WEEK #9: Differential Equations, Anti-Derivatives, Integration
by Substitution
Goals:
• Introduction and Definitions in Differential Equations
• Euler’s Method
• Anti-Derivatives
• Solving Pure-Time Differential Equations
Textbook reading for Week #9: Read Sections 4.1, 4.2, 4.3
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Differential Equations
From Section 4.1
Current Model
In a battery-powered electrical circuit (e.g. cordless power tools), it is difficult to
measure the amount of charge in a battery, but straightforward to measure the
amount of current flowing in the circuit.
If I is the amount of charge in a battery, how could we represent the current?
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
3
We measure the current flowing out of a battery:
dI
= 20t − 400
dt
in units of mC per hour. Note the derivative is negative as the battery is
losing charge. Based on the derivative, can you guess what the function I(t),
the net charge in the battery, was?
4
Does your answer fit with the additional information that the battery originally
contained 560 mC at t = 0? If not, how could it be changed so it does?
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
5
Population Model
We have previously considered discrete-time models for population, like
bt+1 = rbt
These models work well when the population has a natural ‘clock’: e.g. synchronized cell division in labs, and animals with annual reproductive cycles.
In other populations, e.g. bacteria, humans, that model is too restrictive, and we
think of the population changing continuously. This naturally leads to a derivative
interpretation, like
db
= sb
dt
Note that the numerical rates (r and s)in the two forms of the model (discrete vs.
continuous time) will generally not be directly comparable.
6
Question: Guess at the solution of the continuous time population model,
db
= 1.5b
dt
1.5 2
A.
b
2
B. e1.5b
C. e1.5t
If you knew the initial population were 1000, how would that be represented
in your model?
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
7
Verify your solution by plugging it back into the original equation, and checking that the left and right sides are equal.
8
Classifying Differential Equations
The last two models are both examples of differential equations, or DEs:
dI
= 20t − 400
(1)
dt
db
= 1.5b
(2)
dt
Comment on the similarities and differences between the two equations.
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
9
Which would be called a “pure time” DE, and which an “autonomous” DE?
We will first study pure-time differential equations (which are generally easier),
before moving on to autonomous and mixed DEs.
10
Graphical Analysis
If we have a graph of the derivative over time, we can develop a decent sketch of
the original function, based on the relationship between functions and their slopes.
Example: There is two-way transfer of hydrogen in a fuel cell. The hydrogen
ions transfer across a membrane at the rate
r(t) = 4t − 8 mmol/min
If a positive rate indicates transfer from the left to right side of the membrane,
sketch a graph of the rate over time, and then the amount of hydrogen on the
right side of the membrane over time.
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
r(t) = 4t − 8 mmol/min
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12
Relation between Rate and Original Graphs
Positive rate → function increasing
Negative rate → function decreasing
Zero rate → function horizontal
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
13
Question: Consider the rate graph shown below:
Which of the following graphs is of the original function?
A
B
C
D
14
These graphical methods just provide sketches. To get more precise graphs of the
original function, we will also explore:
• integration or anti-derivatives (later), and
• numerical solutions (next)
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
Numerical Solutions and Euler’s Method
Example: You are given two pieces of information about a function f (t):
df
6t
=
and f (0) = 3
dt 1 + t
Sketch
df
and then f (t) on the axes below.
dt
From the sketch, can you tell when f (t) exceeds 4?
15
16
We will now use a numerical method, called Euler’s Method, to generate a
more accurate sketch of the original function, f (t).
6t
df
=
,
• The DE is
dt 1 + t
• the initial condition is f (0) = 3, and
• we will break up time into small intervals of ∆t = 0.2.
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
Euler’s Method
• Start at a given point, (t, y)
• Choose a time interval, ∆t
dy
• Compute the slope there, , using the DE
dt
• Compute the coordinates of the next point:
next t = t + ∆t
dy
∆t
dt
• Repeat until either t or y reaches a value of interest.
next y = y + ∆y = y +
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Solving Pure-Time Differential Equations
From Section 4.2
We now consider how to get a formula for F (t), starting from a differential equation,
dF
= f (t)
dt
We will refer to the process of reversing the derivative as anti-differentiation
or integration. In symbols,
if
dF
= f (t)
dt
then
Z
F (t) = f (t) dt
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
Example: Solve the following differential equation for y(t)
dy
= t3
dt
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Example: Solve the following differential equation for y(t)
dy
= t5
dt
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
Power Rule for Integrals
The general rule for integrating a power of t will be
Z
tn dt =
What is the one value of n for which this rule does not work?
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Question: The integral
evaluates to
5
3 3/2
A. 3 + x + C
3x
2
−5 2 3/2
B. 3 + x + C
3x
3
−5 2 3/2
C. 1 + x + C
x
3
−5
D. 1 − 2x−1/2 + C
x
−5 1
E. 1 − x−1/2 + C
x
2
Z 5
+
2
x
√
x dx
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
23
In this example we implicitly used the Sum rule and Constant Product Rules.
Sum Rule
Z
(f (x) + g(x)) dx =
Constant Product Rule
Z
Z
f (x) dx +
kf (x) dx = k
Z
Z
f (x) dx
g(x) dx
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Example: The acceleration near the Earth’s surface is 9.8 m/s2. Derive the
classic velocity and position formulas for the height of a thrown object over
time, given an initial height y(0) = y0 and initial vertical velocity v(0) = v0.
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
25
Other Integral Rules
From Section 4.3
We know how to integrate polynomial functions, or any sum of powers of t or x.
Our next step is to enlarge our toolkit to determine the integrals of other commonly
encountered functions.
d
ln(x)
Find the derivative
dx
Evaluate
Z
1
dx
x
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d 4x
Evaluate the derivative
e
dx
Evaluate the integral
Z
e4x dx
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
Evaluate the integral
Z
cos(t) dt
Evaluate the integral
Z
sin(t) dt
Evaluate the integral
Z
√
1
dx
2
1−x
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Question: Evaluate the integral
Z
4
3
+ 2 − 5e−0.1x dx
x x
A. 4 ln(x) +3(ln(x))2 +0.5e−0.1x + C
B. 4 ln(x) +3(ln(x))2 + 50e−0.1x + C
1
C. 4 ln(x) −3
x
1
D. 4 ln(x) −3
x
1
1
E. −4 2 −6 3
x
x
+ 0.5e−0.1x + C
+ 50e−0.1x + C
+0.5e−0.1x + C
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
29
Integration by Substitution
Z
√
1 4
3
4
Show that
x x + 5 dx = (x + 5)3/2 + C.
6
We note the structure of the derivative, as a result of the chain rule:
√
3
x x4 + 5
Integrals with this form are classic problems which can be handled by integration
by substitution.
30
Evaluate the integral
Z
x3(x4 + 5)1/2 dx using substitution.
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
Example: Evaluate the integral
Z
2 x3
xe
dx.
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32
Example:
Evaluate the integral
Z
tan(x) dx.
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
Confirm your answer.
Note that you can always check your integration steps by taking a derivative.
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Sometimes substitutions makes the booking keeping of integration easier.
Example: Use a substitution to help evaluate the integral
Z
40 + 5 cos (2π(t + 1.5)) dt
Week 9 – Differential Equations, Anti-Derivatives, Integration by Substitution
35
Sometimes a substitution will still simplify the integral, even if you don’t see the
cue of “function and its derivative” in the integrand.
Example: Consider the integral
Z
1
√
dx
x+1
√
Why isn’t the anti-derivative simply ln( x + 1)?
36
Find the true formula for the anti-derivative/integral
Z
1
√
dx
x+1