Summary Difference- & Differential Equations for EOR (RUG)
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Course
Difference- & Differential Equations (EBB812A05)
Institution
Rijksuniversiteit Groningen (RuG)
Book
Further Mathematics For Economic Analysi
Summary for the course Difference- & Differential Equation for the bachelor programme Econometrics & Operations Research containing all important concepts discussed in the lectures. Lecture slides made by A. van der Made. 2nd year course taught by D. Vullings.
1 Week 1
1.1 Lecture 1
A first order differential equation (first order ODE) is an equation of the form
F (t, x(t), x′ (t)) = 0, t∈T, (1)
where F is a function of (at most) 3 variables and T ⊆ R is connected.
NB: we confine attention to the ODEs that can be written as a recurrence relation:
x′ (t) = H(t, x(t)), t∈T,
for some function H.
Some important concepts:
− Ordinary: the function x is only differentiated with respect to one variable. This
variable is often time and denoted t.
− A first-order ODE that does not depend on t explicitly, i.e. that can be written as
F (x(t), x′ (t)) = 0, is called autonomous.
− If (y, z) 7→ F (t, y, z) is affine for all t ∈ T , then (1) is linear.
− A solution of the ODE (1) is a differentiable function x : T → C that satisfies (1).
− The general solution of (1) is the set containing all solutions of (1). So, an element of
the general solution is a function.
− An ODE like (1) together with an initial condition x(t0 ) = x0 is called an initial value
problem. A solution of the ODE that also satisfies the initial condition is a solution of
the initial value problem.
Let f and g be continuous functions. Four types of first order ODEs for which a general
method can be used to find solutions are:
1) x′ (t) = g(t) (type I ODE)
2) x′ (t) = f (t)g(x(t)) (seperable ODE)
3) x′ (t) = f (t)x(t) (homogeneous linear ODE)
4) x′ (t) = f (t)x(t) + g(t), g ̸≡ 0 (inhomogenous linear ODE)
NB: The symbol ≡ is used for constant functions, i.e. functions that attain only one value
over their entire domain.
Solutions of Type I ODEs
Consider an ODE of the following form:
x′ (t) = g(t), t∈T.
Solutions of this type can be found by integrating both sides:
Z t Z t Z t
′
x(t) − x(t0 ) = x (s)ds = g(s) ⇒ x(t) = g(s)ds + x(t0 ).
t0 t0 t0
So, the general solution of a type I equation reads
Z t
x(t) = g(s)ds + c, t ∈ T , c ∈ C.
t0
Solutions of Seperable ODEs
Consider an ODE of the following form:
x′ (t) = f (t)g(x(t)), t∈T.
Suppose g(x) ̸= 0 for all x. The ODE can then be written as follows:
x′ (t)
= f (t), t∈T.
g(x(t))
Suppose we can find a primitive P of 1/g and a primitive F of f . Then by the Chain Rule:
Z t ′ Z t
x (t)
ds = P (x(t)) − P (x(t0 )) = f (s)ds = F (t) − F (t0 ).
t0 g(x(t)) t0
yielding the implicit general solution
P (x(t)) = F (t) + c, t∈T, c ∈ C.
Solutions of Homogeneous Linear ODEs
Consider an ODE of the following form:
x′ (t) = f (t)x(t), t∈T
Because this ODE is a special case of a seperable equation (with g : x 7→ x), we can again
apply the method of seperation of variables:
Z t ′ Z t
x (s)
ds = f (s)ds ⇒ log|x(t)| = F (t) + c, c ∈ R,
t0 x(s) t0
where F is a primitive of f .
So, |x(t)| = eF (t)+c and the general solution is consequently
x(t) = DeF (t) , t∈T, D ∈ R.
The General Solution of Inhomogeneous Linear ODEs
Consider an ODE of the form
x′ (t) = f (t)x(t) + g(t), t∈T, (2)
with g ̸≡ 0.
To find solutions of this ODE we use the following result:
Theorem:
Let x∗ be a particular solution of (2). Then every solution of (2) can be written as the
sum of x∗ and a solution of the homogeneous equation x′ (t) = f (t)x(t). Conversely, any
function that can be written as the sum of x∗ and a solution of x′ (t) = f (t)x(t) is a solution
of (2). The proof is as follows:
So, x1 − x∗ is a solution of the homogeneous equation x′ = f x. The first claim now
follows by noting that x1 = x∗ + (x1 − x∗ ).
• Let y be a solution of the homogeneous equation x′ = f x. Then:
(x∗ + y)′ = x∗′ + y ′ = (f x∗ + g) + f y = f (x∗ + y) + g.
We conclude that x∗ + y is a solution of (2).
Example
Consider the following ODE:
2 +t
x′ (t) = 2tx(t) + et , t ∈ R.
We first determine the general solution of x′ (t) = 2tx(t):
x′ (t) 2
= 2t ⇒ log|x(t)| = t2 + c̃, c̃ ∈ R ⇒ x(t) = cet , c ∈ R.
x(t)
2 +t
Next, we ”figure out” that x∗ (t) = et is a particular solution of the inhomogeneous ODE.
So, the general solution reads:
2 2 +t
x(t) = cet + et , t ∈ R, c ∈ R.
4
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