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Differential Equations with Boundary-Value Problems, Exercise 7.1, 7 to 10
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WTW 256 – Differential EquationsStudy Notes
Source: Class notes & Textbook
Reference tables
Theme 1: First order differential equations
Theme 2: Second and higher order linear DE’s
Theme 3: The Laplace transform and linear DE’s
Theme 4: Systems of linear DE’s
Python code
Source: Class notes, Textbook and Study guide
Textbook: Zill/Wright - Differential Equations with Boundary-Value
Problems, 9th edition, International Metric Edition ISBN 9781337559881
, Laplace transforms
1
ℒ{1} =
𝑠
𝑛!
ℒ{𝑡𝑛 } =
𝑠𝑛+1
1
ℒ{𝑒𝑎𝑡 } =
𝑠−𝑎
𝑒−𝑎𝑠
ℒ{𝒰(𝑡 − 𝑎)} =
𝑠
𝑘
ℒ{sin(𝑘𝑡)} =
𝑠2 + 𝑘2
𝑠
ℒ{cos(𝑘𝑡)} =
𝑠2 + 𝑘2
𝑘
ℒ{sinh(𝑘𝑡)} =
𝑠2 − 𝑘2
𝑠
ℒ{cosh(𝑘𝑡)} =
𝑠2 − 𝑘2
𝑑𝑓
ℒ{ } = 𝑠 ∙ ℱ(𝑠) − 𝑓(0)
𝑑𝑡
𝑑2 𝑓
ℒ{ } = 𝑠2 ∙ ℱ(𝑠) − 𝑠 ∙ 𝑓(0) − 𝑓′ (0)
𝑑𝑡2
ℒ{𝑓(𝑡)𝑒 𝑎𝑡 } = ℱ(𝑠 − 𝑎)
ℒ{𝒰(𝑡 − 𝑎) ∙ 𝑓(𝑡 − 𝑎)} = 𝑒−𝑎𝑠 ∙ ℱ(𝑠)
ℒ{𝑔(𝑡) ∙ 𝒰(𝑡 − 𝑎)} = 𝑒−𝑎𝑠 ∙ ℒ{𝑔(𝑡 − 𝑎)}
ℒ{𝑡𝑛 ∙ 𝑓(𝑡)} = (−1)𝑛 ∙ ℱ [𝑛] (𝑠)
ℒ{𝑡 ∙ 𝑓(𝑡)} = −ℱ ′ (𝑠)
ℒ{𝑓(𝑡) ∗ 𝑔(𝑡)} = 𝐺(𝑠) ∙ ℱ(𝑠)
𝑡
ℱ(𝑠)
ℒ {∫ 𝑓(𝜏)𝑑𝜏} =
0 𝑠
This table will be given in exams and tests.
,Standard derivatives
𝑑𝑦
(𝑘) = 0
𝑑𝑥
𝑑𝑦 𝑛
(𝑥 ) = 𝑛𝑥 𝑛−1
𝑑𝑥
𝑑𝑦 𝑥
(𝑒 ) = 𝑒 𝑥
𝑑𝑥
𝑑𝑦 𝑥
(𝑎 ) = ln(𝑎) ∙ 𝑎 𝑥
𝑑𝑥
𝑑𝑦
(sin(𝑥)) = cos(𝑥)
𝑑𝑥
𝑑𝑦
(cos(𝑥)) = − sin(𝑥)
𝑑𝑥
𝑑𝑦
(tan(𝑥)) = sec 2 (𝑥)
𝑑𝑥
𝑑𝑦
(𝑐𝑜𝑠𝑒𝑐(𝑥)) = −𝑐𝑜𝑠𝑒𝑐(𝑥) ∙ cot(𝑥)
𝑑𝑥
𝑑𝑦
(sec(𝑥)) = sec(𝑥) tan(𝑥)
𝑑𝑥
𝑑𝑦
(cot(𝑥)) = −𝑐𝑜𝑠𝑒𝑐 2 (𝑥)
𝑑𝑥
𝑑𝑦 1
(ln(𝑥)) =
𝑑𝑥 𝑥
𝑑𝑦 1
(log 𝑎 𝑥) =
𝑑𝑥 𝑥 ∙ ln(𝑎)
𝑑
(𝑓(𝑥) ∙ 𝑔(𝑥)) = 𝑓 ′ (𝑥) ∙ 𝑔(𝑥) + 𝑓(𝑥) ∙ 𝑔′ (𝑥)
𝑑𝑥
𝑑 𝑓(𝑥) 𝑓 ′ (𝑥) ∙ 𝑔(𝑥) − 𝑔′ (𝑥) ∙ 𝑓(𝑥)
( )= 2
𝑑𝑥 𝑔(𝑥) (𝑔(𝑥))
𝑑
(𝑓(𝑥) 𝑜 𝑔(𝑥)) = 𝑓 ′ (𝑔(𝑥)) ∙ 𝑔′ (𝑥)
𝑑𝑥
, Standard integrals
∫ 𝑘 ∙ 𝑑𝑥 = 𝑘𝑥 + 𝐶
1
∫ 𝑥 𝑛 ∙ 𝑑𝑥 = 𝑥 𝑛+1 + 𝐶
𝑛+1
∫ 𝑒 𝑥 ∙ 𝑑𝑥 = 𝑒 𝑥 + 𝐶
1
∫ 𝑎 𝑥 ∙ 𝑑𝑥 = 𝑎𝑥 + 𝐶
ln(𝑎)
∫ sin(𝑥) ∙ 𝑑𝑥 = − cos(𝑥) + 𝐶
∫ cos(𝑥) ∙ 𝑑𝑥 = sin(𝑥) + 𝐶
∫ sec 2(𝑥) ∙ 𝑑𝑥 = tan(𝑥) + 𝐶
∫ 𝑐𝑜𝑠𝑒𝑐(𝑥) cot(𝑥) ∙ 𝑑𝑥 = −𝑐𝑜𝑠𝑒𝑐(𝑥) + 𝐶
∫ sec(𝑥) tan(𝑥) ∙ 𝑑𝑥 = sec(𝑥) + 𝐶
∫ 𝑐𝑜𝑠𝑒𝑐 2 (𝑥) ∙ 𝑑𝑥 = − cot(𝑥) + 𝐶
1
∫ ∙ 𝑑𝑥 = ln(|𝑥|) + 𝐶
𝑥
∫ 𝑓 ′ (𝑥) ∙ 𝑔(𝑥) ∙ 𝑑𝑥 = 𝑓(𝑥) ∙ 𝑔(𝑥) − ∫ 𝑓(𝑥) ∙ 𝑔′ (𝑥) ∙ 𝑑𝑥 + 𝐶
,Theme 1: First order differential equations
7 Lectures [week 1 – 4]
This theme will cover the basic ideas of differential equations and more
specifically look at special forms of first order differential equations as well as
solution methods and applications. This theme will be separated into 5 sub-
themes:
Theme 1.1: Basic ideas
Theme 1.2: Separable differential equations and their applications
Theme 1.3: Linear first order differential equations
Theme 1.4: Substitution methods
Theme 1.5: Autonomous differential equations
Pre-knowledge to this theme consists of integration and differentiation
techniques as well as partial fractions.
,Theme 1.1 – Basic ideas
What is a differential equation?
A differential equation (DE) is an equation where the functions and its
derivatives are present within the equation, where the solution will be the
function. The solution is no more only a constant (although still possible) but a
time dependant function.
Consider a mass-spring system, let y be the displacement (in m), M, the mass (in
kg) and the spring stiffness, k (in N/m). Time in seconds. Neglect damping.
Using force equilibrium on the mass:
𝐹 = 𝑀𝑦̈
𝑑2𝑦
𝑀∙ 2 +𝑘∙𝑦 =0
𝑑𝑡
And there you have it, a DE, where the displacement (the function) and
acceleration (a derivative of the function) is present.
Using KVL in the circuit:
𝑉(𝑡) = 𝑉𝑐 + 𝑉𝑅
1
𝑉(𝑡) = 𝑖(𝑡) ∙ 𝑅 + ∙ ∫ 𝑖(𝑡) 𝑑𝑡
𝐶
𝑑𝑉 𝑑𝑖(𝑡) 1
= ∙ 𝑅 + ∙ 𝑖(𝑡)
𝑑𝑡 𝑑𝑡 𝐶
The current, i(t), is present in this equation as both a function and its derivative.
,What is an initial value and an initial value problem?
An initial value is some value that is known of a system at the start. This value
must also be satisfied by the solution.
Considering the spring-mass system. It is known that the initial velocity, y’(t),
should be zero and the initial displacement of the mass (compressed/stretched
the spring), y(0) = -d.
𝑑2𝑦
𝑀 ∙ 2 + 𝑘 ∙ 𝑦 = 0; 𝑦(0) = −𝑑; 𝑦 ′ (0) = 0
𝑑𝑡
This is known as an initial value problem (IVP), basically a DE with initial
values. Solving this IVP will give us a solution that will be time dependant, that
is, the displacement. The general solution to this DE is:
𝑘 𝑘
𝑦(𝑡) = 𝐶1 cos (√ ∙ 𝑡 ) + 𝐶2 sin (√ ∙ 𝑡)
𝑀 𝑀
The specific solution, however, to the IVP is:
𝑘
𝑦(𝑡) = −𝑑 ∙ cos (√ ∙𝑡)
𝑀
This is a specific solution. This only applies for the problems with the same
initial values.
Validating the solution:
Let d = 0.02m, k = 50N/m, M = 0.5kg
𝑚
𝑦(𝑡) = −0.02 cos(10𝑡) : [𝑦(0) = 0.02𝑚]; 𝑦 ′ (𝑡) = 0.2 sin(10𝑡) : [𝑦 ′ (0) = 0 ]
𝑠
,What is the order of a DE? What does it mean?
The order of a DE is the same as its order with respect to its derivatives. The
order of a DE is the highest order derivative present within a DE. The spring-
mass system is a second order DE as it contains the second derivative of a
function, whereas the RC-circuit only contains the first derivative of the current
and therefore called a first order DE. Both examples are linear DE’s.
Linear vs non-linear DE’s.
Linear DE’s presents the function to a linear exponent, that is, the function is
always and only raised to the power of one and never an argument of a non-
linear function such as logarithms or trigonometric functions. Non-linear DE’s
may be raised to powers greater than one and may be arguments of non-linear
functions or both. Both examples are linear DE’s.
General vs specific solutions.
Re-consider the two solutions of the spring mass system:
𝑘 𝑘
𝑦(𝑡) = 𝐶1 cos (√ ∙ 𝑡 ) + 𝐶2 sin (√ ∙ 𝑡)
𝑀 𝑀
And
𝑘
𝑦(𝑡) = 𝑑 ∙ cos (√ ∙𝑡)
𝑀
The first solution is obtained without considering the IVP, this results in the
constants C1 and C2 to be determined. This is a general solution which will
belong to a family of solutions. The difference only results in the different initial
values that needs to be satisfied. In the second case the IVP is considered that
led to the specific solution where C1 and C2 could be determined.
Implicit vs explicit solutions.
Explicit solutions present the function as the subject of an equation whereas
implicit has a messy solution where the function is not the subject of the
equation.
,Theme 1.2 – Separable DE’s and applications
When is a DE separable?
Separable DE’s can be written in the form:
𝑑𝑦
= 𝑓(𝑦) ∙ 𝑔(𝑡)
𝑑𝑡
For example:
𝑑𝑦 𝑑𝑦
= 𝑒 𝑦+𝑡 ⇒ = 𝑒𝑡 ∙ 𝑒𝑦
𝑑𝑡 𝑟𝑒−𝑤𝑟𝑖𝑡𝑒 𝑑𝑡
𝑓(𝑦) = 𝑒 𝑦 ; 𝑔(𝑡) = 𝑒 𝑡
How does one go about solving a separable DE?
Now solving this DE:
𝑒 −𝑦 𝑑𝑦 = 𝑒 𝑡 𝑑𝑡
∫ 𝑒 −𝑦 𝑑𝑦 = ∫ 𝑒 𝑡 𝑑𝑡
𝑒 −𝑦 = −𝑒 𝑡 − 𝐶 → 𝐼𝑚𝑝𝑙𝑖𝑐𝑖𝑡 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
0
𝑦(𝑡) = − ln(−𝑒 𝑡 − 𝐶) → 𝐸𝑥𝑝𝑙𝑖𝑐𝑖𝑡 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
0
Now solving the IVP:
𝑦(0) = 0
0 = − ln(−1 − 𝐶) → 𝐶 = −2
𝐶
𝑦(𝑡) = −ln (−𝑒 𝑡 + 2)
, Uniqueness theorem for separable DE’s.
Suppose the function f’(y) and g(t) are continuous on the intervals J and I
respectively. If I contains ‘a’ and J contains ‘b’, then the IVP will have at most
one solution. (Basically saying, if there is a solution, there will only be one. This
does not confirm that there is a solution)
𝑑𝑦
= 𝑓(𝑦) ∙ 𝑔(𝑡) ; 𝑦(𝑎) = 𝑏
𝑑𝑡
Considering the previous example, f’ and g where both continuous on the
intervals J and I, J and I both being all real numbers. Both a and b are contained
inside the respected intervals.
What are the applications of separable DE’s?
Exponential growth or decay usually deals with a population growth (people,
micro-organisms and wildlife), and decay such as half-life of a substance. The
exponential growth takes the form:
𝑑𝑃(𝑡)
= 𝑘 ∙ 𝑃(𝑡)
𝑑𝑡
Suppose a town has an initial population of 100 people. After 10 years the
population increased by 20 people. How large will the population be after 20
years?
𝑃(0) = 100; 𝑃(10) = 120; 𝑃(20) =?
𝑑𝑃(𝑡)
= 𝑘 ∙ 𝑃(𝑡)
𝑑𝑡
1
∫ 𝑑𝑃 = ∫ 𝑘 𝑑𝑡
𝑃(𝑡)
ln(𝑃(𝑡)) = 𝑘𝑡 + 𝐶
𝑃(𝑡) = 𝐴𝑒 𝑘𝑡
𝑃(0) = 100: 100 = 𝐴
120
ln (100)
𝑃(10) = 120: 𝑘 = ( ) = 0.0182
10
𝑃(20) =? : 𝑃(20) = 100𝑒 0.0182(20) = 144
After 20 years the total population is expected to be 144 people.
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