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) .
Solution. This is a constant coefficient, nonhomogeneous, linear equation. Itscharacteristic polynomial is
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 .
Solution. This is a constant coefficient, nonhomogeneous, linear equation. Itscharacteristic polynomial is
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.
Therefore a general solution is
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.
Solution. This is a constant coefficient, nonhomogeneous, linear equation. Itscharacteristic polynomial is
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 .
Therefore a general solution is
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
L[f ](s) = limT→∞
[(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
L[f ](s) = limT→∞
∫ 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
L[f ](s) = limT→∞
[− 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
L[f ](s) = limT→∞
∫ 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
L[f ](s) = limT→∞
−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.
Therefore we conclude that
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.
Therefore we conclude that
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.
Therefore we conclude that
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 .