Math 2280 - Assignment 6

Dylan Zwick

Summer 2013

Section 3.7 - 1, 5, 10, 17, 19

Section 3.8 - 1, 3, 5, 8, 13

Section 4.1 - 1, 2, 13, 15, 22

Section 4.2 - 1, 10, 19, 28

1

Section 3.7 - Electrical Circuits

3.7.1 This problem deals with the RL circuit pictured below. It is a series circuit containing an inductor with an inductance of L henries,a resistor with a resistance of R ohms, and a source of electromotiveforce (emf), but no capacitor. In this case the equation governing oursystem is the first-order equation

LI’ + RI = E(t).

C

Suppose that L = 511, R = 25Q, and the source E of emf is a batterysupplying 1OOl to the circuit. Suppose also that the switch has beenin position 1 for a long time, so that a steady current of 4A is flowingin the circuit. At time t = 0, the switch is thrown to position 2, sothatl(0)=r4andE=Ofort>0. FindI(t).

Solution - The ODE that describes this system is:

We can rewrite this as:

51’ + 251 = 0.

I’ + SI = 0.

2

This is a first-order linear ODE, and the corresponding integrating

factor is ρ(t) = eR

5dt = e5t. Multiplying both sides of the ODE by thisintegrating factor we get:

e5tI ′ + 5e5tI = 0 ⇒ d

dt

(

e5tI)

= 0.

Integrating both sides of the above equation we get:

e5tI = C ⇒ I(t) = Ce−5t.

If we plug in the initial condition I(0) = 4 = C we get the finalsolution:

I(t) = 4e−5t.

3

3.7.5 - In the circuit from Problem 3.7.1, with the switch in position 1, sup-pose that E(t) = 100e−10t cos 60t, R = 20, L = 2, and I(0) = 0. FindI(t).

Solution - The differential equation governing this system will be:

2I ′ + 20I = 100e−10t cos (60t).

If we divide both sides by 2 we get the ODE:

I ′ + 10I = 50e−10t cos (60t).

The integrating factor for this first-order linear ODE will be ρ(t) =e

R

10dt = e10t. Multiplying both sides of our equation by this integrat-ing factor we get:

d

dt

(

e10tI)

= 50 cos (60t).

Integrating both sides of this equation gives us:

e10tI =5

6sin (60t) + C

⇒ I(t) =5

6e−10t sin (60t) + Ce−10t.

If we plug in the initial condition I(0) = 0 we get C = 0, and oursolution is:

I(t) =5

6e−10t sin (60t).

4

3.7.10 - This problem deals with an RC circuit pictured below, containinga resistor (R ohms), a capacitor (C farads), a switch, a source of emf,but no inductor. This system is governed by the linear first-orderdifferential equation

dQ 1R-- + = E(t).

for the charge Q = Q(t) on the capacitor at time t. Note that 1(t)Q’(t).

C

Suppose an emf of voltage E(t) = B0 cos wt is applied to the RCcircuit at time t = 0 (with the switch closed), and Q(0) = 0. SubstituteQ5(t) = A cos wt + B sin wt in the differential equation to show thatthe steady periodic charge on the capacitor is

E0CQ5 (t) =

__________

cos (wt—

3)p

where /3 = tan’ (wRC).

5

Solution - Our steady periodic charge will be of the form:

Qsp = A cos (ωt) + B sin (ωt).

Its derivative will is:

Q′sp = −ωA sin (ωt) + ωB cos (ωt).

If we plug these values into the ODE:

RQ′sp +

1

CQsp = E0 cos (ωt),

we get:

−ARω sin (ωt) + BRω cos (ωt) +A

Ccos (ωt) +

B

Csin (ωt) = E0 cos (ωt).

Grouping like terms together and multiplying both sides by C wehave:

(B − AωRC) sin (ωt) + (A + BωRC) cos (ωt) = CE0 cos (ωt).

Equating coefficients we get B − AωRC = 0, which means B =AωRC. Plugging these into the above equation gives us:

A(1 + ω2R2C2) cos (ωt) = CE0 cos (ωt).

So, A =E0C

1 + ω2R2C2, and B =

E0C2ωR

1 + ω2R2C2. From this we get:

6

Qsp =E0C

1 + ω2R2C2(cos (ωt) + ωRC sin (ωt)).

If we define α = tan−1(ωRC) then

sin α =ωRC√

1 + ω2R2C2,

cos α =1√

1 + ω2R2C2.

Using these we can rewrite Qsp as:

Qsp =E0C√

1 + ω2R2C2(cos (α) cos (ωt) + sin (α) sin (ωt)).

If we use the trigonometric identity

cos (θ − φ) = cos (θ) cos (φ) + sin (θ) sin (φ)

we can rewrite the above equation as:

Qsp =E0C√

1 + ω2R2C2cos (ωt− α).

This is our desired form.

7

3.7.17 For the RLC circuit pictured below find the current 1(t) using thegiven values of R, L, C and V(t), and the given initial values.

L

(7C

R = 16Q, L = 2H, C = .02F;

E(t) = 100V; 1(0) = 0, Q(0) 5.

Solution - The differential equation that governs this system is:

We can rewrite this as:

21’+161-i-50Q= 100.

I’ + 81 + 25Q 50.

The corresponding homogeneous equation is:

I’ + 81 + 25Q = 0.

The characteristic polynomial for this linear homogeneous ODE is:

r+8r+25,

8

which has roots:

−8 ±√

82 − 4(1)(25)

2= −4 ± 3i.

So, the corresponding homogeneous solution is:

Qh(t) = c1e−4t cos (3t) + c2e

−4t sin (3t).

As for a particular solution, we guess our particular solution is of theform Qp(t) = A. Plugging this into our ODE we get:

25A = 50,

so A = 2. Therefore the solution to our differential equation is:

Q(t) = c1e−4t cos (3t) + c2e

−4t sin (3t) + 2.

Its derivative is:

I(t) =−3c1e

−4t sin (3t) − 4c1e−4t cos (3t) + 3c2e

−4t cos (3t) − 4c2e−4t sin (3t).

Plugging in our initial conditions gives us:

Q(0) = c1 + 2 = 5,

I(0) = −4c1 + 3c2 = 0.

Solving this system of equations gives us c1 = 3, c2 = 4. So, oursolution is:

9

Q(t) = 3e−4t cos (3t) + 4e−4t sin (3t) + 2,

and

I(t) = −25e−4t sin (3t).

10

3.7.19 Same instructions as Problem 3.7.17, but with the values:

R = 60Ω, L = 2H , C = .0025F ;

E(t) = 100e−10tV ; I(0) = 0, Q(0) = 1.

Solution - The differential equation that governs this circuit is:

2I ′ + 60I + 400Q = 100e−10t.

We can rewrite this ODE as:

I ′ + 30I + 200Q = 50e−10t.

The corresponding homogeneous equation is:

I ′ + 30I + 200Q = 0.

This has characteristic polynomial:

r2 + 30r + 200 = (r + 20)(r + 10).

The roots of this polynomial are: r = −10,−20, and so the homoge-neous solution is:

Qh = c1e−20t + c2e

−10t.

11

As for a particular solution, our first guess is the particular solutionwill be of the form Ae−10t, but this would not be linearly independentof our homogeneous solution, so we need to use the guess Ate−10t.With this guess we get:

Qp = Ate−10t,

Q′p = A(−10te−10t + e−10t),

Q′′p = A(100te−10t − 20e−10t).

Plugging these into the ODE we get:

100Ate−10t − 20Ae−10t − 300Ate−10t + 30Ae−10t + 200Ate−10t =10Ae−10t = 50e−10t

⇒ A = 5.

So, our solution is:

Q(t) = 5te−10t + c1e−20t + c2e

−10t.

Its derivative is:

Q′(t) = I(t) = −50te−10t + 5e−10t − 20c1e−20t − 10c2e

−10t.

Plugging in our initial conditions we get:

Q(0) = c1 + c2 = 1,

I(0) = 5 − 20c1 − 10c2 = 0.

Solving this system we get c1 =3

2, c2 = −1

2, and our current will be:

I(t) = 10e−20t − 10e−10t − 50te−10t.

12

Section 3.8 - Endpoint Problems and Eigenvalues

3.8.1 For the eigenvalue problem

y′′ + λy = 0; y′(0) = 0, y(1) = 0,

first determine whether λ = 0 is an eigenvalue; then find the positiveeigenvalues and associated eigenfunctions.

Solution - First, if λ = 0 then the solution to the differential equation

y′′ = 0

is

y = Ax + B.

From this we get y′ = A, and so if y′(0) = 0 we must have A = 0. Thiswould mean y = B, and if y(1) = 0 then B = 0. So, only the trivialsolution A = B = 0 works, and therefore λ = 0 is not an eigenvalue.

For λ > 0 the characteristic polynomial for our linear differentialequation is:

r2 + λ = 0,

which has roots r = ±√−λ. The corresponding solution to our ODE

will be:

y = A cos (√

λx) + B sin (√

λx).

with derivative

y′ = −A√

λ sin (√

λx) + B√

λ cos (√

λx).

13

So, y′(0) = B√

λ, and therefore if y′(0) = 0 then we must have B = 0,as λ > 0. So, our solution must be of the form:

y = A cos (√

λx).

If we plug in y(1) = 0 we get:

y(1) = A cos (√

λ) = 0.

If A 6= 0 we must have cos(√

λ) = 0, which is true only if√

λ =π

2+ nπ. So, the eigenvalues are:

λn =

(

π

(

1

2+ n

))2

, with n ∈ N,

and corresponding eigenfunctions

yn = cos((π

2+ nπ

)

x)

.

14

3.8.3 Same instructions as Problem 3.8.1, but for the eigenvalue problem:

y′′ + λy = 0; y(−π) = 0, y(π) = 0.

Solution - If λ = 0 then, as in Problem 3.8.1, our solution will be ofthe form:

y = Ax + B.

This means y(π) = Aπ + B = 0, and y(−π) = −Aπ + B = 0. Addingthese two equations we get 2B = 0, which means B = 0. If B = 0then Aπ = 0, which means A = 0. So, the only solution is the trivialsolution A = B = 0, and therefore λ = 0 is not an eigenvalue.

Now if λ > 0 then again just as in Problem 3.8.1 we’ll have a solutionof the form:

y(x) = A cos (√

λx) + B sin (√

λx).

If we plug in our endpoint values we get:

y(π) = A cos (√

λπ) + B sin (√

λπ) = 0,

y(−π) = A cos (−√

λπ) + B sin (−√

λπ) =

A cos (√

λπ) − B sin (√

λπ) = 0,

where in the second line above we use that cos is an even function,while sin is odd.

If we add these two equations together we get:

2A cos (√

λπ) = 0.

15

This is true if either A = 0 or√

λ =

(

1

2+ n

)

. If√

λ =

(

1

2+ n

)

then

y(π) = B sin

((

1

2+ n

)

π

)

= 0.

As sin

((

1

2+ n

)

π

)

= ±1 we must have B = 0.

On the other hand, if A = 0 above then we have:

y(π) = B sin (√

λπ).

If B 6= 0 then we must have√

λ = n. Combining our two results weget that the possible eigenvalues are:

λn =n2

4,

for n ∈ N, and n > 0, with corresponding eigenfunctions:

yn(x) =

cos(

n2x)

n oddsin

(

n2x)

n even

16

3.8.5 Same instructions as Problem 3.8.1, but for the eigenvalue problem:

y′′ + λy = 0; y(−2) = 0, y′(2) = 0.

Solution - If λ = 0 then, just as in Problem 3.8.1, the solution to theODE will be:

y(x) = Ax + B,

y′(x) = A.

If we plug in our endpoint conditions we get y(−2) = −2A + B = 0and y′(2) = A = 0. These equations are satisfied if and only if A =B = 0, which is the trivial solution. So, λ = 0 is not an eigenvalue.

If λ > 0 then, just as in Problem 3.8.1, the solution to the ODE will beof the form:

y(x) = A cos (√

λx) + B sin (√

λx),

with

y′(x) = −A√

λ sin (√

λx) + B√

λ cos (√

λx).

Plugging in the endpoint conditions, and using that cos is even andsin is odd, we get:

y(−2) = A cos (−2√

λ) + B sin (−2√

λ) =

A cos (2√

λ) − B sin (2√

λ) = 0,

y′(2) = −A√

λ sin (2√

λ) + B√

λ cos (2√

λ) = 0.

If we divide both sides of the second equality by√

λ we get

17

−A sin (2√

λ) + B cos (2√

λ) = 0.

From these equations we get:

A cos (2√

λ) = B sin (2√

λ) ⇒ A

B= tan (2

√λ),

B cos (2√

λ) = A sin (2√

λ) ⇒ B

A= tan (2

√λ).

So,

A

B=

B

A⇒ A2 = B2.

So, either A = B or A = −B.

If A = B then tan (2√

λ) = 1, which means 2√

λ =π

4+ nπ, and

therefore

λ =

((

1 + 4n

8

)

π

)2

.

If A = −B then tan (2√

λ) = −1, which means 2√

λ =3π

4+ nπ, and

therefore

λ =

((

3 + 4n

8

)

π

)2

.

So, the eigenvalues are:

18

λn =

((

1 + 2n

8

)

π

)2

with n ∈ N and n > 0, with corresponding eigenfunctions:

yn =

cos((

1+2n8

)

πx)

+ sin((

1+2n8

)

πx)

n evencos

((

1+2n

8

)

πx)

− sin((

1+2n

8

)

πx)

n odd

19

3.8.8 - Consider the eigenvalue problem

y′′ + λy = 0; y(0) = 0 y(1) = y′(1) (not a typo).;

all its eigenvalues are nonnegative.

(a) Show that λ = 0 is an eigenvalue with associated eigenfunctiony0(x) = x.

(b) Show that the remaining eigenfunctions are given by yn(x) =sin βnx, where βn is the nth positive root of the equation tan z =z. Draw a sketch showing these roots. Deduce from this sketchthat βn ≈ (2n + 1)π/2 when n is large.

Solution -

(a) - If λ = 0 then the solution to the ODE will be of the form:

y(x) = Ax + B,

with

y′(x) = A.

So, y(0) = B = 0, and y(1) = A = y′(1). So, any function of theform y(x) = Ax will work, and our eigenfunction for λ = 0 is:

y0 = x.

(b) - For λ > 0 the solutions will all be of the form:

y(x) = A cos (λx) + B sin (λx).

If we plug in y(0) = A = 0 we get the solutions are of the form:

20

y(x) = Bsin(\/5x),

with

= Bvcos (\/I).

If we plug in the other endpoint values we get:

y(l) = Bsin(v) = BVcos(v’) = y’(l).

If B 0 then we must have:

tan(/)=

So, v”X works if it’s a root of the equation tan z = z, and if 8, is

the iith such root, then the associated eigenfunction is:

A sketchintersect:

sin (/3x).

of z and tan z are below. The roots are where they

As ii gets large it occurs at approximately

L

/277 + iN2

J7F.

21

3.8.13 - Consider the eigenvalue problem

y′′ + 2y′ + λy = 0; y(0) = y(1) = 0.

(a) Show that λ = 1 is not an eigenvalue.

(b) Show that there is no eigenvalue λ such that λ < 1.

(c) Show that the nth positive eigenvalue is λn = n2π2 + 1, withassociated eigenfunction yn(x) = e−x sin (nπx).

Solution -

(a) - If λ = 1 then the characteristic polynomial is:

r2 + 2r + 1 = (r + 1)2,

which has roots r = −1,−1. So, −1 is a root with multiplicity 2. Thecorresponding solution to the ODE will be:

y(x) = Ae−x + Bxe−x.

If we plug in the endpoint values we get:

y(0) = A = 0,

y(1) = Ae−1 + Be−1 = Be−1 = 0.

From these we see the only solution is the trivial solution A = B = 0,so λ = 1 is not an eigenvalue.

22

(b) - If λ < 1 then the characteristic polynomial will be:

r2 + 2r + λ,

which has roots

r =−2 ±

√

22 − 4(1)λ

2= −1 ±

√1 − λ.

If λ < 1 then√

1 − λ will be real, and the solution to our ODE will beof the form:

y(x) = Ae(−1+√

1−λ)x + Be(−1−√

1−λ)x.

Plugging in our endpoint values we get:

y(0) = A + B = 0,

y(1) = Ae−1+√

1−λ + Be−1−√

1−λ = 0.

From these we get, after a little algebra:

A(1 − e−2√

1−λ) = 0.

If λ < 1 then e−2√

1−λ < 1, and therefore 1 − e2√

1−λ > 0. So, for theabove equality to be true we must have A = 0, which means B = 0,and so the only solution is the trivial solution A = B = 0. Therefore,no value λ < 1 is an eigenvalue.

(c) - If λ > 1 then again using the roots from the quadratic equation inpart (b) we get that our solutions will be of the form:

23

y(x) = Ae−x cos (√

λ − 1x) + Be−x sin (√

λ − 1x).

If we plug in the endpoint values we get:

y(0) = A = 0,

and so

y(x) = Be−x sin (√

λ − 1x).

If we plug in our other endpoint value we get:

y(1) = Be−1 sin (√

λ − 1) = 0.

If B 6= 0 then we must have sin (√

λ − 1) = 0, which is only possibleif

√λ − 1 = nπ,

⇒ λn = n2π2 + 1.

So, the eigenvalues are given above, and the corresponding eigen-functions are:

yn = e−x sin (nπx),

for n ∈ N, n > 0.

24

Section 4.1 - First-Order Systems and Applications

4.1.1 - Transform the given differential equation into an equivalent systemof first-order differential equations.

x′′ + 3x′ + 7x = t2.

Solution - If we define x = x1 then define:

x′1 = x2,

x′2 = t2 − 3x2 − 7x1.

So, the system is:

x′1 = x2

x′2 = −7x1 − 3x2 + t2

.

25

4.1.2 - Transform the given differential equation into an equivalent systemof first-order differential equations.

x(4) + 6x′′ − 3x′ + x = cos 3t.

Solution - Define x = x1. Then the equivalent system is:

x′1 = x2

x′2 = x3

x′3 = x4

x′4 = −6x3 + 3x2 − x1 + cos (3t)

.

26

4.1.13 - Find the particular solution to the system of differential equationsbelow. Use a computer system or graphing calculator to construct adirection field and typical solution curves for the given system.

x′ = −2y, y′ = 2x; x(0) = 1, y(0) = 0.

Solution - If we differentiate y′ = 2x, we get y′′ = 2x′ = −4y. So, wehave the differential equation:

y′′ + 4y = 0.

The solution to this ODE is:

y(t) = A cos (2t) + B sin (2t).

Now,

x(t) =1

2y′ =

1

2(−2A sin (2t) + 2B cos (2t)) = −A sin (2t) + B cos (2t).

If we plug in x(0) = B = 1 and y(0) = A = 0 we get:

x(t) = cos (2t)

y(t) = sin (2t).

27

More room, if necessary, for Problem 4.1.13.

Pirec

/‘4

o(

lviV

7’Ndfe; 5hoI) beCI(cIJ 40

A 1• Ibei

19

4.1.15 - Find the general solution to the system of differential equationsbelow. Use a computer system or graphing calculator to construct adirection field and typical solution curves for the given system.

x′ =1

2y, y′ = −8x.

Solution - If we differentiate y′ = −8x we get y′′ = −8x′ = −4y. So,our ODE is:

y′′ + 4y = 0.

The solution to this ODE is:

y(t) = A cos (2t) + B sin (2t).

The function x(t) is:

x(t) = −1

8y′(t) = −A

4sin (2t) +

B

4cos (2t).

So, the general solution to this system of ODEs is:

x(t) = −A

4sin (2t) +

B

4cos (2t)

y(t) = A cos (2t) + B sin (2t).

29

More room, if necessary, for Problem 4.1.15.

Pirecio

21

4.1.22 (a) - Beginning with the general solution of the system from Prob-lem 13, calculate x2 + y2 to show that the trajectories are circles.

(b) - Show similarly that the trajectories of the system from Problem15 are ellipses of the form 16x2 + y2 = C2.

(a) - The general solution to the system of ODEs from Problem 4.1.13 is:

x(t) = −A sin (2t) + B cos (2t)

y(t) = A cos (2t) + B sin (2t).

From these we get:

x(t)2 + y(t)2 = (−A sin (2t) + B cos (2t))2 + (A cos (2t) + B sin (2t))2

= A2 sin2 (2t) − 2AB sin (2t) cos (2t) + B2 cos2 (2t) + A2 cos2 (2t) +2AB sin (2t) cos (2t) + B2 sin2 (2t)

= A2 + B2.

So, circles.

(b) - The general solution to the system of ODEs from Problem 4.1.15 is:

x(t) = −A

4sin (2t) +

B

4cos (2t)

y(t) = A cos (2t) + B sin (2t).

So,

16x(t)2 = A2 sin2 (2t) − 2AB sin (2t) cos (2t) + B2 cos2 (2t),

y(t)2 = A2 cos2 (2t) + 2AB sin (2t) cos (2t) + B2 sin2 (2t).

Combining these we get 16x(t)2 + y(t)2 = A2 + B2 = C2. So, ellipses.

31

Section 4.2 - The Method of Elimination

4.2.1 - Find a general solution to the linear system below. Use a computersystem or graphing calculator to construct a direction field and typi-cal solution curves for the system.

x′ = −x + 3yy′ = 2y

Solution - The differential equation

y′ = 2y

has the solution

y(t) = Ae2t.

So,

x′ = −x + 3Ae2t ⇒ x′ + x = 3Ae2t.

This is a first-order linear ODE. Its integrating factor is:

ρ(t) = eR

1dt = et.

Multiplying both sides by this integrating factor our linear ODE be-comes:

d

dt

(

etx)

= 3Ae3t.

32

Integrating both sides we get:

etx = Ae3t + B

x Ae2 + Be.

So, the general solution to this system is:

x(t) = Ae2t + Be_t,

j(t) = Ae2t.

We can write this in vector form as:

(I)4(i)2t+B(i)t

looks kind of like this:The direction field

33

4.2.10 Find a particular solution to the given system of differential equa-tions that satisfies the given initial conditions.

x′ + 2y′ = 4x + 5y,

2x′ − y′ = 3x;

x(0) = 1, y(0) = −1.

Solution - If we add 2 times the second equation to the first we get:

5x′ = 10x + 5y.

If we subtract 2 times the first equation from the second we get:

−5y′ = −5x − 10y ⇒ 5y′ = 5x + 10y.

Differentiating 5x′ = 10x+5y and plugging in 5y′ = 5x+10y we get:

5x′′ = 10x′ + 5y′ = 10x′ + (5x + 10y)

⇒ 5x′′ = 10x′ + (5x + 10x′ − 20x)

⇒ 5x′′ = 20x′ − 15x

⇒ x′′ = 4x′ − 3x.

The linear homogeneous differential equation x′′ − 4x′ + 3x = 0 hascharacteristic equation:

r2 − 4r + 3 = (r − 3)(r − 1).

34

So, the roots are r = 3, 1, and the general solution to the ODE is:

x(t) = c1e3t + c2e

t.

From the equation 5y′ = 5x + 10y we get y′ = x + 2y, and therefore:

y′ − 2y = c1e3t + c2e

t.

If we multiply both sides by the integrating factor e−2t we get:

d

dt

(

e−2ty)

= c1et + c2e

−t.

Integrating both sides we get:

e−2ty = c1et − c2e

−t + C,

and so:

y(t) = c1e3t − c2e

t + Ce2t.

Plugging this into any of the equations in our system gives us C = 0.So,

y(t) = c1e3t − c2e

t.

We can write this solution in matrix form as:

x(t) = c1

(

11

)

e3t + c2

(

1−1

)

et.

35

If we plug in

x(0) =

(

1−1

)

= c1

(

11

)

+ c2

(

1−1

)

we can see immediately that c1 = 0 and c2 = 1. So, the solution toour initial value problem is:

x(t) =

(

1−1

)

et.

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4.2.19 Find a general solution to the given system of differential equations.

x′ = 4x − 2y,

y′ = −4x + 4y − 2z,

z′ = −4y + 4z.

Solution - If we differentiate the first equation we get:

x′′ = 4x′ − 2y′ = 4x′ − 2(−4x + 4y − 2z)

⇒ x′′ = 4x′ + 8x − 8y + 4z.

Differentiating again we get:

x(3) = 4x′′ + 8x′ − 8y′ + 4z′ = 4x′′ + 8x′ − 8y′ + 4(−4y + 4z)

⇒ x(3) = 4x′′ + 8x′ − 8y′ − 16y + 16z

⇒ x(3) = 4x′′ + 8x′ − 8y′ − 16y + 8(−y′ − 4x + 4y)

⇒ x(3) = 4x′′ + 8x′ − 16y′ + 16y − 32x

⇒ x(3) = 4x′′ + 8x′ − 8(4x′ − x′′) + 8(4x − x′) − 32x

⇒ x(3) = 12x′′ − 32x′ ⇒ x(3) − 12x′′ + 32x′ = 0.

The characteristic equation for this ODE is

r3 − 12r2 + 32r = r(r − 8)(r − 4).

So,

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x(t) = c1 + c2e8t + c3e

4t.

From this we get:

x′(t) = 8c2e8t + 4c3e

4t

and

y(t) = 2x − 1

2x′ = 2c1 − 2c2e

8t.

Finally,

z(t) = −2x′(t) + 2y(t) − 1

2y′(t) = 2c1 + 2c2e

8t − 2c3e4t.

So,

x(t) = c1 + c2e8t + c3e

4t,

y(t) = 2c1 − 2c2e8t,

z(t) = 2c1 + 2c2e8t − 2c3e

4t.

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4.2.28 For the system below first calculate the operational determinant todetermine how many arbitrary constants should appear in a generalsolution. Then attempt to solve the system explicitly so as to findsuch a general solution.

(D2 + D)x + D2y = 2e−t

(D2 − 1)x + (D2 − D)y = 0

Solution - The operational determinant of the system above is:

(D2 +D)(D2 −D)−D2(D2 − 1) = D4 −D3 +D3 −D2 −D4 +D2 = 0.

So, there are 0(!) arbitrary constants. How is this possible? Well, ifwe subtract the second relation from the first we get:

(D + 1)x + Dy = 2e−t

⇒ Dy = 2e−t − (D + 1)x

⇒ D2y = −2e−t − (D2 + D)x

⇒ (D2 + D)x + D2y = −2e−t.

However, this cannot be, as our first relation above is:

(D2 + D)x + D2y = 2e−t,

and 2e−t 6= −2e−t. So, there is no solution to the system.

39