c1 + 1c2 − 1c3 = 0

We can solve the equations as c1 = −2c3

c2 = c3.

Now all that remains is to choose a value for c3. Recall that we want at least one of the c1, c2, orc3 to be nonzero. So, as a general rule, when I get to choose a value for a constant, I just choose1. (Note that choosing c3 = 1 means that c3 is nonzero!) If we choose c3 = 1, we get c1 = −2 andc2 = 1. Looking way back to our original equation, we can fill in our constants:

c1x(1) + c2x

(2) + . . .+ cnx(n) = 0

=⇒ 1x(1) − 2x(2) + 1x(n) = 0

Okay - let’s do the problems now.

2

(13) We follow the steps from above to make an augmented matrix and row reduce it:(e−t e−t 3e−t 02e−t e−t 0 0

)r1=e

tr1−→(

1 1 3 02e−t e−t 0 0

)r2=r2−2e−tr1−→

(1 1 3 00 −e−t −6e−t 0

)r2=−etr2−→

(1 1 3 00 1 6 0

)r1=r1−r2−→

(1 0 −3 00 1 6 0

)

Since the left part of the matrix is not an identity matrix, this is case (ii) from our steps above. Inother words, we determined that the vectors are linearly dependent. Moreover, we can read off twoequations from the matrix we got:

1c1 + 0c2 − 3c3 = 0

0c1 + 1c2 + 6c3 = 0

=⇒

c1 = 3c3

c2 = −6c3[Next, choose c3 = 1]

=⇒

c1 = 3

c2 = −6

c3 = 1

Thus, we can write the following linear relation:

3x(1) − 6x(2) + x(3) = 0.

(14) Once again, we follow the steps from above to make an augmented matrix and row reduce it:(2 sin t sin t 0sin t 2 sin t 0

)r1=

12 sin t r1−→

(1 1/2 0

sin t 2 sin t 0

)r2=r2−(sin t)r1−→

(1 1/2 00 (3 sin t)/2 0

)r2=

23 sin t r2−→

(1 1/2 00 1 0

)r1=r1− 1

2 r2−→(

1 0 00 1 0

)

This is case (i) from our steps above because the left part of the matrix is an identity matrix. In otherwords, the vectors are linearly independent.

3

7.3 #16,17,18,21 In each problem, find all eigenvalues and eigenvectors of the given matrix.

(16)

(5 −13 1

)

(17)

(3 −24 −1

)

(18)

(−2 11 −2

)

(21)

(−3 3/4−5 1

)

Quick reminder: det

(a bc d

)= ad− bc.

To find the eigenvalues of a matrix A, we simply solve the equation

det(A− λI) = 0

for λ. [Note that A−λI is the same as A, except we subtract λ from each entry along the diagonal of A.] Wename these solutions λ1, λ2, etc.. Next, to find the eigenvector corresponding to λi (we call this eigenvectorx(1)), we row reduce the augmented matrix (

A− λiI 0)

and use the resulting matrix to spit out a system of equations, just like we did to find linear relations.Let’s do some problems - I think examples are clearer for this than general explanations.

4

(16) We solve the equation det(A− λI) = 0:

det(A− λI) = 0

=⇒ det

(5− λ −1

3 1− λ

)= 0

=⇒ (5− λ)(1− λ)− (−1)(3) = 0

=⇒ λ2 − 6λ+ 8 = 0

=⇒ (λ− 4)(λ− 2) = 0

=⇒ λ = 4, 2

Thus, we have the eigenvalues λ1 = 4 and λ2 = 2. Now we row reduce the appropriate augmentedmatrices.

(i) λ1 = 4: (5− 4 −1 0

3 1− 4 0

)=

(1 −1 03 −3 0

)r2=r2−3r1−→

(1 −1 00 0 0

)

Thus, the first row gives us the equation x1 − x2 = 0, which we can solve to get x1 = x2. I will

select x2 = 1, which yields x1 = 1. Thus, we get the eigenvector x(1) =

(x1x2

)=

(11

).

(ii) λ2 = 2: (5− 2 −1 0

3 1− 2 0

)=

(3 −1 03 −1 0

)r2=r2−1r1−→

(3 −1 00 0 0

)

Thus, the first row gives us the equation 3x1 − x2 = 0, which we can solve to get x2 = 3x1. I will


(x1x2

)=

(13

).

5


det(A− λI) = 0

=⇒ det

(3− λ −2

4 −1− λ

)= 0

=⇒ (3− λ)(−1− λ)− (−2)(4) = 0

=⇒ λ2 − 2λ+ 5 = 0

=⇒ λ =−(−2)±

√(−2)2 − 4(1)(5)

2(1)

=⇒ λ = 1± 2i

Thus, we have the eigenvalues λ1 = 1 + 2i and λ2 = 1 − 2i. Now we row reduce the appropriateaugmented matrices.

(i) λ1 = 1 + 2i:(3− (1 + 2i) −2 0

4 −1− (1 + 2i) 0

)=

(2− 2i −2 0

4 −2− 2i 0

)r1=

12−2i r1−→

(1 −2

2−2i 0

4 −2− 2i 0

)[We need to rationalize that denominator]

=

(1 −2

2−2i ·2+2i2+2i 0

4 −2− 2i 0

)=

(1 −1−i

2 04 −2− 2i 0

)r2=r2−4r1−→

(1 −1−i

2 00 0 0

)

Thus, the first row gives us the equation x1 + −1−i2 x2 = 0, which we can solve to get x1 = 1+i

2 x2.

I will select x2 = 1, which yields x1 = 1+i2 . Thus, we get the eigenvector x(1) =

(x1x2

)=

(1+i21

).

(ii) λ2 = 1− 2i:(3− (1− 2i) −2 0

4 −1− (1− 2i) 0

)=

(2 + 2i −2 0

4 −2 + 2i 0

)r1=

12+2i r1−→

(1 −2

2+2i 0

4 −2 + 2i 0

)[We need to rationalize that denominator]

=

(1 −2

2+2i ·2−2i2−2i 0

4 −2 + 2i 0

)=

(1 −1+i

2 04 −2 + 2i 0

)r2=r2−4r1−→

(1 −1+i

2 00 0 0

)

Thus, the first row gives us the equation x1 + −1+i2 x2 = 0, which we can solve to get x1 = 1−i

2 x2.

I will select x2 = 1, which yields x1 = 1−i2 . Thus, we get the eigenvector x(2) =

(x1x2

)=

(1−i21

).

6


det(A− λI) = 0

=⇒ det

(−2− λ 1

1 −2− λ

)= 0

=⇒ (−2− λ)(−2− λ)− (1)(1) = 0

=⇒ λ2 + 4λ+ 3 = 0

=⇒ (λ+ 3)(λ+ 1) = 0

=⇒ λ = −3,−1

Thus, we have the eigenvalues λ1 = −3 and λ2 = −1. Now we row reduce the appropriate augmentedmatrices.

(i) λ1 = −3: (−2 + 3 1 0

1 −2 + 3 0

)=

(1 1 01 1 0

)r2=r2−r1−→

(1 1 00 0 0

)

Thus, the first row gives us the equation x1 + x2 = 0, which we can solve to get x1 = −x2. I will

select x2 = 1, which yields x1 = −1. Thus, we get the eigenvector x(1) =

(x1x2

)=

(−11

).

(ii) λ2 = −1: (−2 + 1 1 0

1 −2 + 1 0

)=

(−1 1 01 −1 0

)r1=(−1)r1−→

(1 −1 01 −1 0

)r2=r2−r1−→

(1 −1 00 0 0

)

Thus, the first row gives us the equation x1 − x2 = 0, which we can solve to get x1 = x2. I will


(x1x2

)=

(11

).

7


det(A− λI) = 0

=⇒ det

(−3− λ 3/4−5 1− λ

)= 0

=⇒ (−3− λ)(1− λ)− (3/4)(−5) = 0

=⇒ λ2 + 2λ+ 3/4 = 0

=⇒ (λ+ 3/2)(λ+ 1/2) = 0

=⇒ λ = −3/2,−1/2

Thus, we have the eigenvalues λ1 = −3/2 and λ2 = −1/2. Now we row reduce the appropriateaugmented matrices.

(i) λ1 = −3/2: (−3 + 3/2 3/4 0−5 1 + 3/2 0

)=

(−3/2 3/4 0−5 5/2 0

)r1=− 2

3 r1−→(

1 −1/2 0−5 5/2 0

)r2=r2+5r1−→

(1 −1/2 00 0 0

)

Thus, the first row gives us the equation x1− 12x2 = 0, which we can solve to get x1 = 1

2x2. I will

select x2 = 1, which yields x1 = 12 . Thus, we get the eigenvector x(1) =

(x1x2

)=

(1/21

).

(ii) λ2 = −1/2: (−3 + 1/2 3/4 0−5 1 + 1/2 0

)=

(−5/2 3/4 0−5 3/2 0

)r1=− 2

5 r1−→(

1 −3/10 0−5 3/2 0

)r2=r2+5r1−→

(1 −3/10 00 0 0

)

Thus, the first row gives us the equation x1 − 310x2 = 0, which we can solve to get x1 = 3

10x2. I

will select x2 = 1, which yields x1 = 310 . Thus, we get the eigenvector x(2) =

(x1x2

)=

(3/10

1

).

8

7.4 #4 If x1 = y and x2 = y′, the the second order equation

y′′ + p(t)y′ + q(t)y = 0 (1)

corresponds to the system x′1 = x2,

x′2 = −q(t)x1 − p(t)x2. (2)

Show that if x(1) and x(2) are a fundamental set of solutions of Equations (2), and if y(1) and y(2) area fundamental set of solutions of Equation (1), then W [y(1), y(2)] = cW [x(1),x(2)], where c is a nonzeroconstant. [Hint: y(1)(t) and y(2)(t) must be linear combinations of x11(t) and x12(t).]

First note that to transform from the single differential equation to the system of differential equations,we use the transformation y = x1.

Now suppose x(1) and x(2) are a fundamental set of solutions of Equations (2), and suppose y(1) and y(2)

are a fundamental set of solutions of Equation (1). Since x(1) =

(x11x21

)and x(2) =

(x12x22

)are solutions, we

know that any solution y(1) could be written as

y(1) = c1x11 + c2x12,

and any different solution y(2) could be written as

y(2) = c3x11 + c4x12.

Next, note that

W [x(1),x(2)] = W

[(x11x21

),

(x12x22

)]= det

(x11 x12x21 x22

)= x11x22 − x12x21

Finally, we piece everything together:

W [y(1), y(2)] = y(1)(y(2))′ − (y(1))′y(2)

= [c1x11 + c2x12][c3x11 + c4x12]′ − [c1x11 + c2x12]′[c3x11 + c4x12]

= [c1x11 + c2x12][c3x′11 + c4x

′12]− [c1x

′11 + c2x

′12][c3x11 + c4x12]

= [c1x11 + c2x12][c3x21 + c4x22]− [c1x21 + c2x22][c3x11 + c4x12] (since x′11 = x21 and x′12 = x22)

= c1c3x11x21 + c1c4x11x22 + c2c3x12x21 + c2c4x12x22 − c1c3x21x11 − c1c4x21x12 − c2c3x22x11 − c2c4x22x12= (c1c3 − c1c3)x11x21 + (c1c4 − c2c3)x11x22 + (c2c3 − c1c4)x12x21 + (c2c4 − c2c4)x12x22

= (c1c4 − c2c3)x11x22 + (c2c3 − c1c4)x12x21

= (c1c4 − c2c3)x11x22 − (c1c4 − c2c3)x12x21

= (c1c4 − c2c3)(x11x22 − x12x21)

= (c1c4 − c2c3)W [x(1),x(2)]

= cW [x(1),x(2)]

Moreover, we know that c is nonzero because we know that W [y(1), y(2)] is not zero (since y(1) and y(2) area fundamental set of solutions and the Wronskian of any fundamental set of solutions is nonzero.)

9

7.4 #6,7 Given vectors x(1)(t) and x(2)(t), complete the following:

(a) Compute the Wronskian of x(1) and x(2).

(b) In what intervals are x(1) and x(2) linearly independent?

(c) What conclusion can be drawn about the coefficients in the system of homogemeous differential equa-tions satisfied by x(1) and x(2)?

(d) Find this system of equations and verify the conclusions of part (c).

Complete the above for the following sets of vectors:

(6) x(1)(t) =

(t1

)and x(2)(t) =

(t2

2t

)

(7) x(1)(t) =

(t2

2t

)and x(2)(t) =

(et

et

).

(6) (a) W [x(1),x(2)] = det(x(1)x(2)

)= det

(t t2

1 2t

)= t(2t)− t2(1) = t2

(b) We know that two solutions are linearly independent if and only if their Wronskian is nonzero.But the Wronskian, t2, is only zero when t = 0. Thus, the solutions are linearly independenteverywhere else, namely (−∞, 0) and (0,∞).

(c) According to Theorem 7.4.3, at least one of the coefficients must be discontinuous at t = 0. (If itweren’t then the Wronskian wouldn’t vanish.)

(d) Essentially, this is asking us to find a matrix A such that x′ = Ax for any solution x. Note thatany solution x could be written as x = c1x

(1) + c2x(2). So let’s simplify from our basic equation:

x′ = Ax

=⇒[c1x

(1) + c2x(2)]′

=

(a11 a12a21 a22

)[c1x

(1) + c2x(2)]

=⇒[c1

(t1

)+ c2

(t2

2t

)]′=

(a11 a12a21 a22

)[c1

(t1

)+ c2

(t2

2t

)]=⇒

(c1t+ c2t

2

c1 + 2c2t

)′=

(a11 a12a21 a22

)(c1t+ c2t

2

c1 + 2c2t

)=⇒

(c1 + 2c2t

2c2

)=

(a11[c1t+ c2t

2] + a12[c1 + 2c2t]a21[c1t+ c2t

2] + a22[c1 + 2c2t]

)=⇒

((1)c1 + (2t)c2(0)c1 + (2)c2

)=

((a11t+ a12)c1 + (a11t

2 + 2a12t)c2(a21t+ a22)c1 + (a21t

2 + 2a22t)c2

)

Looking at the coefficients of c1 and c2 from the top entry, we see that1 = a11t+ a12

2t = a11t2 + 2a12t

.

We can solve the first equation to get

a12 = 1− a11t.

Substituting this into the second equation, we get 2t = a11t2 + 2(1− a11t)t = 2− a11t2, which we

can solve as

a11 =2− 2t

t2.

10

Substituting this back into the expression for a12, we get

a12 = 1− 2− 2t

t2· t =

t

t− 2− 2t

t=

3t− 2

t

Looking at the coefficients of c1 and c2 from the bottom entry, we see that0 = a21t+ a22

2 = a21t2 + 2a22t

.

Solving the first equation, we geta22 = −a21t.

Substituting this into the second equation, we get 2 = a21t2 + 2(−a21t)t = −a21t2, which we can

solve to get

a21 = − 2

t2.

Plugging this back into the expression for a22, we get

a22 = −(− 2

t2

)· t =

2

t.

Now we have all entries from the matrix A. So, we rewrite the original matrix equation x′ = Axas

x′ =

(2−2tt2

3t−2t

− 2t2

2t

)x.

This confirms our assessment from part (c) that at least one of these coefficients is discontinuousat t = 0. (In fact, all of them are discontinuous at t = 0!)

(7) (a) W [x(1),x(2)] = det(x(1)x(2)

)= det

(t2 et

2t et

)= t2et − 2tet = (t2 − 2t)et = t(t− 2)et

(b) We know that two solutions are linearly independent if and only if their Wronskian is nonzero.But the Wronskian, t(t− 2)et, is only zero when t = 0 or t = 2. Thus, the solutions are linearlyindependent everywhere else, namely (−∞, 0), (0, 2) and (2,∞).

(c) According to Theorem 7.4.3, at least one of the coefficients must be discontinuous at t = 0, andat least one of the coefficients must be discontinuous at t = 2. (If it weren’t then the Wronskianwouldn’t vanish.)

(d) Essentially, this is asking us to find a matrix A such that x′ = Ax for any solution x. Note thatany solution x could be written as x = c1x

(1) + c2x(2). So let’s simplify from our basic equation:

x′ = Ax

=⇒[c1x

(1) + c2x(2)]′

=

(a11 a12a21 a22

)[c1x

(1) + c2x(2)]

=⇒[c1

(t2

2t

)+ c2

(et

et

)]′=

(a11 a12a21 a22

)[c1

(t2

2t

)+ c2

(et

et

)]=⇒

(c1t

2 + c2et

2c1t+ c2et

)′=

(a11 a12a21 a22

)(c1t

2 + c2et

2c1t+ c2et

)=⇒

(2c1t+ c2e

t

2c1 + c2et

)=

(a11[c1t

2 + c2et] + a12[2c1t+ c2e

t]a21[c1t

2 + c2et] + a22[2c1t+ c2e

t]

)=⇒

((2t)c1 + (et)c2(2)c1 + (et)c2

)=

((a11t

2 + 2a12t)c1 + (a11et + a12e

t)c2(a21t

2 + 2a22t)c1 + (a21et + a22e

t)c2

)

11

Looking at the coefficients of c1 and c2 from the top entry, we see that2t = a11t

2 + 2a12t

et = a11et + a12e

t.

We can solve the first equation to get

a11 =t(2− 2a12)

t2.

Substituting this into the second equation, we get et =(t(2−2a12)

t2

)et + a12e

t, which we can solveas

a12 =t2 − 2t

t2 − 2t.

Substituting this back into the expression for a12, we get

a11 =2− 2

(t2−2tt2−2t

)t

= 0.

Looking at the coefficients of c1 and c2 from the bottom entry, we see that2 = a21t

2 + 2a22t

et = a21et + a22e

t.

Solving the first equation, we get

a22 =2− a21t2

2t.

Substituting this into the second equation, we get et = a21et +

(2−a21t2

2t

)et, which we can solve

to get

a21 =2t− 2

2t− t2.

Plugging this back into the expression for a22, we get

a22 =2− a21t2

2t=

2−(

2t−22t−t2

)· t2

2t=

2(2t−t2)−(2t−2)t22t−t2

2t=

4t−2t2−2t3+2t2

2t−t2

2t=

2− t2

2t− t2

Now we have all entries from the matrix A. So, we rewrite the original matrix equation x′ = Axas

x′ =

(0 t2−2t

t2−2t2t−22t−t2

2−t22t−t2

)x.

This confirms our assessment from part (c) that at least one of these coefficients is discontinuousat t = 0, and at least one of these coefficients is discontinuous at t = 2. (In fact, all of them arediscontinuous at t = 0!)

12

HOW TO SOLVE EQUATIONS OF THE FORM x′ = Ax (if A has distinct, real eigenvalues):

(i) Find the eigenvalues of A. We will call the eigenvalues r1, r2, . . . , rn

(ii) Find the eigenvectors corresponding to r1, r2, . . . , rn; we will call these eigenvectors ξ(1), ξ(2), . . . , ξ(n),respectively.

(iii) The general solution is x = c1ξ(1)er1t + c2ξ

(2)er2t + . . .+ cnξ(n)ernt.

Note that equations of the form x′ = Ax are referred to as homogeneous linear systems [as opposed tononhomogeneous linear systems, which look like x′ = Ax + g(t)]

7.5 #1a,2a,3a For each problem, find the general solution of the given system of equations and describethe behavior of the solution as t→∞.

(1a) x′ =

(3 −22 −2

)x

(2a) x′ =

(1 −23 −4

)x

(3a) x′ =

(2 −13 −2

)x

13

(1a) I will follow the steps outlined above:

(i) We first find the eigenvalues by solving the equation below:

det(A− λI) = 0

=⇒ det

(3− λ −2

2 −2− λ

)= 0

=⇒ (3− λ)(−2− λ)− (−2)(2) = 0

=⇒ λ2 − λ− 2 = 0

=⇒ (λ− 2)(λ+ 1) = 0

Thus, we get eigenvalues r1 = 2 and r2 = −1.

(ii) We find the eigenvectors, starting with ξ(1), which corresponds to eigenvalue r1 = 2. To do this,we row reduce the augmented matrix below:

(A− r1I 0

)=

(3− 2 −2 0

2 −2− 2 0

)=

(1 −2 02 −4 0

)r2=r2−2r1−→

(1 −2 00 0 0

)

The first row gives us the equation x1− 2x2 = 0, which we can solve to get x1 = 2x2. If we select

x2 = 1, we get x1 = 2. Thus, we have the eigenvector ξ(1) =

(x1x2

)=

(21

).

Next, we find ξ(2), the eigenvector associated with eigenvalue r2 = −1. To do this, we row reducethe augmented matrix below:

(A− r2I 0

)=

(3− (−1) −2 0

2 −2− (−1) 0

)=

(4 −2 02 −1 0

)r1=

14 r1−→(

1 −1/2 02 −1 0

)r2=r2−2r1−→

(1 −1/2 00 0 0

)



(x1x2

)=

(12

).

(iii) Thus, the general solution is

x = c1ξ(1)er1t + c2ξ

(2)er2t = c1

(21

)e2t + c2

(12

)e−t

By creating a phase portrait, we can see that x approaches the line formed along the vector ξ(1), i.e.x2 = 1

2x1.

14



det(A− λI) = 0

=⇒ det

(1− λ −2

3 −4− λ

)= 0

=⇒ (1− λ)(−4− λ)− (−2)(3) = 0

=⇒ λ2 + 3λ+ 2 = 0

=⇒ (λ+ 2)(λ+ 1) = 0

Thus, we get eigenvalues r1 = −2 and r2 = −1.

(ii) We find the eigenvectors, starting with ξ(1), which corresponds to eigenvalue r1 = −2. To do this,we row reduce the augmented matrix below:

(A− r1I 0

)=

(1− (−2) −2 0

3 −4− (−2) 0

)=

(3 −2 03 −2 0

)r1=

13 r1−→(

1 −2/3 03 −2 0

)r2=r2−3r1−→

(1 −2/3 00 0 0

)

The first row gives us the equation x1− 23x2 = 0, which we can solve to get x1 = 2

3x2. If we select

x2 = 1, we get x1 = 23 . Thus, we have the eigenvector ξ(1) =

(x1x2

)=

(2/31

).


(A− r2I 0

)=

(1− (−1) −2 0

3 −4− (−1) 0

)=

(2 −2 03 −3 0

)r1=

12 r1−→(

1 −1 02 −1 0

)r2=r2−3r1−→

(1 −1 00 0 0

)

The first row gives us the equation x1 − x2 = 0, which we can solve to get x1 = x2. If we select


(x1x2

)=

(11

).



(2)er2t = c1

(2/31

)e−2t + c2

(11

)e−t

By creating a phase portrait, we can see that x approaches the line formed along the vector ξ(1), i.e.x2 = x1.

15



det(A− λI) = 0

=⇒ det

(2− λ −1

3 −2− λ

)= 0

=⇒ (2− λ)(−2− λ)− (−1)(3) = 0

=⇒ λ2 − 1 = 0

=⇒ (λ− 1)(λ+ 1) = 0



(A− r1I 0

)=

(2− 1 −1 0

3 −2− 1 0

)=

(1 −1 03 −3 0

)r2=r2−3r1−→

(1 −1 00 0 0

)



(x1x2

)=

(11

).


(A− r2I 0

)=

(2− (−1) −1 0

3 −2− (−1) 0

)=

(3 −1 03 −1 0

)r1=

13 r1−→(

1 −1/3 03 −1 0

)r2=r2−3r1−→

(1 −1/3 00 0 0

)

The first row gives us the equation x1− 13x2 = 0, which we can solve to get x1 = 1

3x2. If we select

x2 = 1, we get x1 = 13 . Thus, we have the eigenvector ξ(2) =

(x1x2

)=

(1/31

).



(2)er2t = c1

(11

)et + c2

(1/31

)e−t

By creating a phase portrait, we can see that x approaches the line formed along the vector ξ(1), i.e.x2 = 3x1.

16

7.5 #15,16 Solve each initial value problem, and describe the behavior of the solution as t→∞.

(15) x′ =

(5 −13 1

)x, x(0) =

(2−1

)

(16) x′ =

(−2 1−5 4

)x, x(0) =

(13

)

(15) I will follow the steps outlined above:


det(A− λI) = 0

=⇒ det

(5− λ −1

3 1− λ

)= 0

=⇒ (5− λ)(1− λ)− (−1)(3) = 0

=⇒ λ2 − 6λ+ 8 = 0

=⇒ (λ− 4)(λ− 2) = 0

Thus, we get eigenvalues r1 = 4 and r2 = 2.


(A− r1I 0

)=

(5− 4 −1 0

3 1− 4 0

)=

(1 −1 03 −3 0

)r2=r2−3r1−→

(1 −1 00 0 0

)



(x1x2

)=

(11

).

Next, we find ξ(2), the eigenvector associated with eigenvalue r2 = 2. To do this, we row reducethe augmented matrix below:

(A− r2I 0

)=

(5− 2 −1 0

3 1− 2 0

)=

(3 −1 03 −1 0

)r1=

13 r1−→(

1 −1/3 03 −1 0

)r2=r2−3r1−→

(1 −1/3 00 0 0

)



(x1x2

)=

(13

).

17



(2)er2t = c1

(11

)e4t + c2

(13

)e2t

Next, we use the initial conditions to find c1 and c2:(2−1

)= x(0) = c1

(11

)e4(0) + c2

(13

)e2(0) =

(c1 + c2c1 + 3c2

)By comparing the top and bottom entries, we get the system of equations

2 = c1 + c2

−1 = c1 + 3c2=⇒

c1 = 7/2

c2 = −3/2.

Thus, our final solution is

x =7

2

(11

)e4t − 3

2

(13

)e2t

This will be dominated by the e4t term, so it will approach the line formed by the vector

(11

), which

is the line x2 = x1.

18

(16) I will follow the steps outlined above:


det(A− λI) = 0

=⇒ det

(−2− λ 1−5 4− λ

)= 0

=⇒ (−2− λ)(4− λ)− (1)(−5) = 0

=⇒ λ2 − 2λ− 3 = 0

=⇒ (λ− 3)(λ+ 1) = 0


(ii) We find the eigenvectors, starting with ξ(1), which corresponds to eigenvalue r1 = 3. To do this,we row reduce the augmented matrix below:(

A− r1I 0)

=

(−2− 3 1 0−5 4− 3 0

)=

(−5 1 0−5 1 0

)r1=− 1

5 r1−→(

1 −1/5 0−5 1 0

)r2=r2+5r1−→

(1 −5 00 0 0

)



(x1x2

)=

(15

).

Next, we find ξ(2), the eigenvector associated with eigenvalue r2 = −1. To do this, we row reducethe augmented matrix below:(

A− r2I 0)

=

(−2− (−1) 1 0−5 4− (−1) 0

)=

(−1 1 0−5 5 0

)r1=−1r1−→

(1 −1 0−5 5 0

)r2=r2+5r1−→

(1 −1 00 0 0

)



(x1x2

)=

(11

).



(2)er2t = c1

(15

)e3t + c2

(11

)e−t

Next, we use the initial conditions to find c1 and c2:(13

)= x(0) = c1

(15

)e3(0) + c2

(11

)e−(0) =

(c1 + c25c1 + c2

)

19

By comparing the top and bottom entries, we get the system of equations1 = c1 + c2

3 = 5c1 + c2=⇒

c1 = 1/2

c2 = 1/2.

Thus, our final solution is

x =1

2

(15

)e3t +

1

2

(11

)e−t

This will be dominated by the e3t term, so it will approach the line formed by the vector

(15

), which

is the line x2 = 5x1.

20

HOW TO MAKE A PHASE PORTRAIT [if A has real, distinct eigenvalues]: If we are given a matrixequation x′ = Ax and we find that A has distinct, real eigenvalues, then we can create a phase diagram byfollowing the steps below:

(i) For each eigenvector, draw the line formed (starting at the origin) by extending the vector infinitely inboth directions.

(ii) For each line from step (i), look at the corresponding ri. If ri is positive, draw arrows on the linepointing outward. If ri is negative, draw arrows onn the line pointing inward.

(iii) Draw in sample curves, approaching the lines and following the directions given by the arrows.

7.5 #24,27 For each problem, the eigenvalues and eigenvectors of a matrix A are given. Consider thecorresponding system x′ = Ax.

(a) Sketch a phase portrait of the system.

(b) Sketch the trajectory passing through the initial point (2, 3).

(c) For the trajectory in part (b), sketch the graphs of x1 versus t and of x2 versus t on the same set ofaxes.

[Problems below]

(24) r1 = −1, ξ(1) =

(−12

); r2 = −2, ξ(2) =

(12

).

(27) r1 = 1, ξ(1) =

(12

); r2 = 2, ξ(2) =

(1−2

).

(24) (a) Following the steps outlined above, we create the phase portrait below.

Note that the extended eigenvectors appear in bold black, the arrows appear in blue, the axesappear in fine black, and several sample curves appear in red.

(b) Here we sketch the trajectory passing through initial point (2, 3):

21

The point (2, 3) is marked as a red circle.

(c) To sketch x1 versus t or x2 versus t, it is useful to think about the general solution to ourdifferential equation. The general solution is

x =

(x1x2

)= c1e

r1tξ(1) + c2er2tξ(2) = c1e

−t(−12

)+ c2e

−2t(

12

)=

(−c1e−t + c2e

−2t

2c1e−t + 2c2e

−2t

).

Next we use our initial condition

[x(0) =

(23

)]:

x(0) =

(−c1e−(0) + c2e

−2(0)

2c1e−(0) + 2c2e

−2(0)

)=

(−c1 + c22c1 + 2c2

)=

(23

).

This gives us the system of equations−c1 + c2 = 2

2c1 + 2c2 = 3=⇒

c1 = −1/4

c2 = 7/4.

Thus, we see that our particular solution is

x =

(x1x2

)=

(−c1e−t + c2e

−2t

2c1e−t + 2c2e

−2t

)=

(14e−t + 7

4e−2t

− 12e−t + 7

2e−2t

).

Now we can graph x1 and x2 versus t.

(27) (a) Following the steps outlined above, we create the phase portrait below.

22

Note that the extended eigenvectors appear in bold black, the arrows appear in blue, the axesappear in fine black, and several sample curves appear in red.

(b) Here we sketch the trajectory passing through initial point (2, 3):

The point (2, 3) is marked as a red circle.

(c) To sketch x1 versus t or x2 versus t, it is useful to think about the general solution to ourdifferential equation. The general solution is

x =

(x1x2

)= c1e

r1tξ(1) + c2er2tξ(2) = c1e

t

(12

)+ c2e

2t

(1−2

)=

(c1e

t + c2e2t

2c1et − 2c2e

2t

).

Next we use our initial condition

[x(0) =

(23

)]:

x(0) =

(c1e

(0) + c2e2(0)

2c1e(0) − 2c2e

2(0)

)=

(c1 + c2

2c1 − 2c2

)=

(23

).

This gives us the system of equationsc1 + c2 = 2

2c1 − 2c2 = 3=⇒

c1 = 7/4

c2 = 1/4.

Thus, we see that our particular solution is

x =

(x1x2

)=

(c1e

t + c2e2t

2c1et − 2c2e

2t

)=

(74et + 1

4e2t

72et − 1

2e2t

).

Now we can graph x1 and x2 versus t.

23

HOW TO SOLVE EQUATIONS OF THE FORM x′ = Ax (if A has two complex eigenvalues):

(i) Find the eigenvalues of A. We will call the eigenvalues r1 = λ+ µi and r2 = λ− µi.

(ii) Find the eigenvectors corresponding to r1, r2; we will call these eigenvectors ξ(1) = a + bi and ξ(2) =a− bi respectively.

(iii) The general solution is x = c1u + c2v, where we define

u = eλt(a cos(µt)− b sin(µt)) and v = eλt(a sin(µt) + b cos(µt)).

7.6 #1a,3a For each problem, express the general solution of the given system of equations in terms ofreal-valued functions.

(1a) x′ =

(3 −24 −1

)x

(3a) x′ =

(2 −51 −2

)x

(1a) We follow the steps outlined above:

(i) We solve the equation det(A− λI) = 0:

det

(3− λ −2

4 −1− λ

)= 0

=⇒ (3− λ)(−1− λ)− (−2)(4) = 0

=⇒ λ2 − 2λ+ 5 = 0

=⇒ λ =−(−2)±

√(−2)2 − 4(1)(5)

2(1)

=⇒ λ = 1± 2i

So, we have eigenvalues r1 = 1 + 2i and r2 = 1− 2i. In particular, we have λ = 1 and µ = 2.

24

(ii) Next, we find the ξ(1) by row reducing the augmented matrix(A− r1I|0

):

(A− r1I|0

)=

(3− (1 + 2i) −2 0

4 −1− (1 + 2i) 0

)=

(2− 2i −2 0

4 −2− 2i 0

)r1=

12−2i r1−→

(1 −2

2+2i 0

4 −2− 2i 0

)=

(1 −2

2+2i ·2−2i2−2i 0

4 −2− 2i 0

)[rationalizing the denominator]

=

(1 −1−i

2 04 −2− 2i 0

)r2=r2−4r1−→

(1 −1−i

2 00 0 0

)

Now we can “read off” the equation from our matrix:

x1 +−1− i

2x2 = 0.

We can solve this equation to get

x1 =1 + i

2x2.

Then we arbitrarily select x2 = 1, and we get x1 = 1+i2 . Thus, we have the eigenvector

ξ(1) =

(x1x2

)=

((1 + i)/2

1

)=

(1/21

)+

(1/20

)i.

In other words, we have a =

(1/21

)and b =

(1/20

).

SHORTCUT: Once we have one eigenvector from a matrix with complex eigenvalues, we get forfree that the other eigenvector is the complex conjugate of that eigenvector. In other words, wenow know that

ξ(2) =

(1/21

)−(

1/20

)i.

(iii) Finally, we use our formula to conclude that the general solution is

x = c1u + c2v

= c1eλt(a cos(µt)− b sin(µt)) + c2e

λt(a sin(µt) + b cos(µt))

= c1et

[(1/21

)cos(2t)−

(1/20

)sin(2t)

]+ c2e

t

[(1/21

)sin(2t) +

(1/20

)cos(2t)

]=

(c1e

t[12 cos(2t)− 1

2 sin(2t)]

+ c2et[12 sin(2t) + 1

2 cos(2t)]

c1et cos(2t) + c2e

t sin(2t)

)

(3a) We follow the steps outlined above:

25


det

(2− λ −5

1 −2− λ

)= 0

=⇒ (2− λ)(−2− λ)− (−5)(1) = 0

=⇒ λ2 + 1 = 0

=⇒ λ = ±i

So, we have eigenvalues r1 = i and r2 = −i. In particular, we have λ = 0 and µ = 1.


):

(A− r1I|0

)=

(2− i −5 0

1 −2− i 0

)r1=

12−i r1−→

(1 −5

2−i 0

1 −2− i 0

)=

(1 −5

2−i ·2+i2+i 0

1 −2− i 0


=

(1 −2− i 01 −2− i 0

)r2=r2−r1−→

(1 −2− i 00 0 0

)


x1 + (−2− i)x2 = 0.

We can solve this equation to getx1 = (2 + i)x2.

Then we arbitrarily select x2 = 1, and we get x1 = 2 + i. Thus, we have the eigenvector

ξ(1) =

(x1x2

)=

(2 + i

1

)=

(21

)+

(10

)i.


(21

)and b =

(10

).


ξ(2) =

(21

)−(

10

)i.


x = c1u + c2v



= c1

[(21

)cos t−

(10

)sin t

]+ c2

[(21

)sin 2t+

(10

)cos t

]=

(c1 [2 cos t− sin t] + c2 [2 sin t+ cos t]

c1 cos t+ c2 sin t

)

26

7.6 #9,10 For each problem, find the solution of the given initial value problem.

(9) x′ =

(1 −51 −3

)x, x(0) =

(11

)

(10) x′ =

(−3 2−1 −1

)x, x(0) =

(1−2

)

(9) We follow the steps outlined above:


det

(1− λ −5

1 −3− λ

)= 0

=⇒ (1− λ)(−3− λ)− (−5)(1) = 0

=⇒ λ2 + 2λ+ 2 = 0

=⇒ λ =−(2)±

√(2)2 − 4(1)(2)

2(1)

=⇒ λ = −1± i

So, we have eigenvalues r1 = −1− i and r2 = −1 + i. In particular, we have λ = −1 and µ = 1.


):

(A− r1I|0

)=

(1− (−1 + i) −5 0

1 −3− (−1 + i) 0

)=

(2− i −5 0

1 −2− i 0

)r1=

12−i r1−→

(1 −5

2−i 0

1 −2− i 0

)=

(1 −5

2−i ·2+i2+i 0

1 −2− i 0


=

(1 −2− i 01 −2− i 0

)r2=r2−r1−→

(1 −2− i 00 0 0

)


x1 + (−2− i)x2 = 0.

We can solve this equation to getx1 = (2 + i)x2.

Then we arbitrarily select x2 = 1, and we get x1 = 2 + i. Thus, we have the eigenvector

ξ(1) =

(x1x2

)=

(2 + i

1

)=

(21

)+

(10

)i.


(21

)and b =

(10

).

27


ξ(2) =

(21

)−(

10

)i.


x = c1u + c2v



= c1e−t[(

21

)cos t−

(10

)sin t

]+ c2e

−t[(

21

)sin t+

(10

)cos t

]=

(c1e−t [2 cos t− sin t] + c2e

−t [2 sin t+ cos t]c1e−t cos t+ c2e

−t sin t

)

Next, we use the initial condition:

x(0) =

(c1e−(0) [2 cos(0)− sin(0)] + c2e

−(0) [2 sin(0) + cos(0)]c1e−(0) cos(0) + c2e

−(0) sin(0)

)=

(2c1 + c2c1

)=

(11

)We can solve this to find c1 = 1, c2 = −1. Thus, we plug this back into our general solution to get

x =

(e−t [2 cos t− sin t]− e−t[2 sin t+ cos t]

e−t cos t− e−t sin t

)=

(e−t[cos t− 3 sin t]e−t[cos t− sin t]

)We can note that both terms of x tend toward 0 (because of the e−t terms), so x will tend toward theorigin as t→∞.

(10) We follow the steps outlined above:


det

(−3− λ 2−1 −1− λ

)= 0

=⇒ (−3− λ)(−1− λ)− (2)(−1) = 0

=⇒ λ2 + 4λ+ 5 = 0

=⇒ λ =−(4)±

√(4)2 − 4(1)(5)

2(1)= −2± i

So, we have eigenvalues r1 = −2 + i and r2 = −2− i. In particular, we have λ = −2 and µ = 1.

28


):(

A− r1I|0)

=

(−3− (−2 + i) 2 0

−1 −1− (−2 + i) 0

)=

(−1− i 2 0−1 1− i 0

)r1=

1−1−i r1−→

(1 2

−1−i 0

−1 1− i 0

)=

(1 2

−1−i ·−1+i−1+i 0

−1 1− i 0


=

(1 −1 + i 0−1 1− i 0

)r2=r2+r1−→

(1 −1 + i 00 0 0

)


x1 + (−1 + i)x2 = 0.

We can solve this equation to getx1 = (1− i)x2.

Then we arbitrarily select x2 = 1, and we get x1 = 1− i. Thus, we have the eigenvector

ξ(1) =

(x1x2

)=

(1− i

1

)=

(11

)+

(−10

)i.


(11

)and b =

(−10

).


ξ(2) =

(11

)+

(10

)i.


x = c1u + c2v



= c1e−2t[(

11

)cos t−

(−10

)sin t

]+ c2e

−2t[(

11

)sin t+

(−10

)cos t

]=

(c1e−2t [cos t+ sin t] + c2e

−2t [sin t− cos t]c1e−2t cos t+ c2e

−2t sin t

)

Next, we use the initial condition:

x(0) =

(c1e−2(0) [cos(0) + sin(0)] + c2e

−2(0) [sin(0)− cos(0)]c1e−2(0) cos(0) + c2e

−2(0) sin(0)

)=

(c1 − c2c1

)=

(1−2

)We can solve this to find c1 = −2, c2 = −3. Thus, we plug this back into our general solution to get

x =

((−2)e−2t [cos t+ sin t] + (−3)e−2t [sin t− cos t]

(−2)e−2t cos t+ (−3)e−2t sin t

)=

(e−2t[cos t− 5 sin t]

e−2t[−2 cos t− 3 sin t]

)We can note that both terms of x tend toward 0 (because of the e−2t terms), so x will tend towardthe origin as t→∞.

29

7.6 #13ab

x′ =

(α 1−1 α

)x

(a) Determine the eigenvalues in terms of α.

(b) Find the critical value or values of α where the qualitative nature of the phase portrait for the systemchanges.

(c) Draw a phase portrait for a value of α slightly below, and for another value slightly above, each criticalvalue.

(a) Let A =

(α 1−1 α

). To find the eigenvalues, we solve the following equation:

det(A− λI) = 0

=⇒ det

(α− λ 1−1 α− λ

)= 0

=⇒ (α− λ)(α− λ)− (1)(−1) = 0

=⇒ λ2 − 2αλ+ (α2 + 1) = 0

=⇒ λ =−(−2α)±

√(−2α)2 − 4(1)(α2 + 1)

2(1)

=⇒ λ =2α±

√−4

2=⇒ λ = α± i

(b) Let’s think about the general solution. Since our eigenvalues are complex, we can think about phaseportrait for complex eigenvalues. We know that if we have eigenvalue λ + µi, the phase portrait willdepend on λ. If λ > 0, then the phase portrait spirals outward. If λ = 0, the phase portrait will looparound in ellipses. If λ < 0, the phase portrait will spiral inward. Thus, we can see that α = 0 is thecritical value of α at which the qualitative nature of the phase portrait changes.

(c) First, consider α = 1. We would see that the phase portrait spirals out. Moreover, we can plug x =

(01

)into our original differential equation to see

x′ =

(1 1−1 1

)(01

)=

(11

),

so the spiral moves clockwise. Thus, the graph below shows a phase portrait for α = 1:

30

Next, consider α = −1. We would see that the phase portrait spirals out. Moreover, we can plug

x =

(01

)into our original differential equation to see

x′ =

(−1 1−1 −1

)(01

)=

(1−1

),

so the spiral moves clockwise. Thus, the graph below shows a phase portrait for α = 1:

7.6 #26

The electric circuit shown in the figure above is described by the system of differential equations

d

dt

(IV

)=

(0 1

L− 1C − 1

RC

)(IV

), (3)

where I is the current through the inductor and V is the voltage drop across the capacitor. These differentialequations were derived in Problem 19 of Section 7.1.

(a) Show that the eigenvalues of the coefficient matrix are real and different if L > 4R2C; show that theyare complex conjugates if L < 4R2C.

(b) Suppose that R = 1Ω, C = 12F , and L = 1H. Find the general solution of the system (3) in this case.

(c) Find I(t) and V (t) if I(0) = 2A and V (0) = 1V .

(d) For the circuit of part (b) determine the limiting values of I(t) and V (t) as t→∞. Do these limitingvalues depend on the initial condition?

(a) Let x =

(IV

), and let A =

(0 1

L− 1C − 1

RC

). Then we see that our differential equation is of the form

31

x′ = Ax. Let’s find the eigenvalues of A:

det(A− λI) = det

(0− λ 1

L− 1C − 1

RC − λ

)= 0

=⇒ (−λ)

(− 1

RC− λ)−(

1

L

)(− 1

C

)= 0

=⇒ λ2 +

(1

RC

)λ+

(1

LC

)= 0

=⇒ λ =−(

1RC

)±√(

1RC

)2 − 4(1)(

1LC

)2(1)

=⇒ λ =−(

1RC

)±√

1R2C2 − 4

LC

2(1)

=⇒ λ =−(

1RC

)±√

L−4R2CLR2C2

2

We see that the radical expression here is real and nonnegative if and only if L− 4R2C is positive. So,the eigenvalues will be real and different if and only if L − 4R2C > 0, which is equivalent to if andonly if L > 4R2C. Similarly, we see that the radical term will be negative (yielding complex conjugateeigenvalues) if and only if L− 4R2C is negative, which is if and only if L < 4R2C.

(b) Plugging in R = 1, C = 1/2, and L = 1, we use our work from part a to get that the eigenvalues are

λ =−(

1RC

)±√

L−4R2CLR2C2

2(1)=−(

11(1/2)

)±√

1−4(1)2(1/2)(1)(1)2(1/2)2

2(1)=−2±

√−4

2= −1± i

So, we have eigenvalues r1 = −1 + i and r2 = −1− i. In particular, we see that λ = −1 and µ = 1.

Next, we find the eigenvector corresponding to r1 = −1 + i. To do this, we row reduce the augmentedmatrix (A− λI|0):(

0− (−1 + i) 1L 0

− 1C − 1

RC − (−1 + i) 0

)=

(1− i 1

L 0− 1C 1− 1

RC − i 0

)=

(1− i 1 0−2 −1− i 0

)r1=

11−i r1−→

(1 1

1−i 0

−2 −1− i 0

)=

(1 1

1−i ·1+i1+i 0

−2 −1− i 0


=

(1 1+i

2 0−2 −1− i 0

)r2=r2+2r1−→

(1 1+i

2 00 0 0

)Next, we “read off” the equation this gives us:

x1 +

(1 + i

2

)x2 = 0.

We can solve this to get x1 = −1−i2 x2. Then, we can choose x2 = 1 and solve to get x1 = −1−i

2 . Thus,we have the eigenvector

ξ(1) =

(x1x2

)=

((−1− i)/2

1

)=

(−1/2

1

)+

(−1/2

0

)i.

32


(−1/2

1

)and b =

(−1/2

0

).

Thus, the general solution is

x = c1u + c2v



= c1e−t[(−1/2

1

)cos t−

(−1/2

0

)sin t

]+ c2e

−t[(−1/2

1

)sin t+

(−1/2

0

)cos t

]=

(c1e−t [− 1

2 cos t+ 12 sin t

]+ c2e

−t [− 12 sin t− 1

2 cos t]

c1e−t cos t+ c2e

−t sin t

)

(c) We use our initial condition to see that(I(0)V (0)

)= x(0) =

(c1e−(0) [− 1

2 cos(0) + 12 sin(0)

]+ c2e

−(0) [− 12 sin(0)− 1

2 cos(0)]

c1e−(0) cos(0) + c2e

−(0) sin(0)

)=

(− 1

2c1 −12c2

c1

)=

(21

)We can solve this system of equations to get c1 = 1 and c2 = −5. Thus, we get the solution

x =

(c1e−t [− 1

2 cos t+ 12 sin t

]+ c2e

−t [− 12 sin t− 1

2 cos t]

c1e−t cos t+ c2e

−t sin t

)=

(1e−t

[− 1

2 cos t+ 12 sin t

]− 5e−t

[− 1

2 sin t− 12 cos t

]1e−t cos t− 5e−t sin t

)=

(e−t[2 cos t+ 3 sin t]e−t[cos t− 5 sin t]

)

(d) Looking back at our general solution for part (b), we see that both I and V tend toward zero (becauseof the ”e−t” terms), regardless of c1 and c2. I other words, I(t) and V (t) tend toward 0 as t→∞, andthis limiting value does not depend on the initial conditions.

7.6 #28ab A mass m on a spring with constant k satisfies the differential equation

mu′′ + ku = 0,

where u(t) is the displacement at time t of the mass from its equilibrium position.

(a) Let x1 = u, x2 = u′, and show that the resulting system is

x′ =

(0 1

−k/m 0

)x.

(b) Find the eigenvalues of the matrix for the system in part (a).

(a) We start by transforming our differential equation into a system of first-order differential equations.Using x1 = u and x2 = u′, we can also note that u′′ = x′2. Thus, we can replace the original differentialequation with mx′2 + kx1 = 0. We also have the relationship x2 = x′1. Let’s rewrite this as a system ofequations:

x′1 = x2

x′2 = (−k/m)x1=⇒

x′1 = 0x1 + 1x2

x′2 = (−k/m)x1 + 0x1.

33

Note that if we let x =

(x1x2

), we can rewrite this system as the following matrix equation:

(x′1x′2

)= x′ =

(0 1

−k/m 0

)x,

as desired.

(b) We find the eigenvalues by solving the following equation:

det(A− λI) = det

(0− λ 1−k/m 0− λ

)= 0

=⇒ (−λ)(−λ)− (1)(−k/m) = 0

=⇒ λ2 + k/m = 0

=⇒ λ = ±i√k/m

34

20d - homework assignment 5

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