solutions to the sample problems for the midterm exam ucla math

Solutions to the Sample Problems for the Midterm ExamUCLA Math 135, Winter 2015, Professor David Levermore

(1) Give the interval of definition for the solution of the initial-value problem

d3x

dt3+

cos(3t)

4− tdx

dt=

e−2t

1 + t, x(2) = x′(2) = x′′(2) = 0 .

Solution. The coefficient and forcing are both continuous over the interval (−1, 4),which contains the initial time t = 2. The coefficient is not defined at t = 4 whilethe forcing is not defined at t = −1. Therefore the interval of definition is (−1, 4).

(2) Suppose that Y1(t) and Y2(t) are solutions of the differential equation

y′′ + 2y′ + (1 + t2)y = 0 .

Suppose you know that W [Y1, Y2](0) = 5. What is W [Y1, Y2](t)?

Solution. The Abel Theorem states that w(t) = W [Y1, Y2](t) satisfies w′ + 2w = 0.It follows that w(t) = w(0)e−2t. Because we are given w(0) = W [Y1, Y2](0) = 5, wesee that w(t) = 5e−2t. Therefore

W [Y1, Y2](t) = 5e−2t .

(3) Show that the functions Y1(t) = cos(t), Y1(t) = sin(t), and Y3(t) = 1 are linearlyindependent.

Solution. The Wronskian of these functions is

W [Y1, Y2, Y3](t) = det

cos(t) sin(t) 1− sin(t) cos(t) 0− cos(t) − sin(t) 0

= 1 · (− sin(t)) · (− sin(t))− (− cos(t)) · cos(t) · 1= sin(t)2 + cos(t)2 = 1 .

Because W [Y1, Y2, Y3](t) 6= 0, the functions are linearly independent.

Alternative Solution. Suppose that

0 = c1Y1(t) + c2Y2(t) + c3Y3(t) = c1 cos(t) + c2 sin(t) + c3 .

To show linear independence we must show that c1 = c2 = c3 = 0. But setting t = 0,t = π, and t = π

2into this relation yields the linear algebraic system

0 = c1 cos(0) + c2 sin(0) + c3 = c1 + c3 ,

0 = c1 cos(π) + c2 sin(π) + c3 = −c1 + c3 ,

0 = c1 cos(π

2) + c2 sin(

π

2) + c3 = c2 + c3 .

By subtracting the second equation from the first we see that c1 = 0. By adding thefirst two equations we see that c3 = 0. By plugging c3 = 0 into the thrid equationwe see that c2 = 0, Therefore c1 = c2 = c3 = 0, whereby Y1(t), Y2(t), and Y3(t) arelinearly independent.

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(4) Let L be a linear ordinary differential operator with constant coefficients. Supposethat all the roots of its characteristic polynomial (listed with their multiplicities) are−2 + i3, −2− i3, i7, i7, −i7, −i7, 5, 5, 5, −3, 0, 0.(a) Give the order of L.

Solution. There are 12 roots listed, so the degree of the characteristic polyno-mial is 12, whereby the order of L is 12.

(b) Give a general real solution of the homogeneous equation Ly = 0.

Solution. A general solution is

y = c1e−2t cos(3t) + c2e

−2t sin(3t)

+ c3 cos(7t) + c4 sin(7t) + c5t cos(7t) + c6t sin(7t)

+ c7e5t + c8t e

5t + c9t2e5t + c10e

−3t + c11 + c12t .

The reasoning is as follows:• the single conjugate pair −2± i3 yields e−2t cos(3t) and e−2t sin(3t);• the double conjugate pair ±i7 yields

cos(7t) , sin(7t) , t cos(7t) , and t sin(7t) ;

• the triple real root 5 yields e5t, t e5t, and t2e5t;• the single real root −3 yields e−3t;• the double real root 0 yields 1 and t.

(5) Let D =d

dt. Solve each of the following initial-value problems.

(a) D2y + 4Dy + 4y = 0 , y(0) = 1 , y′(0) = 0 .

Solution. This is a constant coefficient, homogeneous, linear equation. Itscharacteristic polynomial is

p(z) = z2 + 4z + 4 = (z + 2)2 .

This has the double real root −2, which yields a general solution

y(t) = c1e−2t + c2t e

−2t .

Because

y′(t) = −2c1e−2t − 2c2t e

−2t + c2e−2t ,

when the initial conditions are imposed, we find that

y(0) = c1 = 1 , y′(0) = −2c1 + c2 = 0 .

These are solved to find c1 = 1 and c2 = 2. Therefore the solution of theinitial-value problem is

y(t) = e−2t + 2t e−2t = (1 + 2t)e−2t .

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(b) D2y + 9y = 20et , y(0) = 0 , y′(0) = 0 .

Solution. This is a constant coefficient, nonhomogeneous, linear equation. Itscharacteristic polynomial is

p(z) = z2 + 9 = z2 + 32 .

This has the conjugate pair of roots ±i3, which yields a general solution of theassociated homogeneous problem

yH(t) = c1 cos(3t) + c2 sin(3t) .

The forcing 20et has characteristic µ + iν = 1, degree d = 0, and multiplicitym = 0. A particular solution yP (t) can be found using either Key Identityevaluations or undetermined coefficients.

Key Indentity Evaluations. Because m + d = 0, we can use the zero degreeformula

L

(et

p(1)

)= et ,

or simply evaluate the Key identity at z = 1, to find

L(et) = p(1)et = (12 + 9)et = 10et .

Multiplying this equation by 2 yields L(2et) = 20et. Hence, yP (t) = 2et.

Undetermined Coefficients. Because m = d = 0, we seek a particular solu-tion of the form

yP (t) = Aet .

Because

y′P (t) = Aet , y′′P (t) = Aet ,

we see that

LyP (t) = y′′P (t) + 9yP (t) = Aet + 9Aet = 10Aet .

Setting LyP (t) = 10Aet = 20et, we see that A = 2. Hence, yP (t) = 2et.

By either approach we find yP (t) = 2et, which yields the general solution

y(t) = c1 cos(3t) + c2 sin(3t) + 2et .

Because

y′(t) = −3c1 sin(3t) + 3c2 cos(3t) + 2et ,

when the initial conditions are imposed we find that

y(0) = c1 + 2 = 0 , y′(0) = 3c2 + 2 = 0 .

These are solved to obtain c1 = −2 and c2 = −23. Therefore the solution of the

initial-value problem is

y(t) = −2 cos(3t)− 23

sin(3t) + 2et .

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(6) The functions cos(2t) and sin(2t) are a fundamental set of solutions to y′′ + 4y = 0.Find the solution Y (t) to the general initial-value problem

y′′ + 4y = 0 , y(0) = y0 , y′(0) = y1 .

Solution. Because we are given that cos(2t) and sin(2t) is a fundamental set ofsolutions to y′′ + 4y = 0, a general solution is y(t) = c1 cos(2t) + c2 sin(2t). Becausey′(t) = −2c1 sin(2t) + 2c2 cos(2t), the initial conditions imply

y0 = y(0) = c1 , y1 = y′(0) = 2c2 .

We solve these equations to obtain

c1 = y0 , c2 = 12y1 .

Therefore the solution to the general initial-value problem is

y(t) = y0 cos(2t) + y112

sin(2t) .

(7) Let D =d

dt. Give a general real solution for each of the following equations.

(a) D2y + 4Dy + 5y = 3 cos(2t) .


p(z) = z2 + 4z + 5 = (z + 2)2 + 1 .

This has the conjugate pair of roots −2 ± i, which yields a general solution ofthe associated homogeneous problem

yH(t) = c1e−2t cos(t) + c2e

−2t sin(t) .

The forcing 3 cos(2t) has characteristic µ+iν = i2, degree d = 0, and multiplicitym = 0. A particular solution yP (t) can be found using either Key Identityevaluations or undetermined coefficients.

Key Indentity Evaluations. Because m + d = 0, we can use the zero degreeformula

L

(ei2t

p(i2)

)= ei2t ,

or simply evaluate the Key identity at z = i2, to find

L(ei2t)

= p(i2)ei2t =((i2)2 + 4(i2) + 5

)ei2t = (1 + i8)ei2t .

Because the forcing 3 cos(2t) = 3 Re(ei2t), we divide the above by 1 + i8 and

multiply by 3 to find

L

(3

1 + i8ei2t)

= 3ei2t .

Hence,

yP (t) = Re

(3

1 + i8ei2t)

= Re

(3(1− i8)

12 + 82ei2t)

= 365

Re((1− i8)ei2t

)= 3

65

(cos(2t) + 8 sin(2t)

)= 3

65cos(2t) + 24

65sin(2t) .

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Therefore a general solution is

y = c1e−2t cos(t) + c2e

−2t sin(t) + 365

cos(2t) + 2465

sin(2t) .

Undetermined Coefficients. Because m = d = 0, we seek a particular solu-tion of the form

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

Because

y′P (t) = −2A sin(2t) + 2B cos(2t) ,

y′′P (t) = −4A cos(2t)− 4B sin(2t) ,

we see that

LyP (t) = y′′P (t) + 4y′P (t) + 5yP (t)

=[− 4A cos(2t)− 4B sin(2t)

]+ 4[− 2A sin(2t) + 2B cos(2t)

]+ 5[A cos(2t) +B sin(2t)

]= (A+ 8B) cos(2t) + (B − 8A) sin(2t) .

Setting LyP (t) = (A+ 8B) cos(2t) + (B − 8A) sin(2t) = 3 cos(2t), we see that

A+ 8B = 3 , B − 8A = 0 .

We find that A = 365

and B = 2465

. Hence, yP (t) = 365

cos(2t) + 2465

sin(2t).Therefore a general solution is

y = c1e−2t cos(t) + c2e

−2t sin(t) + 365

cos(2t) + 2465

sin(2t) .

(b) D2y − y = t et .


p(z) = z2 − 1 = (z + 1)(z − 1) .

This has the real roots −1 and 1, which yields a general solution of the associatedhomogeneous problem

yH(t) = c1e−t + c2e

t .

The forcing t et has characteristic µ + iν = 1, degree d = 1, and multiplicitym = 1. A particular solution yP (t) can be found using either Key Identityevaluations or undetermined coefficients.

Key Indentity Evaluations. Because m + d = 2, we need the Key identityand its first two derivatives

L(ezt)

= (z2 − 1)ezt ,

L(t ezt

)= (z2 − 1)t ezt + 2zezt ,

L(t2ezt

)= (z2 − 1)t2ezt + 4zt ezt + 2ezt .

Evaluate these at z = 1 to find

L(et)

= 0 , L(t et)

= 2et , L(t2et)

= 4t et + 2et .

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Subtracting the second equation from the third yields

L(t2et − t et

)= 4t et .

Dividing this equation by 4 gives L(14(t2−t) et

)= t et. Hence, yP (t) = 1

4(t2−t) et.


y = c1e−t + c2e

t + 14(t2 − t) et .

Undetermined Coefficients. Because m = 1 and d = 1, we seek a particularsolution of the form

yP (t) =(A0t

2 + A1t)et ,

Because

y′P (t) =(A0t

2 + A1t)et +

(2A0t+ A1

)et

=(A0t

2 + (2A0 + A1)t+ A1

)et ,

y′′P (t) =(A0t

2 + (2A0 + A1)t+ A1

)et +

(2A0t+ (2A0 + A1)

)et

=(A0t

2 + (4A0 + A1)t+ 2A0 + 2A1

)et ,

we see that

LyP (t) = y′′P (t)− yP (t)

=(A0t

2 + (4A0 + A1)t+ 2A0 + 2A1

)et −

(A0t

2 + A1t)et

=(4A0t+ 2A0 + 2A1

)et = 4A0t e

t + 2(A0 + A1)et .

Setting LyP (t) = 4A0t et+2(A0+A1)e

t = t et, we obtain 4A0 = 1 andA0+A1 = 0.It follows that A0 = 1

4and A1 = −1

4. Hence, yP (t) = 1

4(t2 − t) et. Therefore a

general solution is

y = c1e−t + c2e

t + 14(t2 − t) et .

(c) D2y − y =1

1 + et.


p(z) = z2 − 1 = (z − 1)(z + 1) .

This has the real roots 1 and −1, which yields a general solution of the associatedhomogeneous problem

yH(t) = c1et + c2e

−t .

The forcing does not have the form needed for undertermined coefficients. There-fore we must use either the Green function method or the variation of parametersmethod.

Green Function. The Green function g(t) satisfies

D2g − g = 0 , g(0) = 0 , g′(0) = 1 .

Set g(t) = c1et + c2e

−t. The first initial condition implies g(0) = c1 + c2 = 0.Because g′(t) = c1e

t−c2e−t, the second initial condition yields g′(0) = c1−c2 = 1.

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It follows that c1 = 12

and c2 = −12, whereby g(t) = 1

2(et − e−t). Hence, a

particular solution is

YP (t) = 12

∫ t

0

(et−s − e−t+s

) 1

1 + esds

= 12et∫ t

0

e−s

1 + esds− 1

2e−t∫ t

0

es

1 + esds .

The definite integrals on the right-hand side can be evaluated as∫ t

0

e−s

1 + esds =

∫ t

0

e−2s

e−s + 1ds =

∫ t

0

e−s − e−s

e−s + 1ds

=[− e−s + log

(e−s + 1

)]∣∣∣∣ts=0

= 1− e−t + log

(e−t + 1

2

),∫ t

0

es

1 + esds = log

(1 + es

)∣∣∣∣ts=0

= log

(1 + et

2

).

Hence, the particular solution YP (t) is given by

YP (t) = 12

[et − 1 + et log

(e−t + 1

2

)]− 1

2e−t log

(1 + et

2

).

Therefore a general solution is y = YH(t) + YP (t) where YH(t) and YP (t) aregiven above. This yields

y = c1et + c2e

−t + 12

[et − 1 + et log

(e−t + 1

2

)]− 1

2e−t log

(1 + et

2

).

Variation of Parameters. Seek a solution in the form

y = u1(t) et + u2(t) e

−t ,

where u1(t) and u2(t) satisfy

u′1(t) et + u′2(t) e

−t = 0 ,

u′1(t) et − u′2(t) e−t =

1

1 + et.

Solve this system to obtain

u′1(t) = 12

e−t

1 + et, u′2(t) = −1

2

et

1 + et.

Integrate these equations to find

u1(t) = 12

∫e−t

1 + etdt = 1

2

∫e−2t

e−t + 1dt

= 12

∫e−t − e−t

e−t + 1dt = −1

2e−t + 1

2log(e−t + 1

)+ c1 ,

u2(t) = −12

∫et

1 + etdt = −1

2log(1 + et

)+ c2 .


y = c1et + c2e

−t − 12

+ 12et log

(e−t + 1

)− 1

2e−t log

(1 + et

).

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(8) Let D =d

dt. Consider the equation

Ly = D2y − 6Dy + 25y = et2

.

(a) Compute the Green function g(t) associated with L.

Solution. The Green function g(t) satisfies

D2g − 6Dg + 25g = 0 , g(0) = 0 , g′(0) = 1 .

The characteristic polynomial of L is p(z) = z2 − 6z + 25 = (z − 3)2 + 42,which has roots 3 ± i4. Set g(t) = c1e

3t cos(4t) + c2e3t sin(4t). The first initial

condition implies g(0) = c1 = 0, whereby g(t) = c2e3t sin(4t). Because g′(t) =

3c2e3t sin(4t)+4c2e

3t cos(4t), the second initial condition implies g′(0) = 4c2 = 1,whereby c2 = 1

4. Therefore the Green function associated with L is given by

g(t) = 14e3t sin(4t) .

(b) Use the Green function to express a particular solution YP (t) in terms of definiteintegrals.

Solution. A particular solution YP (t) is given by

YP (t) =

∫ t

0

g(t− s)es2 ds = 14

∫ t

0

e3(t−s) sin(4(t− s)

)es

2

ds .

Because sin(4(t− s)

)= sin(4t) cos(4s)− cos(4t) sin(4s), this particular solution

is given in terms of definite integrals as

YP (t) = 14e3t sin(4t)

∫ t

0

e−3s cos(4s)es2

ds− 14e3t cos(4t)

∫ t

0

e−3s sin(4s)es2

ds .

Remark. The above definite integrals cannot be evaluated analytically.

(9) Compute the Green function g(t) associated with the differential operator

L = D2 − 4D + 4 , where D =d

dt.

Solution. Because the differential operator L has constant coefficients, the Greenfunction g(t) associated with it satisfies the initial-value problem

Lg = D2g − 4Dg + 4g = 0 , g(0) = 0 , g′(0) = 1 .

The characteristic polynomial is

p(z) = z2 − 4z + 4 = (z − 2)2 ,

which has the double real root 2. Hence, the Green function has the form

g(t) = c1e2t + c2t e

2t .

The initial condition g(0) = 0 implies that c1 = 0. Because

g′(t) = 2c2t e2t + c2e

2t ,

the initial condition g′(0) = 1 implies that c2 = 1. Therefore the Green function is

g(t) = t e2t .

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(10) Solve the initial-value problem

w′′ − 4w′ + 4w =2e2t

1 + t2, w(0) = w′(0) = 0 .

Solution. By the previous problem the Green function for this problem is g(t) = t e2t.Because this equation is in normal form, the solution to this inital-value problem isgiven by the Green function formula

w(t) =

∫ t

0

g(t− s)f(s) ds =

∫ t

0

(t− s)e2(t−s) 2e2s

1 + s2ds

= 2t e2t∫ t

0

1

1 + s2ds− e2t

∫ t

0

2s

1 + s2ds .

Because ∫ t

0

1

1 + s2ds = tan−1(s)

∣∣∣ts=0

= tan−1(t) ,∫ t

0

2s

1 + s2ds = log(1 + s2)

∣∣∣ts=0

= log(1 + t2) ,

we find that

w(t) = 2t e2t tan−1(t)− e2t log(1 + t2) .

Remark. This problem can also be solved by using variation of parameters. Howeverthat approach is not as efficient because it does not directly solve the initial-valueproblem. Rather, after finding a particular solution the constants c1 and c2 in YH(t)must be determined to satisfy the initial conditions.

(11) Compute the Laplace transform of f(t) = t e3t from its definition.

Solution. The definition of the Laplace transform gives

L[f ](s) = limT→∞

∫ T

0

e−stt e3t dt = limT→∞

∫ T

0

t e(3−s)t dt .

This limit diverges to +∞ for s ≤ 3 because in that case∫ T

0

t e(3−s)t dt ≥∫ T

0

t dt =T 2

2,

which clearly diverges to +∞ as T →∞.For s > 3 an integration by parts shows that∫ T

0

t e(3−s)t dt = te(3−s)t

3− s

∣∣∣∣T0

−∫ T

0

e(3−s)t

3− sdt

=

(te(3−s)t

3− s− e(3−s)t

(3− s)2

)∣∣∣∣T0

=

(Te(3−s)T

3− s− e(3−s)T

(3− s)2

)+

1

(3− s)2.

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Hence, for s > 3 we have that


[(Te(3−s)T

3− s− e(3−s)T

(3− s)2

)+

1

(3− s)2

]=

1

(3− s)2+ lim

T→∞

(Te(3−s)T

3− s− e(3−s)T

(3− s)2

)=

1

(3− s)2.

(12) Compute the Laplace transform of f(t) = u(t− 5) e2t from its definition.(Here u is the unit step function.)

Solution. The definition of Laplace transform gives


∫ T

0

e−stu(t− 5) e2t dt = limT→∞

∫ T

5

e−(s−2)t dt .

When s ≤ 2 this limit diverges to +∞ because in that case we have for every T > 5∫ T

5

e−(s−2)t dt ≥∫ T

5

dt = T − 5 ,

which clearly diverges to +∞ as T →∞.

When s > 2 we have for every T > 5∫ T

5

e−(s−2)t dt = −e−(s−2)t

s− 2

∣∣∣∣T5

= −e−(s−2)T

s− 2+e−(s−2)5

s− 2,

whereby


[− e−(s−2)T

s− 2+e−(s−2)5

s− 2

]=e−(s−2)5

s− 2.

(13) Compute the Laplace transform of cos(t)δ(t− π) from its definition.(Here δ is the unit impulse.)

Solution. The definition of Laplace transform gives


∫ T

0

e−st cos(t)δ(t− π) dt .

But by properties of the unit impulse, for T ≥ π we have∫ T

0

e−st cos(t)δ(t− π) dt =

∫ T−π

−πe−s(t+π) cos(t+ π)δ(t) dt = e−sπ cos(π) = −e−sπ .

Therefore


−e−sπ = −e−sπ .

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(14) Find the Laplace transform Y (s) of the solution y(t) of the initial-value problem

y′′ + 4y′ + 13y = f(t) , y(0) = 4 , y′(0) = 1 ,

where

f(t) =

{cos(t) for 0 ≤ t < 2π ,

t− 2π for t ≥ 2π .

You may refer to the table on the last page. DO NOT take the inverse Laplacetransform to find y(t), just solve for Y (s)!

Solution. The Laplace transform of the initial-value problem is

L[y′′](s) + 4L[y′](s) + 13L[y](s) = L[f ](s) ,

where

L[y](s) = Y (s) ,

L[y′](s) = sY (s)− y(0) = sY (s)− 4 ,

L[y′′](s) = s2Y (s)− sy(0)− y′(0) = s2Y (s)− 4s− 1 .

To compute L[f ](s), first write f as

f(t) =(1− u(t− 2π)

)cos(t) + u(t− 2π)(t− 2π)

= cos(t)− u(t− 2π) cos(t) + u(t− 2π)(t− 2π)

= cos(t)− u(t− 2π) cos(t− 2π) + u(t− 2π)(t− 2π) .

Referring to the table on the last page, item 6 with c = 2π and j(t) = cos(t), item 6with c = 2π and j(t) = t, item 2 with a = 0 and b = 1, and item 1 with n = 1 anda = 1 then show that

L[f ](s) = L[cos(t)](s)− L[u(t− 2π) cos(t− 2π)](s) + L[u(t− 2π)(t− 2π)](s)

= L[cos(t)](s)− e−2πsL[cos(t)](s) + e−2πsL[t](s)

=(1− e−2πs

) s

s2 + 1+ e−2πs

1

s2.

The Laplace transform of the initial-value problem then becomes(s2Y (s)− 4s− 1

)+ 4(sY (s)− 4

)+ 13Y (s) =

(1− e−2πs

) s

s2 + 1+ e−2πs

1

s2,

which becomes

(s2 + 4s+ 13)Y (s)− 4s− 1− 16 =(1− e−2πs

) s

s2 + 1+ e−2πs

1

s2.

Therefore Y (s) is given by

Y (s) =1

s2 + 4s+ 13

(4s+ 17 +

(1− e−2πs

) s

s2 + 1+ e−2πs

1

s2

).

(15) Find the inverse Laplace transforms of the following functions.You may refer to the table on the last page.

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(a) F (s) =2

(s+ 5)2,

Solution. Referring to the table on the last page, item 1 with n = 1 and a = −5gives

L[t e−5t](s) =1

(s+ 5)2.

Therefore we conclude that

L−1[

2

(s+ 5)2

](t) = 2L−1

[1

(s+ 5)2

](t) = 2t e−5t .

(b) F (s) =3s

s2 − s− 6,

Solution. Because the denominator factors as (s−3)(s+2), we have the partialfraction identity

3s

s2 − s− 6=

3s

(s− 3)(s+ 2)=

95

s− 3+

65

s+ 2.

Referring to the table on the last page, item 1 with n = 0 and a = 3, and withn = 0 and a = −2 gives

L[e3t](s) =1

s− 3, L[e−2t](s) =

1

s+ 2.


L−1[

3s

s2 − s− 6

](t) = L−1

[ 95

s− 3+

65

s+ 2

](t)

= 95L−1

[1

s− 3

](t) + 6

5L−1

[1

s+ 2

](t)

= 95e3t + 6

5e−2t .

(c) F (s) =(s− 2)e−3s

s2 − 4s+ 5.

Solution. Complete the square in the denominator to get (s−2)2+1. Referringto the table on the last page, item 2 with a = 2 and b = 1 gives

L[e2t cos(t)](s) =s− 2

(s− 2)2 + 1.

Item 6 with c = 3 and j(t) = e2t cos(t) then gives

L[u(t− 3)e2(t−3) cos(t− 3)](s) = e−3ss− 2

(s− 2)2 + 1.


L−1[e−3s

s− 2

s2 − 4s+ 5

](t) = u(t− 3)e2(t−3) cos(t− 3) .

13

(16) Find the inverse Laplace transform of Y (s) =3

(s2 − 4s+ 5)(s2 − 4s+ 8).

You may refer to the table on the last page.

Solution. By completing squares we obtain

Y (s) =3

(s2 − 4s+ 5)(s2 − 4s+ 8)=

3

((s− 2)2 + 1)((s− 2)2 + 4).

By setting z = (s− 2)2 in the partial fraction identity

3

(z + 1)(z + 4)=

1

z + 1+−1

z + 4,

we see that

Y (s) =1

(s− 2)2 + 12− 1

2

2

(s− 2)2 + 22.

Referring to the table on the last page, item 3 with a = 2 and b = 1 gives

L[e2t sin(t)](s) =1

(s− 2)2 + 1,

while item 3 with a = 2 and b = 2 gives

L[e2t sin(2t)](s) =2

(s− 2)2 + 22.


L−1[Y ](t) = e2t sin(t)− 12e2t sin(2t) .

(17) Let f(t) be continuous over [0,∞) and be of exponential order α as t→∞ for someα ≥ 0. Define h(t) by

h(t) =

∫ t

0

f(t′) dt′ for every t ∈ [0,∞) .

(a) Show that h(t) is of exponential order α as t→∞.

Solution. We have to show that for every σ > α there exists Kσ < ∞ andTσ <∞ such that

|h(t)| ≤ Kσeσt for every t > Tσ .

Let σ > α. Because f(t) is of exponential order α as t → ∞ and σ > α, weknow that there exists K ′σ <∞ and Tσ <∞ such that

|f(t)| ≤ K ′σeσt for every t > Tσ .

Because f(t) continuous over the closed bounded interval [0, Tσ], there existsMσ <∞ such that

|f(t)| ≤Mσ for every t ∈ [0, Tσ] .

14

Then, because σ > α ≥ 0, for every t > Tσ we have

|h(t)| ≤∫ Tσ

0

|f(t′)| dt′ +∫ t

Tσ

|f(t′)| dt′

≤Mσ

∫ Tσ

0

dt′ +K ′σ

∫ t

Tσ

eσt′dt′

= MσTσ +K ′σσ

(eσt − eσTσ

)=

[MσTσe

−σTσeσ(Tσ−t) +K ′σσ

(1− eσ(Tσ−t)

)]eσt

≤ Kσeσt ,

where

Kσ = max

{MσTσe

−σTσ ,K ′σσ

}.

Therefore h(t) is of exponential order α as t→∞. �

(b) Express H(s) = L[h](s) in terms of F (s) = L[f ](s).

Solution. Because h is differentiable with h′(t) = f(t) and h(0) = 0 we have

F (s) = L[f ](s) = L[h′](s) = sL[h](s)− h(0) = sH(s) .

Therefore H(s) = F (s)/s. �

(18) Solve the initial-value problem

y′′ + y′ − 6y = 7δ(t− 4) , y(0) = 0 , y′(0) = 5

Here δ is the unit impulse. You may refer to the table on the last page.

Solution. The Laplace transform of the initial-value problem is

L[y′′](s) + L[y′](s)− 6L[y](s) = 7L[δ(t− 4)](s) ,

whereL[y](s) = Y (s) ,

L[y′](s) = sL[y](s)− y(0) = s Y (s) ,

L[y′′](s) = sL[y′](s)− y′(0) = s2Y (s)− 5 ,

and by properties of the unit impulse

L[δ(t− 4)](s) =

∫ ∞0

e−stδ(t− 4) dt = e−4s .

The Laplace transform of the initial-value problem then becomes(s2Y (s)− 5

)+ s Y (s)− 6Y (s) = 7e−4s ,

which becomes(s2 + s− 6)Y (s) = 5 + 7e−4s .

Therefore

Y (s) =5 + 7e−4s

s2 + s− 6.

15

Because the denominator factors as (s − 2)(s + 3), we have the partial fractionidentity

5

(s− 2)(s+ 3)=

1

s− 2+−1

s+ 3.

Referring to the table on the last page, item 1 with n = 0 and a = 2, and with n = 0and a = −3 gives

L[e2t](s) =1

s− 2, L[e−3t](s) =

1

s+ 3.

Therefore we see that

L−1[

5

s2 + s− 6

](t) = L−1

[1

s− 2

](t)− L−1

[1

s+ 3

](t) = e2t − e−3t .

Item 6 with c = 4 and j(t) = e2t − e−3t then gives

L[u(t− 4)

(e2(t−4) − e−3(t−4)

)](s) = e−4s

5

s2 + s− 6.

Therefore the solution of the initial-value problem is

y(t) = L−1[Y ](t) = L−1[

5

s2 + s− 6

](t) + 7

5L−1

[5e−4s

s2 + s− 6

](t)

= e2t − e−3t + 75u(t− 4)e2(t−4) − 7

5u(t− 4)e−3(t−4) .

(19) Let f(t) = cos(2t) and g(t) = e3t. Compute f ∗ g(t).You may refer to the table on the last page.

Solution. Referring to the table on the last page, item 2 with a = 0 and b = 2, anditem 1 with n = 0 and a = 3 give

L[cos(2t)](s) =s

s2 + 4, L[e3t](s) =

1

s− 3.

The Convolution Theorem gives

L[f ∗ g](s) = L[f ](s)L[g](s) =s

s2 + 4

1

s− 3.

We have the partial fraction identity

s

(s2 + 4)(s− 3)=

313

s− 3+− 3

13s

s2 + 4+

413

s2 + 4.

Referring to the table on the last page, item 3 with a = 0 and b = 2 gives

L[sin(2t)](s) =2

s2 + 4.

By combining this entry with the two given eariler, we see that

f ∗ g(t) = L−1[

s

(s2 + 4)(s− 3)

](t)

= 313L−1

[1

s− 3

](t)− 3

13L−1

[s

s2 + 4

](t) + 2

13L−1

[2

s2 + 4

](t)

= 313e3t − 3

13cos(2t) + 2

13sin(2t) .

16

Remark. The most efficient way to compute the convolution directly starts

f ∗ g(t) = g ∗ f(t) =

∫ t

0

g(t− t′)f(t′) dt′

=

∫ t

0

e3(t−t′) cos(2t′) dt′ = e3t

∫ t

0

e−3t′cos(2t′) dt′ .

However computing the last integral above could take some time!

(20) Solve the integral equation e−t = y(t) + 2 cos ∗y(t).You may refer to the table on the last page.

Solution. The Laplace transform of the integral equation is

L[e−t](s) = L[y](s) + 2L[cos ∗y](s) ,

Referring to the table on the last page, item 1 with n = 0 and a = −1 and item 2with a = 0 and b = 1 give

L[e−t](s) =1

s+ 1, L[cos](s) =

s

s2 + 1.

Let L[y](s) = Y (s). Then by the Convolution Theorem we have

L[cos ∗y](s) = L[cos](s)L[y](s) =s

s2 + 1Y (s) .

The Laplace transform of the integral equation then becomes

1

s+ 1= Y (s) +

2s

s2 + 1Y (s) =

s2 + 2s+ 1

s2 + 1Y (s) .

Therefore Y (s) is given by

Y (s) =s2 + 1

(s+ 1)3.

Because s2 + 1 = (s+ 1)2 − 2(s+ 1) + 2, we have the partial fraction identity

s2 + 1

(s+ 1)3=

1

s+ 1− 2

(s+ 1)2+

2

(s+ 1)3.

Referring to the table on the last page, item 1 with n = 0 and a = −1. and withn = 1 and a = −1, and with n = 2 and a = −1 gives

L[

1

s+ 1

](t) = e−t , L

[1

(s+ 1)2

](t) = t e−t , L

[2

(s+ 1)3

](t) = t2e−t .

Therefore the solution of the integral equation is

y(t) = L−1[Y ](t) = L−1[s2 + 1

(s+ 1)3

](t)

= L−1[

1

s+ 1

](t)− 2L−1

[1

(s+ 1)2

](t) + L−1

[2

(s+ 1)3

](t)

= e−t − 2t e−t + t2e−t = (1− t)2e−t .

17

(21) Consider the initial-value problem

y′ = 3y23 , y(0) = −8 .

(a) What is the largest time interval over which this problem has a unique solution?

Solution. The largest time interval over which this problem has a unique solu-tion is (−∞, 2]. The work is shown below.

(b) Find the unique solution over this time interval.

Solution. The unique solution over the time interval (−∞, 2] is y = (t − 2)3.The work is shown below.

(c) Find two solutions that extend beyond this time interval.

Solution. Two solutions that extend beyond this time interval are

y1(t) = (t− 2)3 for all t ∈ R , y2(t) =

{(t− 2)3 for t ≤ 2 ,

0 for t > 2 .

The work is shown below.

Work. The differential equation is autonomous, and thereby is separable. Theright-hand side is continuously differentiable everywhere except at y = 0, its onlystationary point. The initial value is −8, so a unique solution that is nonstationarywill exist for so long as the solution avoids either blowing up or hitting y = 0. Wemust solve the equation to compute this time interval.

The separated differential form of the equation is 13y−

23 dy = dt, whereby∫

13y−

23 dy =

∫dt .

Upon integrating both sides we find that y13 = t + c. The initial condition then

implies that (−8)13 = 0 + c, whereby c = (−8)

13 = −2. The solution is thereby given

implicitly by y13 = t− 2, which yields the unique explicit solution

y = (t− 2)3 .

This solution never blows up, but does hit the stationary point y = 0 at t = 2.

Because the solution hits the stationary point at a finite time, it can be extendednonuniquely. It is shown above that all nonstationary solutions of the differentialequation have the form (t− b)3 for some b ∈ R. The fact that these are solutions canalso be checked directly. This fact can be used to construct many extensions of thesolution found above to t > 2. For example, it can be extended to t > 2 either bysetting y(t) = (t− 2)3, by setting y(t) = 0, or by picking any b > 2 and setting

y(t) =

{0 for 2 < t ≤ b ,

(t− b)3 for b < t .

The answer given above used the first two extensions, but any two could be used. �

18

(22) Let f(t, y) = cos(t)y + y3. For each r > 0 consider the set

Sr ={

(t, y) : t ∈ R , y ∈ [−r, r]}.

(a) Find Mr, the smallest upper bound on |f(t, y)| over Sr.

Solution. Because | cos(t)| ≤ 1 and |y| ≤ r for every (t, y) ∈ Sr, we have theupper bound

|f(t, y)| = | cos(t)y + y3| ≤ | cos(t)||y|+ |y|3 ≤ r + r3 .

This upper bound is attained when (t, y) = (2kπ,±r) for every integer k, so itis the smallest upper bound. Therefore Mr = r + r3. �

(b) Find Lr, the smallest Lipschitz constant for the y variable of f(t, y) over Sr.

Solution. Because f(t, y) is periodic in t and continuously differentiable over Sr,it has a Lipschitz constant. The smallest Lipschitz constant will be the smallestupper bound of |∂yf(t, y)|. Because ∂yf(t, y) = cos(t) + 3y2, we have the upperbound

|∂yf(t, y)| = | cos(t) + 3y2| ≤ | cos(t)|+ 3|y|2 ≤ 1 + 3r2 .

This upper bound is attained when (t, y) = (2kπ,±r) for every integer k, so itis the smallest upper bound. Therefore Lr = 1 + 3r2. �

Alternative Solution. Let (t, y) and (t, z) be in Sr. Then

|f(t, y)− f(t, z)| = | cos(t)(y − z) + y3 − z3|= | cos(t) + y2 + yz + z2||y − z| .

Therefore the smallest Lipschitz constant will be the smallest upper bound of| cos(t) + y2 + yz + z2|. Because

| cos(t) + y2 + yz + z2| ≤ | cos(t)|+ |y|2 + |y||z|+ |z|2

≤ 1 + r2 + r2 + r2 = 1 + 3r2 .

Therefore we have the Lipschitz constant 1 + 3r2. That this is the smallestpossible Lipschitz constant is seen by setting t = 2kπ for any intrger k, y = r,and z = r − η for an arbitrarily small η > 0. Therefore Lr = 1 + 3r2. �

(23) Consider the initial-value problem

y′ = cos(t)y + y3 , y(0) = 1 .

(a) If the initial Picard iterate is y0(t) = 1, compute the first Picard iterate y1(t).

Solution. In general, the first Picard iterate y1(t) is given in terms of the initialPicard iterate y0(t) by

y1(t) = y(0) +

∫ t

0

f(s, y0(s)) ds .

19

Because f(t, y) = cos(t)y + y3, y(0) = 1, and y0(t) = 1, this becomes

y1(t) = 1 +

∫ t

0

(cos(s) · 1 + 13

)ds

= 1 +

∫ t

0

(cos(s) + 1

)ds = 1 +

(sin(s) + s

)∣∣∣ts=0

= 1 + t+ sin(t) .

(b) Use the Picard Theorem to show that this problem has a unique solution over atime interval [0, h] for some h > 0.

Solution. Because f(t, y) = cos(t)y + y3 is continuously differentiable over R2,the Picard Theorem asserts that there exists a time interval (a, b) containing theinitial time t = 0 over which a unique classical solution y(t) of the initial-valueproblem exists. Just pick h > 0 small enough so that [0, h] ⊂ (a, b). �

(c) Show that the solution of this problem blows up at a finite positive time.

Solution. The differential equation may be written as

y′ − cos(t)y = y3 .

Upon multiplying this by e− sin(t), the integrating factor for the associated non-homogeneous linear equation, we have

d

dt

(e− sin(t)y

)= e− sin(t)y3 .

Now set z = e− sin(t)y and recast the initial-value problem as

dz

dt= e2 sin(t)z3 , z(0) = 1 .

This equation is separable. Its separated differential form is

−2

z3dz = −2e2 sin(t) dt .

Upon integration we find that the solution is given implicitly by

1

z2− 1 = −2

∫ t

0

e2 sin(t′) dt′ .

Here the integral on the right-hand side cannot be evaluated analytically. Uponsolving this for z we find

z(t) =

(1− 2

∫ t

0

e2 sin(t′) dt′

)− 12

,

whereby

y(t) = esin(t)(

1− 2

∫ t

0

e2 sin(t′) dt′

)− 12

.

Because e2 sin(t) ≥ e−2 for every t ∈ R, for every t ≥ 0 we have the bound∫ t

0

e2 sin(t′) dt′ ≥

∫ t

0

e−2 dt′ =t

e2.

20

Therefore for every t ≥ 0 we have

1− 2t

e2≥ 1− 2

∫ t

0

e2 sin(t′) dt′ ,

which implies that y(t) can be bounded below as

y(t) ≥ esin(t)(

1− 2t

e2

)− 12

.

Beacuse this lower bound on y(t) blows up at t = 12e2, we conclude that y(t)

must blow up a some positive time before t = 12e2. �

Remark. Better estimates on the blow up time can be made. For example, itcan be shown using the Jensen inequality (which was not covered) that∫ t

0

e2 sin(t′) dt′ ≥ t exp

(2

t

∫ t

0

sin(t′) dt′)

= t exp

(2

t

(1− cos(t)

))≥ t .

Then proceeding as above, this implies that y(t) can be bounded below as

y(t) ≥ esin(t)(1− 2t)−12 .

Beacuse this lower bound on y(t) blows up at t = 12, we now conclude that y(t)

must blow up a some positive time before t = 12.

(24) Transform the equation v′′′+t2v′−3v3 = sinh(2t) into a first-order system of ordinarydifferential equations.

Solution. Because the equation is third order, the first-order system must havedimension three. The simplest such first-order system is

d

dt

x1x2x3

=

x2x3

sinh(2t) + 3x 31 − t2x2

, where

x1x2x3

=

vv′

v′′

.

A Short Table of Laplace Transforms

L[tneat](s) =n!

(s− a)n+1for s > a .

L[eat cos(bt)](s) =s− a

(s− a)2 + b2for s > a .

L[eat sin(bt)](s) =b

(s− a)2 + b2for s > a .

L[tnj(t)](s) = (−1)nJ (n)(s) where J(s) = L[j(t)](s) .

L[eatj(t)](s) = J(s− a) where J(s) = L[j(t)](s) .

L[u(t− c)j(t− c)](s) = e−csJ(s) where J(s) = L[j(t)](s)

and u is the unit step function .

solutions to the sample problems for the midterm exam ucla math

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