Lecture 8 - Lamia Jaafar Belaid

Chapter 2
Higher order differential equations
Lecture8
ODE (Lecture8)
Introduction to ODE-Math214
Dr Lamia Jaafar
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General cases
1
Generalization of the auxiliary equation associated to an n−order
linear differential equation with constant coefficients.
2
Reduction of the order (The coefficients of the ODE are not
necessary constant).
3
Cauchy Euler equation
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Introduction to ODE-Math214
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Generalization of the auxiliary equation associated to
an n−order linear differential equation with constant
coefficients
ODE (Lecture8)
Introduction to ODE-Math214
Dr Lamia Jaafar
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Recall
Let
(E) : an y (n) + an−1 y (n−1) + .. + a1 y 0 + a0 = f (x),
a linear differential equation of order n.
We recall that if y1 , .., yn are n linearly independent solutions of (Eh ) :
the homogeneous equation associated to (E), then by the
superposition principle, the homogeneous solution is given by
yh (x) = α1 y1 + .. + αn yn .
Generalized method for the auxiliary equation
We propose to generalize the method applied to second order linear
differential equations, by solving the auxiliary equation
an r n + an−1 r n−1 + ... + a1 r + a0 = 0
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Introduction to ODE-Math214
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Generalization
The solution of the n-order homogeneous linear differential equation
(Eh ) depends on the nature of the roots r1 , .., rn of the associated
auxiliary equation.
Example1: Third order differential equation
Find the solution of the ODE defined by y (3) + y 00 = 0
Answer: The solutions of the auxiliary equation r 3 + r 2 = 0 are given
by r1 = −1 and r2 = r3 = 0. We deduce that the solution is
y = α + βx + γe−x , α, β, γ ∈ R.
Deduce the general solution of y (3) + y 00 = ex cos x.
1 x
solution yp = − 10
e cos x + 51 ex sin x.
ODE (Lecture8)
Introduction to ODE-Math214
Dr Lamia Jaafar
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Example2: Third order differential equation
Solve y 000 + 3y 00 − 4y = 0.
Answer: The solutions of the auxiliary equation r 3 + 3r 2 − 4 = 0 are
given by r1 = 1 and r2 = r3 = −2. We deduce that the solution is
y = αex + βe−2x + γxe−2x , α, β, γ ∈ R.
Deduce the general solution of y 000 + 3y 00 − 4y = ex + x 2 + 1.
solution yp = 19 xex − 14 x 2 − 85 .
ODE (Lecture8)
Introduction to ODE-Math214
Dr Lamia Jaafar
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Example3: Fourth order differential equation
Solve y (4) + y (3) = 0.
Answer: The solutions of the auxiliary equation r 4 + r 3 = 0 are
r1 = −1 and r2 = r3 = r4 = 0. deduce that the solution is given by
y = αe−x + β + γx + δx 2 , α, β, γ, δ ∈ R.
Solve y (4) + y (3) = 1 − x 2 e−x .
Hint:
yp = Ax 3 + Bx 3 e−x + Cx 2 e−x + Dxe−x , A, B, C, D ∈ R.
Solve y (4) + 2y (2) + y = 0.
Answer: y = α cos x + β sin x + γx cos x + δx sin x, α, β, γ, δ ∈ R.
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Introduction to ODE-Math214
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Reduction of the order
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Introduction to ODE-Math214
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Proposed Method
Let (E) : y 00 + P(x)y 0 + Q(x)y = 0, a given ODE, where P, Q are
continuous functions on a given interval I. We want to solve (E)
which has not necessary constant coefficients.
Let y1 a given solution of (E). By considering y = uy1 , we prove
that (E) is equivalent to u 00 y1 + 2u 0 y10 + Pu 0 y1 = 0.
Let w = u 0 . By replacing in (E), we obtain a separating variables
equation such that
R its integration gives
2
wy1 = λ exp(− Pdx), λ ∈ R.
The second solution is deduced by integrating u 0 = w.
The general solution is deduced using the superposition principle.
ODE (Lecture8)
Introduction to ODE-Math214
Dr Lamia Jaafar
9 / 15
By considering x 2 a first solution of (E) : x 2 y 00 − 3xy 0 + 4y = 0,
deduce the general solution of (E), onR?+ .
Answer:
Let y = x 2 u ⇒ u 00 x + u 0 = 0.
Let w = u 0 ⇒ w 0 x = −w a separating variables first order
differential equation.
α
w = , α ∈ R.
x
u = α ln x + β, α, β ∈ R.
The general solution (superposition principle) is given by
y = αx 2 + βx 2 ln x, α, β ∈ R.
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Introduction to ODE-Math214
Dr Lamia Jaafar
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Cauchy Euler Equation
ODE (Lecture8)
Introduction to ODE-Math214
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Definition
A Cauchy Euler equation is a non homogeneous n− order differential
linear equation having the general form
(E) : an x n y (n) + an−1 x n−1 y (n−1) + .. + a1 xy 0 + a0 y = f (x).
Proposed method
We search for a solution y = x m .
m verifies an auxiliary equation associated to (E).
Three cases can be discussed:
2
distinct roots mi :⇒ y = α1 x m1 + .. + αn x mn , αi ∈ R,
repeated roots: let m1 a root of multiplicity
X
k ⇒ y = α0 x m1 + α1 x1m ln x + .. + αk−1 x1m (ln x)k−1 +
β i x mi , m i :
3
distinct roots, αi , βi ∈ R,
conjugate
X complex roots: mk = αk + iβk , mk = αk − iβk ⇒
y=
x αk (ck1 cos βk ln x + ck2 sin βk ln x), ck1 , ck2 ∈ R.
1
i
k
ODE (Lecture8)
Introduction to ODE-Math214
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Exercise
Solve the following Cauchy Euler ODE’s
1
x 2 y 00 − 2xy 0 − 4y = 0,
2
4x 2 y 00 + 8xy 0 + y = 0,
3
2x 2 y 00 + y = 0.
ODE (Lecture8)
Introduction to ODE-Math214
Dr Lamia Jaafar
13 / 15
Let us summarize this chapter!
To solve a given ODE or a given IVP/BVP, we need to classify the
considered ODE. 4 kinds have been studied. Two of them are with
constant coefficients:
A second ODE with constant coefficients:
ay 00 + by 0 + cy = f (x) ⇒ y(x) = yh (x) + yp (x) and yh depends on
the sign of ∆ = b 2 − 4ac.
A higher ODE with constant coefficients:
an y (n) + an−1 y (n−1) + .. + a0 y = f (x) ⇒ y(x) = yh (x) + yp (x) and
yh depends on the solutions of the characteristic equation
an r n + an−1 r n−1 + .. + a0 = 0, as a generalization of the previous
case.
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Introduction to ODE-Math214
Dr Lamia Jaafar
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Let us summarize this chapter!
Two of them are with variable constant coefficients:
A second ODE with variable coefficients:
y 00 + P(x)y 0 + Q(x)y = 0 ⇒ we change to y = uy0 where y0 is a
particular solution and then we change to v = u 0 . We obtain a first
order ODE.
A particular higher ODE with variable coefficients, which is the
Cauchy Euler equation ⇒ we use y = x m and deduce the solution
with respect to m (3 cases are possible.)
ODE (Lecture8)
Introduction to ODE-Math214
Dr Lamia Jaafar
15 / 15

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