Contents
1
Differentiation
1.1
1.2
1.3
1.4
2
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General Power Rule Integrals . . . . .
Logarithmic and Exponential Integrals
Trigonometric Integrals . . . . . . . .
Inverse Trigonometric Forms . . . . .
Trigonometric Substitution . . . . . .
Integration by Parts . . . . . . . . . .
Integration of Rational Functions . . .
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Solutions to Differential Equations . . . .
Separation of Variables . . . . . . . . . . .
First-Order Linear Differential Equations
Applications of Differential Equations . .
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Higher-Order Homogeneous Differential Equations . . . . . .
Auxiliary Equations . . . . . . . . . . . . . . . . . . . . . . .
Non-homogeneous Differential Equations . . . . . . . . . . . .
Applications of Second-Order Equations . . . . . . . . . . . .
Computing the Laplace Transformation . . . . . . . . . . . .
Computing the Inverse Laplace Transformation . . . . . . . .
Solving Differential Equations Using Laplace Transformations
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Geometric Series . . . . .
Tests for Convergence . .
Maclaurin Series . . . . .
Operations with Series . .
Computations with Series
Fourier Series . . . . . . .
6
7
8
10
11
12
13
14
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14
15
16
17
18
19
20
Higher Order Differential Equations
5.1
5.2
5.3
5.4
5.5
5.6
5.7
2
3
4
5
6
First-Order Differential Equations
4.1
4.2
4.3
4.4
5
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Series
3.1
3.2
3.3
3.4
3.5
3.6
4
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Integration
2.1
2.2
2.3
2.4
2.5
2.6
2.7
3
2
Review from Math 22100 .
L’Hospital’s Rule . . . . . .
Applications of Derivatives
Newton’s Method . . . . . .
20
21
22
23
24
1
24
25
26
27
28
29
31
Department of Mathematical Sciences @IUPUI
1
Differentiation
1.1
Review from Math 22100
Find the derivative of each of the following functions
p
3
1. y = 3 ln t2 + 2
2. y = ln
3. y =
x
2x − 1
ex
x2
Solution
Use logarithm rules to simplify the first two. Use quotient rule on the third.
1. y 0 =
2t
t2 + 2
2. y 0 =
2
1
−
x 2x − 1
3. y 0 =
ex (x − 2)
x3
22200 Exam Jam
Department of Mathematical Sciences @IUPUI
1.2
L’Hospital’s Rule
Evaluate the following limits using L’Hospital’s rule
1 − ex
x→0
2x
1. lim
2. lim
x→0
x − sin x
x
Solution
Both are already in the form 00 , so apply L’Hospital’s rule directly.
1 − ex H
−ex
−1
= lim
=
x→0
x→0 2
2x
2
1. lim
2. lim
x→0
x − sin x H
1 − cos x
= lim
=0
x→0
x
1
22200 Exam Jam
Department of Mathematical Sciences @IUPUI
1.3
22200 Exam Jam
Applications of Derivatives
Find the minima and maxima, the points of inflection, and sketch the graph of the curve below.
y = xe−x
Solution
To calculate the extremum, find the zeros of the first derivative.
y 0 (x) = (x − 1)e−x = 0
Since e−x > 0, we only have the solution x = 1. Therefore, the only possible relative extremum is at x = 1.
To determine whether this is a maximum of minimum, use the second derivative test.
y 00 (1) = −e−1 < 0
Therefore, (1, e−1 ) is a local maximum.
To determine any points of inflection, find the zeros of the second derivative.
y 00 (x) = (x − 2)e−x = 0
Since e−x > 0, we only have the solution x = 2. Since the second derivatives changes sign here (check values
less than 2 and greater than 2), this value is an inflection point.
To sketch the curve, it helps to find any asymptotes and where the function is increasing/decreasing. In
doing so, we see the curve has a horizontal asymptote of y = 0, increasing on (−∞, 1), and decreasing on
(1, ∞). We can then see our graph has the following shape.
y
x
x=1
x=2
Department of Mathematical Sciences @IUPUI
1.4
22200 Exam Jam
Newton’s Method
Find (approximately) a positive root of the following equation.
4 sin x − x = 0
(Note: Not all classes cover this material.)
Solution
Make a table of values using the relation
xn+1 = xn −
f (xn )
f 0 (xn )
Where f (x) = 4 sin x − x. We may choose (almost) any value for x0 . We choose x0 = 2. The table is the
following
n
0
1
2
3
4
xn
2
2.6144
2.4793
2.4746
2.4746
The values remain the same (up to 4 decimal points) after n = 4. Therefore, x = 2.4746 is an approximate
solution to our equation.
Department of Mathematical Sciences @IUPUI
2
22200 Exam Jam
Integration
2.1
General Power Rule Integrals
Using the general power rule to evaluate the following integrals
Z
p
1.
e2x 1 + e2x dx
Z
2.
Z
(1 − cos 5x)3 sin 5x dx
e
√
3.
1
ln x
dx
x
Solution
1. Use the substitution u = 1 + e2x .
Z
Z
p
1√
1
1
e2x 1 + e2x dx =
u du = u3/2 + c = (1 + e2x )3/2 + c
2
3
3
2. Use the subsitution u = 1 − cos 5x.
Z
Z
1 4
1
1 3
u du =
u +c=
(1 − cos 5x)4 + c
(1 − cos 5x)3 sin 5x dx =
5
20
20
3. Use the substitution u = ln x.
Z
1
e
√
ln x
dx =
x
Z
0
1
√
1
2
2 3/2 u du = u =
3
3
0
Department of Mathematical Sciences @IUPUI
2.2
22200 Exam Jam
Logarithmic and Exponential Integrals
Evaluate the following integrals
Z
2
1.
tet dx
Z
2.
Z
3.
2x dx
1
dx
x ln x
Solution
1. Use the substitution u = t2 .
Z
2
tet dt =
Z
1 u
1
1 2
e du = eu + c = et + c
2
2
2
2. Use the formula for integration of exponential functions.
Z
1 x
2 +c
2x dx =
ln 2
3. Use the substitution u = ln x.
Z
1
dx =
x ln x
Z
1
du = ln |u| + c = ln | ln x| + c
u
Department of Mathematical Sciences @IUPUI
2.3
22200 Exam Jam
Trigonometric Integrals
Evaluate the following integrals
Z
1.
x2 sec x3 tan x3 dx
Z
2.
x3 sec x4 dx
Solution
1. Use the substitution u = x3 .
Z
Z
1
1
1
2
3
3
sec u tan u du = sec u + c = sec x3 + c
x sec x tan x dx =
3
3
3
2. Use the substitution u = x4 .
Z
Z
1
1
1
3
4
x sec x dx =
sec u du = ln | sec u + tan u| + c = ln | sec x4 + tan x4 | + c
4
4
4
Department of Mathematical Sciences @IUPUI
22200 Exam Jam
Evaluate the following integrals
Z
1.
sin5 x cos6 x dx
Z
2.
sin2 x cos2 x dx
Solution
1. Separate one of the sin x terms and write the rest in terms of cos x.
Z
Z
Z
sin5 x cos6 x dx = (sin2 x)2 cos6 x sin x dx = (1 − cos2 x)2 cos6 x sin x dx
Now use the substitution u = cos x.
Z
Z
Z (1 − cos2 x)2 cos6 x sin x dx = −(1 − u2 )2 u6 du =
−u6 + 2u8 − u10 du
−1 7 2 9
1
u + u − u11 + c
7
9
11
2
1
−1
cos7 x + cos9 x −
cos1 1x + c
=
7
9
11
=
2. Use the trigonometric identities
sin 2θ = 2 sin θ cos θ, sin2 θ =
1
(1 − cos 2θ)
2
to simplify as follows.
Z
Z
Z
Z
1
1
2
2
2
2
sin x cos x dx = (sin x cos x) dx =
sin 2x dx =
(1 − cos 4x) dx
4
8
Now integrate as normal.
Z 1 1
− cos 4x
8 8
dx =
1
1
x−
sin 4x + c
8
32
Department of Mathematical Sciences @IUPUI
2.4
22200 Exam Jam
Inverse Trigonometric Forms
Use trigonometric substitution to evaluate the following integral
Z
1
√
dx
5 − 3x2
Solution
First, simplify the problem.
Z
Now use the substitution x =
1
√
3
Z
p
1
1
√
dx = √
2
3
5 − 3x
5/3 sin θ (since a =
1
p
dx = √
2
3
5/3 − x
Z
1
=√
3
Z
1
p
Z
1
p
dx
5/3).
r
1
p
5/3 − x2
5/3 cos2
θ
5
cos θ dθ
3
1
1
dθ = √ θ + c = √ sin−1
3
3
(Note: In the last step, to back substitute, we use the fact that x =
the inverse.)
√
x
p
5/3
!
+c
53 sin θ, then solved for sin θ and took
Department of Mathematical Sciences @IUPUI
2.5
22200 Exam Jam
Trigonometric Substitution
Use a trigonometric substitution to evaluate the following integral.
Z √ 2
x −9
dx
x
Solution
Use the substitution x = 3 sec θ.
Z √ 2
Z √
Z
x −9
9 tan2 θ
dx =
3 sec θ tan θ dθ = 3 tan2 θ dθ
x
3 sec θ
Now use the identity tan2 θ = sec2 θ − 1.
Z
Z
2
3 tan θ dθ = 3 sec2 θ − 3 dθ = 3 tan θ − 3θ + c
Now make a triangle to back substitute.
√
x
θ
x2 − 9
3
This gives us
tan θ =
1p 2
x − 9, θ = sec−1
3
x
3
Therefore, we have
Z √
p
x2 − 9
dx = x2 − 9 − 3 sec−1
x
x
+c
3
Department of Mathematical Sciences @IUPUI
2.6
22200 Exam Jam
Integration by Parts
Utilize integration by parts to evaluate the following integrals
Z
1.
xe−x dx
Z
2.
cot−1 x dx
Solution
1. Choose u = x and dv = e−x dx to obtain the following.
Z
Z
xe−x dx = −xe−x − −e−x dx = −xe−x − e−x + c
2. Choose u = cot−1 x and dv = dx to obtain the following.
Z
Z
−1
−1
cot x dx = x cot x −
−x
dx
x2 + 1
Now apply a u-substitution to evaluate the new integral to get the following.
Z
1
cot−1 x dx = x cot−1 x + ln |x2 + 1| + c
2
Department of Mathematical Sciences @IUPUI
2.7
22200 Exam Jam
Integration of Rational Functions
Use either long division or partial fractions to evaluate the following integral.
Z
x3 + 3x
dx
(x2 + 1)2
Solution
Begin by applying partial fractions.
x3 + 3x
Ax + B
Cx + D
= 2
+ 2
(x2 + 1)2
x +1
(x + 1)2
Multiplying both sides by (x2 + 1)2 gives
x3 + 3x = (Ax + B)(x2 + 1) + (Cx + D)
x3 + 3x = Ax3 + Bx2 + (A + C)x + (B + D)
Since we would like the left side to be exactly the same as the right side, we must have the following.
A=1
B=0
A+C =3
B+D =0
Which has the solution of
A = 1, B = 0, C = 2, D = 0
Therefore, we have the following.
x
2x
x3 + 3x
= 2
+
(x2 + 1)2
x + 1 (x2 + 1)2
So we have
Z
x3 + 3x
dx =
(x2 + 1)2
Z
x
2x
1
1
+
dx = ln |x2 + 1| − 2
+c
x2 + 1 (x2 + 1)2
2
x +1
(Note: We skipped the integration of the two functions. They can be done using u-substitutions.)
Department of Mathematical Sciences @IUPUI
3
3.1
22200 Exam Jam
Series
Geometric Series
Find the sum of the following series.
∞
X
(−1)n−1
2n−1
n=1
Solution
First, rewrite the series to look a bit nicer.
n−1
∞ ∞
X
X
−1
(−1)n−1
=
2n−1
2
n=1
n=1
From this, we have r = −1/2. Since |r| < 1, the series converges. Then using the formula
∞
X
arn−1 =
n=1
a
1−r
We have the following.
n−1
∞ X
−1
n=1
2
=
1
2
=
1 − (−1/2)
3
Department of Mathematical Sciences @IUPUI
3.2
22200 Exam Jam
Tests for Convergence
Determine whether the following series converges or diverges.
∞
X
n
2+1
n
n=1
(Note: Not all classes cover this material.)
Solution
Use limit comparison test and compare n/(n2 + 1) to 1/n.
n
n2 +1
1
n→∞
n
lim
= lim
1
n2
= lim
= 1 6= 0
+ 1 n→∞ 1 + n12
n→∞ n2
P1
P1
Therefore,
n converges and diverges if
n diverges. Since the
P 1 the series in the problem converges if
series
diverges,
the
series
of
the
problem
diverges
as
well.
n
Department of Mathematical Sciences @IUPUI
3.3
Maclaurin Series
Find the first three nonzero terms in the Maclaurin series for the following functions
1. y = cos x
2. y = ln(1 + x)
Solution
1. Use the formula for the MacLaurin series. The first few derivatives are as follows.
f (0) = 1, f 0 (0) = 0, f 00 (0) = −1, f 000 (0) = 0, f 0000 (0) = 1
So we have the following.
1
1
cos x ≈ 1 − x2 + x4
2
24
2. Use the formula for the MacLaurin series. The first few derivatives are as follows.
f (0) = 0, f 0 (0) = 1, f 00 (0) = −1, f 000 (0) = 2
So we have the following.
1
1
ln(1 + x) ≈ x − x2 + x3
2
3
22200 Exam Jam
Department of Mathematical Sciences @IUPUI
3.4
22200 Exam Jam
Operations with Series
Find the Maclaurin series of the following function.
y = ln(1 + x2 )
Solution 1
The following series should be familiar to us already.
ln(1 + z) =
∞
X
(−1)n−1 z n
n
n=1
If we let z = x2 , we arrive at the following result.
ln(1 + x2 ) =
∞
X
(−1)n−1 x2n
n
n=1
Solution 2
Differentiating the function gives us the following.
y0 =
2x
2x
=
1 + x2
1 − (−x2 )
Using the formula
∞
X
a
arn−1
=
1 − r n=1
we have
y0 =
∞
∞
n−1 X
X
2x
2
2(−1)n−1 x2n−1
2x
−x
=
=
1 − (−x2 ) n=1
n=1
To get back to our function y, we now integrate.
Z
y=
y 0 dx =
Z X
∞
2(−1)n−1 x2n−1 dx =
n=1
∞
X
(−1)n−1 x2n
+c
n
n=1
Plugging in x = 0 shows us that
y(0) = ln(1 + 02 ) = 0,
∞
X
(−1)n−1 02n
+c=0+c=c
n
n=1
So we must have c = 0. Therefore, we have
ln(1 + x2 ) =
∞
X
(−1)n−1 x2n
n
n=1
Department of Mathematical Sciences @IUPUI
3.5
22200 Exam Jam
Computations with Series
Use the first 4 terms of the Maclaurin series for y = e−x to approximate the value of e−0.2 . Determine the
error of your approximation.
Solution
The first four terms of the MacLaurin series for y = e−x is the following.
1
1
y = e−x ≈ 1 − x + x2 − x3
2
6
Plugging in x = 0.2 gives us
1
1
y(−0.2) = e−0.2 ≈ 1 − .2 + (.2)2 − (.2)3 = .81867
2
6
The exact value of e−0.2 is approximately .81873. Therefore, the error given by our estimate is approximately
.00006.
Department of Mathematical Sciences @IUPUI
3.6
22200 Exam Jam
Fourier Series
Determine the Fourier series for the following function on the given interval
(
0 if − 1 < t ≤ 0
f (t) =
t if 0 < t < 1
Solution
First, determine the a0 term.
1
a0 =
1
Z
1
Z
f (t) dt =
−1
0
1
1
1
1 2 t dt = t =
2 2
0
Now determine the an terms.
1
an =
1
Z
1
Z
f (t) cos(nt) dt =
−1
0
1
1
sin(n) cos(n)
1
1
1
t cos(nt) dt = t sin(nt) + 2 cos(nt) =
+
− 2
n
n
n
n2
n
0
Similarly, determine the bn terms.
1
bn =
1
Z
1
Z
f (t) sin(nt) =
−1
0
1
1
− cos(n) sin(n)
1
1
−1
t cos(nt) + 2 sin(nt) =
+
+
t sin(nt) dt =
n
n
n
n2
n
0
Then the Fourier series is the following.
∞
f (t) =
a0 X
an cos(nt) + bn sin(nt)
+
2
n=1
Where an and bn are the values found above.
Department of Mathematical Sciences @IUPUI
4
4.1
22200 Exam Jam
First-Order Differential Equations
Solutions to Differential Equations
Show that the function
y = xe−2x + 3e−2x
is a solution to the given differential equation.
dy
+ 2y = e−2x
dx
Solution
To show that the function given is a solution, we first compute its derivative.
y = xe−2x + 3e−2x , y 0 = −2xe−2x − 5e−2x
Now we plug this values (y and y 0 ) in our differential equation for y and its derivative.
−2xe−2x − 5e−2x + 2 xe−2x + 3e−2x = −2xe−2x − 5e−2x + +2xe−2x + 6e−2x = e−2x
Therefore, our function satisfies the differential equation. This means that the function is a solution.
Department of Mathematical Sciences @IUPUI
4.2
22200 Exam Jam
Separation of Variables
Find the general solution to the given differential equations
1. dx + (2 cos2 x − y cos2 x) dy = 0
2. xey dx + e−x dy = 0
Solution
1. First, simplify the equation.
dx + (2 cos2 x − y cos2 x) dy = 0
dx + cos2 x(2 − y) dy = 0
sec2 x dx + (2 − y) dy = 0
Now that the variables are separated (there are no terms with both x and y) we can integrate.
Z
Z
2
sec x dx + (2 − y) dy = c
1
tan x + 2y − y 2 = c
2
This is the general solution to the differential equation.
2. Simplify the equation as before.
xey dx + e−x dy = 0
xex dx + e−y dy = 0
(Divide by e−x and then by ey ) Now integrate.
Z
Z
xex dx + e−y dy = c
(x − 1)ex − e−y = c
(We skipped the integration by parts in the first integral) This is the general solution to the differential
equation.
Department of Mathematical Sciences @IUPUI
4.3
22200 Exam Jam
First-Order Linear Differential Equations
Find the solution to the following differential equation.
2
2
dy
− 8xy = e2x
dx
Solution
First, we need the equation to have no term in front of
2
dy
dx ,
so we simplify.
2
dy
− 8xy = e2x
dx
dy
1 2
− 4xy = e2x
dx
2
Now calculate the integrating factor.
µ(x) = e
R
−4x dx
= e−2x
2
Now multiply this to both sides of the equation.
e−2x
2
2
1
dy
− 4xe−2x y =
dx
2
The reason for this is that now the left side of the equation can be simplified to the derivative of y times our
integrating factor.
e−2x
2
2
dy
− 4xe−2x y =
dx
d −2x2 ye
=
dx
Now integrate both sides and simplify.
Z
d −2x2 ye
dx =
dx
Z
1
2
1
2
1
dx
2
1
x+c
2
2
2
1
y = xe2x + ce2x
2
2
ye−2x =
This is the general solution to the differential equation.
Department of Mathematical Sciences @IUPUI
4.4
22200 Exam Jam
Applications of Differential Equations
A bacteria culture is known to increase at a rate proportional to the number of bacteria present. It is
observed that the size of the culture triples in 3 hours. After how many hours should it be 10 times as large?
Solution
The information of the problem tells us that
dP
= kP
dt
Where P is the number of bacteria. Let the initial amount of bacteria be P0 (at t = 0). Separating variables
and integrating shows that we have
P (t) = Cekt
Now plug in t = 0 to get
P (0) = P0 , P (0) = Ce0 = C
So we have
P (t) = P0 ekt
The problem states that when t = 3, P = 3P0 (triples in 3 hours). So we have
P (3) = 3P0 = P0 e3k
Solving for k shows that
k=
1
ln 3
3
So we have
1
P (t) = P0 e 3 t ln 3
The problem now is to determine for what value of t gives P (t) = 10P0 . Setting P (t) = 10P0 gives us
1
10P0 = P0 e 3 t ln 3
1
10 = e 3 t ln 3
1
ln 10 = t ln 3
3
3 ln 10
=t
ln 3
Department of Mathematical Sciences @IUPUI
5
22200 Exam Jam
Higher Order Differential Equations
5.1
Higher-Order Homogeneous Differential Equations
Find the general solution to the given differential equations
1. 6
dy
d2 y
−
− 2y = 0
2
dx
dx
2. 2D2 y − 3Dy + y = 0
Solution
1. Set up and solve the auxiliary equation.
6r2 − r − 2 = 0
(2r + 1)(3r − 2) = 0
2
−1
, r=
r=
2
3
Then our general solution is the following.
1
2
y(t) = c1 e− 2 t + c2 e 3 t
2. Set up and solve the auxiliary equation as before.
2r2 − 3r + 1 = 0
(2r − 1)(r − 1) = 0
1
r= , r=1
2
Then our general solution is the following.
1
y(t) = c1 e 2 t + c2 et
Department of Mathematical Sciences @IUPUI
5.2
22200 Exam Jam
Auxiliary Equations
Solve the following differential equations.
1. (D2 + 25)y = 0
2. (D2 − 3D + 5)y = 0
Solution
1. The auxiliary equation is
r2 + 25 = 0
Which has roots r = ±5i. Then since these values are complex, the general solution is
y(t) = c1 sin(5t) + c2 cos(5t)
2. The auxiliary equation is
r2 − 3r + 5 = 0
r=
3±
p
√
9 − 4(1)(5)
3 ± i 11
=
2
2
Since these values are complex, the imaginary parts turn into sines and cosines as follows.
!
!
√
√
3
3
11
11
x
x
y(x) = c1 e 2 sin
x + c2 e 2 cos
x
2
2
Department of Mathematical Sciences @IUPUI
5.3
22200 Exam Jam
Non-homogeneous Differential Equations
Find the general solution to the given differential equations.
(D2 − D + 2)y = 4e3x
Solution
First solve the auxiliary equation to determine the homogeneous solution.
r2 − r + 2 = 0
r=
1±
p
√
1 − 4(1)(2)
1±i 7
=
2
2
So the homogeneous equation is the following.
√
yh (x) = c1 e
1
2x
sin
7
x
2
√
!
+ c2 e
1
2x
cos
7
x
2
!
To determine the particular solution, we guess that it has the form yp (t) = Ae3x . Then we have
yp (x) = Ae3x
Dyp (x) = 3Ae3x
D2 yp (x) = 9Ae3x
Plugging these into our equation, we get the following.
9Ae3x − 3Ae3x + 2Ae3x = 4e3x
8Ae3x = 4e3x
A=2
Therefore, our particular equation is yp (t) = 2e3x . Our general solution has the form yg = yh + yp . So we
have
√ !
√ !
1
1
7
7
x
x
x + c2 e 2 cos
x + 2e3x
yg (t) = c1 e 2 sin
2
2
Department of Mathematical Sciences @IUPUI
5.4
22200 Exam Jam
Applications of Second-Order Equations
A 2 lb weight stretches a spring 6 in. The weight is pushed 7 in above the equilibrium position and released.
Find the motion of the weight as a function of time, assuming no damping.
Solution
Using the equations
F = kx, F = mg
and using the information of the problem, we have if x = 0, the force of the spring is equal to the force of
gravity, so we can see
6k = 2(386.09), k = 128.697
Then using the equation for spring position, we have
2D2 y + 128.697y = 0
(Where y is distance from equilibrium) Solving this gives us
y(t) = c1 sin(64.35t) + c2 cos(64.35t)
Our initial conditions are when t = 0, y = 7, Dy = 0. Plugging these in, we arrive at
y(t) = 7 cos(64.35t)
Department of Mathematical Sciences @IUPUI
5.5
22200 Exam Jam
Computing the Laplace Transformation
Verify the identity.
L{sin at} =
a
s2 + a2
Solution
Begin with the definition of Laplace transformation.
Z t
Z ∞
e−sx sin ax dx
L{sin ax} =
e−sx sin ax dx = lim
0
t→∞
0
This integral takes a bit of work to solve. Integration by parts is the usual way of doing it, and it will be
omitted.
t
Z t
−ae−sx cos ax − se−sx sin ax −sx
lim
e
sin ax dx = lim
t→∞ 0
t→∞
s2 + a2
0
−ae−st cos at − se−st sin at
a
= lim
+ 2
t→∞
s2 + a2
s + a2
a
= 2
s + a2
(You may need to use L’Hospital’s rule to evaluate the limit!)
Department of Mathematical Sciences @IUPUI
5.6
22200 Exam Jam
Computing the Inverse Laplace Transformation
Compute the inverse Laplace transformation of the function.
F (s) =
5s
+6
s2
Solution
We know the following.
L−1
a
s2 + a2
= sin at, L−1
s
s2 + a2
= cos at
So we would like to rewrite what we have in terms of these formulas. Since there is an s term in the
numerator, we will use the second equation. Note that the 5 constant does not affect the inverse Laplace,
so we have
!
√ 5s
s
−1
−1
√
L
= 5L
6t
= 5 cos
2
s +6
s2 + ( 6)2
Department of Mathematical Sciences @IUPUI
22200 Exam Jam
Compute the inverse Laplace transformation of the function.
F (s) =
s
(s − 1)(s + 3)
Solution
We will first do partial fractions to make the problem simpler.
s
1/4
3/4
=
+
(s − 1)(s + 3)
s−1 s+3
Then use the rule
−1
1
s−a
1
s−1
L
= eat
To obtain
−1
L
s
(s − 1)(s + 3)
1
= L−1
4
3
+ L−1
4
1
s+3
=
1 t 3 −3t
e + e
4
4
Department of Mathematical Sciences @IUPUI
5.7
22200 Exam Jam
Solving Differential Equations Using Laplace Transformations
Use Laplace transformations to solve the following differential equation
y 00 − 4y 0 + 4y = e3t , y(0) = 0, y 0 (0) = −2
Solution
First, apply the Laplace transformation to both sides of the equation and simplify.
L(y 00 ) − 4L(y 0 ) + 4L(y) = L(e3t )
1
s2 L(y) − sy(0) − y 0 (0) − 4 sL(y) − y(0) + 4L(y) =
s−3
1
2
L(y) s − 4s + 4 + 2 =
s−3
1
L(y)(s − 2)2 =
−2
s−3
2
1
−
L(y) =
(s − 3)(s − 2)2
(s − 2)2
Now we must apply partial fractions to the right side to simplify further.
1
1
−1
−1
=
+
+
2
(s − 3)(s − 2)
s − 3 s − 2 (s − 2)2
Then use the rules
L−1
1
s−a
= eat , L−1
n!
(s − a)n
= tn eat
To obtain the final answer.
1
1!
1
− L−1
− 3L−1
= e3t − e2t − 3te2t
y = L−1
s−3
s−2
(s − 2)2