MAT 412: Ordinary Differential Equation II:
Introduction to Second Order ODE
Imaga Ogbu Famous
Department of Mathematics
Covenant University
Imaga O. F. (Dept. of Mathematics)
Ordinary Differential Equation II
Second-Order ODE
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Objectives
Learning Objectives
1
Identify linear and nonlinear differential equations
2
Identify homogeneous and nonhomogenous linear equation
3
Identify linearly dependent and independent functions
Imaga O. F. (Dept. of Mathematics)
Ordinary Differential Equation II
Second-Order ODE
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The general form of second order linear differential equation is
Definition
P(x)y′′ + Q(x)y′ + R(x)y = G(x)
(1)
where P(x), Q(x), R(x) and G(x) are real valued functions and P(x) ̸=
0. x is the independent variable while y is the dependent variable.
Equation (1) is a homogeneous linear equation if G(x) = 0. If G(x) ̸= 0
then, it is a non homogeneous linear equation.
Equation (1) is linear because y and its derivatives are raised to the first
power and are not multiplied by one another.
Imaga O. F. (Dept. of Mathematics)
Ordinary Differential Equation II
Second-Order ODE
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Example
Classify the following equations:
y′′ + 5x3 y′ + xy3 = 3x
(sin 2x)y′′ + (cos x)y′ + 6y = 0
4u2 x′′ + 3uxx′ + 4x = 0
xy′′ + xy = 4x6
(sin t)y′′ − sin y′ + (cos t)y − sin t = 0
(sin x2 )y′′ − (cos x)y′ + x3 y = y − 2
7xy′′ − 5x3 y′ + 3xy − 6x3 = 0
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Ordinary Differential Equation II
Second-Order ODE
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Solution of a Differential Equation
A solution of a differential equation is a relation between the dependent
and independent variable when substituted into the differential equation
reduces it to an identity.
Show that
y = x3 is a solution of x2 y − xy − 3y = 0
y = 2x2 is a solution of 12 x2 y′′ − xy′ + y = 0.
Imaga O. F. (Dept. of Mathematics)
Ordinary Differential Equation II
Second-Order ODE
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Superposition Principle
The linear combination of two functions y1 (x) and y2 (x) is a function
y(x) = c1 y1 (x) + c2 y2 (x)
where c1 and c2 are two arbitrary constants.
Theorem
If y1 (x) and y2 (x) are solutions to a linear homogeneous differential
equation, then the function
y(x) = c1 y1 (x) + c2 y2 (x),
where c1 and c2 are constants is also a solution.
Imaga O. F. (Dept. of Mathematics)
Ordinary Differential Equation II
Second-Order ODE
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Exercises
Prove the Superposition Principle.
If e−x and e5x are solutions to y′′ − 4y′ − 5y = 0, show that
3e−x + 2e5x is also a solution
If e−2x and e−3x are solutions to y′′ + 5y′ + 6y = 0, show that
2e−2x + 3e−3x is also a solution.
Imaga O. F. (Dept. of Mathematics)
Ordinary Differential Equation II
Second-Order ODE
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Linearly Dependent and Independent Functions
Definitions: Linearly Dependent functions
A set of functions f1 (x), f2 (x), . . . , fn (x) is said to be linearly dependent if there are constants c1 , c2 , . . . , cn not all zero such that
c1 f1 (x) + c2 f2 (x) + · · · + cn fn (x) = 0
for all x over the interval of interest. A set of functions that is not
linearly dependent is said to be linearly independent i.e.
c1 f1 (x) + c2 f2 (x) + · · · + cn fn (x) = 0
implies that c1 = c2 = · · · = cn = 0
Imaga O. F. (Dept. of Mathematics)
Ordinary Differential Equation II
Second-Order ODE
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Proof: Linearly Dependent I
Suppose f2 (x) ≡ 0, then choose c1 = 0 and c2 = 1 then the condition of
linear dependence is met i.e.
c1 f1 (x) + c2 f2 (x) = 0 + 0 = 0.
Conversely, if f1 (x) ̸= 0 and f2 (x) ̸= 0 but f1 (x) = Cf2 (x) for some
constant C, then choose c1 = C and c2 = −1, then the condition of linear
dependence is also met i.e.
c1 f1 (x) + c2 f2 (x) = 0 + 0 = 0.
Next we show that if two functions are linearly dependent, then either one
is identically zero or they are constant multiples of one another. Assume
Imaga O. F. (Dept. of Mathematics)
Ordinary Differential Equation II
Second-Order ODE
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Proof: Linearly Dependent II
f1 (x) and f2 (x) are linearly dependent. Then, there are constants, c1 and
c2 , not both zero, such that
c1 f1 (x) + c2 f2 (x) = 0
for all x over the interval of interest. Then,
c1 f1 (x) = −c2 f(x).
Since, c1 , c2 can’t both be zero, assume c2 ̸= 0. Then, there are two cases:
either c1 = 0 or c1 ̸= 0. If c1 = 0, then
0 = −c2 f2 (x)
0 = f2 (x),
Imaga O. F. (Dept. of Mathematics)
Ordinary Differential Equation II
Second-Order ODE
10 / 16
Proof: Linearly Dependent III
so one of the functions is identically zero. Now suppose c1 ̸= 0. Then,
(
)
c2
f1 (x) = −
f2 (x)
c1
and we see that the function are constant multiples of one another.
Imaga O. F. (Dept. of Mathematics)
Ordinary Differential Equation II
Second-Order ODE
11 / 16
Linear Dependence of Two Functions
Theorem: Linearly Dependence of Two Functions
Two functions, f1 (x) and f2 (x), are said to be linearly dependent if
either one of them is identically zero or if f1 (x) = Cf2 (x) for some
constant C and for all x over the interval of interest. Functions that
are not linearly dependent are said to be linearly independent.
Determine if the following pairs are linearly dependent or linearly
independent
f1 (x) = x2 and f2 (x) = 5x2
f1 (x) = sin x and f2 (x) = cos x
f1 (x) = e3x and f2 (x) = e−3x
f1 (x) = 3x and f2 (x) = 3x + 1
Imaga O. F. (Dept. of Mathematics)
Ordinary Differential Equation II
Second-Order ODE
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General Solution to a Homogeneous Equation
Theorem: Linearly Dependence of Two Functions
If y1 (x) and y2 (x) are linearly independent solutions to a second-order,
linear homogeneous differential equations, then the general solution
is given by
y(x) = c1 y1 (x) + c2 y2 (x),
where c1 and c2 are constants.
If y1 (t) = e3t and y2 (t) = e−3t are solutions to y′′ − 9y = 0, what is the
general solution.
Imaga O. F. (Dept. of Mathematics)
Ordinary Differential Equation II
Second-Order ODE
13 / 16
Second-Order Equations with Constant Coefficients
General Form
ay′′ + by′ + cy = 0,
where a, b, c are constants.
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Ordinary Differential Equation II
Second-Order ODE
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Characteristic Equation
Characteristic Equation
The characteristic equation of the second-order differential equation
ay′′ + by′ + cy = 0 is aλ2 + bλ + c = 0.
Proof:
Let y = eλx , then y′ = λeλx , y′′ = λ2 eλx . Substituting it into the
second-order equation gives
ay′′ + by′ + c = a(λ2 eλx ) + b(λeλx ) + ceλx
= eλx (aλ2 + bλ + c).
Since eλx ̸= 0, then
Imaga O. F. (Dept. of Mathematics)
(aλ2 + bλ + c) = 0.
Ordinary Differential Equation II
Second-Order ODE
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The solution of the characteristic equation using the quadratic formula is
√
−b ± b2 − 4ac
λ=
.
2a
This gives three cases. The characteristic equation has
distinct real roots;
repeated real root; or
complex conjugate.
Imaga O. F. (Dept. of Mathematics)
Ordinary Differential Equation II
Second-Order ODE
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