Linear Algebra and ODE
(MAT228)
Chapter - IV
Second Order Differential Equations
Prof. Zakir H. Ahmed
Office# 2030 Phone# 2582172
E-mail: zaahmed@imamu.edu.sa
10/19/2022
Prepared by Prof. Zakir H. Ahmed
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This Chapter Contains the following
Topics:
4.2 Homogeneous Linear Equations
4.3 Auxiliary Equations with Complex Roots
4.4 Nonhomogeneous Equations
4.5 The Superposition Principle etc. Revisited
4.6 Variation of Parameters
4.7 Variable-Coefficient Equations
Slides are prepared from the following textbook:
• Fundamentals of Differential Equations: R. Kent
Nagle, Edward B. Saff, Arthur David Snider, 2012,
8th Edition.
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4.2 Homogeneous Linear Equations
 Consider the following linear second-order constantcoefficient differential equation
ay’’ + by’ + cy = f(t) (a ≠ 0) ------------- (1)
with the special case where the function f(t) is zero:
ay’’ + by’ + cy = 0 ----------------------- (2)
Equation (2) is called the homogeneous form of equation
(1); f(t) is the “nonhomogeneity” in (1).
If we substitute y = ert into (2), we obtain
ar2ert + brert + cert = 0
=> ert (ar2 + br + c) = 0
=> ar2 + br + c = 0 -------------------------- (3)
So, y = ert is a solution to (2) iff. r satisfies eq. (3),
which is called the auxiliary (characteristic) equation.
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4.2 Homogeneous Linear Equations
 Now the auxiliary equation is just a quadratic, and its
roots are:
If the discriminant, b2 - 4ac > 0, the roots r1 and r2 are
real and distinct.
If b2 - 4ac = 0, the roots are real and equal.
If b2 - 4ac < 0, the roots are complex conjugate
numbers.
We consider the first two cases.
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4.2 Homogeneous Linear Equations
 Distinct Real Roots: If the auxiliary equation (3) has
distinct real roots r1 and r2,
y1(t) = er1t and y2(t) = er2t are solutions to (2) and
y(t) = c1er1t + c2er2t is a general solution.
 Repeated Root: If the auxiliary equation (3) has a
repeated root r,
y1(t) = ert and y2(t) = tert are solutions to (2) and
y(t) = c1ert + c2tert is a general solution.
 Initial-value problem: For a linear DE, we define 2nd
order initial-value problem (IVP) as:
Solve: ay’’ + by’ + cy = f(t)
Subject to: y(t0) = y0, y’(t0) = y1.
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4.2 Homogeneous Linear Equations
 Theorem 4.1 (Existence and Uniqueness: Homogeneous
Case): For any real numbers (a ≠ 0), b, c, t0, y0, and y1,
there exists a unique solution to the IVP:
ay’ + by’ + cy = 0;
y(t0) = y0, y’(t0) = y1 -------- (4)
The solution is valid for all t in (- ∞, ∞).
 Linear Independence of Two Functions: A pair of
functions y1(t) and y2(t) is said to be linearly independent
(LI) on the interval I iff. neither of them is a constant
multiple of the other on all of I. We say that y1 and y2 are
linearly dependent on I if one of them is a constant
multiple of the other on all of I.
 Theorem 4.2 (Solutions to IVP): If y1(t) and y2(t) are any
two solutions to the DE (2) that are LI on (-∞, ∞), then
unique constants c1 and c2 can always be found so that
c1y1(t) + c2y2(t) satisfies the IV problem (4) on (-∞, ∞).
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4.2 Homogeneous Linear Equations
 Boundary-value problem: For a 2nd order linear DE, we
define boundary-value problem (BVP) as:
Solve:
a2y’’ + a1y’ + a0y = f(t)
Subject to: y(a) = y0, y(b) = y1.
 Example 4.1: Find a pair of solutions to
y’’ + 5y’ - 6y = 0 ------------- (i)
 Solution: The auxiliary equation associated with (i) is
r2 + 5r - 6 = (r – 1)(r + 6) = 0
=> r1 = 1, r2 = -6 are roots
=> y1(t) = et and y2(t) = e-6t are solutions to (i).
So, y(t) = c1et + c2e-6t is a general solution.
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4.2 Homogeneous Linear Equations
 Example 4.2: Solve the initial value problem
y’’ + 2y’ - y = 0; y(0) = 0, y’(0) = -1 ---------- (i)
 Solution: The auxiliary equation associated with (i) is
r2 + 2r - 1 = 0
=> r1 = -1 + √2, r2 = -1 - √2 are roots using the
quadratic formula
=> y(t) = c1e(-1 + √2)t + c2e(-1 - √2)t is a solution.
Using initial condition, y(0) = c1e0 + c2e0
=> 0 = c1 + c2 => c1 = -c2
and y’(0) = (-1 + √2)c1e0 + (-1 - √2)c2e0
=> -1 = (-1 + √2)c1 + (-1 - √2)c2.
=> c1 = -√2/4 and c2 = √2/4.
So, y(t) = (-√2/4)e(-1 + √2)t + (√2/4)e(-1 - √2)t is the solution.
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4.2 Homogeneous Linear Equations
 Example 4.3: Solve the initial value problem
y’’ + 4y’ + 4y = 0; y(0) = 1, y’(0) = 3 ---------- (i)
 Solution: The auxiliary equation associated with (i) is
r2 + 4r + 4 = (r + 2)2 = 0
=> r = -2 is a double root.
=> y1(t) = e-2t and y2(t) = te-2t are solutions to (i).
and y(t) = c1e-2t + c2te-2t is a general solution.
Using initial condition, y(0) = c1e0 + c2(0)e0
=> 1 = c1 => c1 = 1 -c2
and y’(t) = -2c1e-2t + c2e-2t -2c2te-2t is
=> y’(0) = -2c1e0 + c2e0 -2c2(0)e0
=> 3 = -2(1)(1) + c2(1) => c2 = 5
So, y(t) = e-2t + 5te-2t is the solution.
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4.2 Homogeneous Linear Equations
 Note: Let’s confirm that and y2(t) = te-2t is a solution.
y'2(t) = e-2t - 2te-2t
y”2(t) = - 2e-2t - 2e-2t + 4te-2t = - 4e-2t + 4te-2t
y”2 + 4y’2 + 4y2 = -4e-2t + 4te-2t + 4(e-2t-2te-2t) +4te-2t = 0
 Example 4.4: Find a general solution to the problem
y’’’ + 3y’’ –y’ - 3y = 0 ---------- (i)
 Solution: The auxiliary equation associated with (i) is
r3 + 3r2 - r - 3 = (r - 1) (r + 1) (r + 3) = 0
=> r1 = 1, r2 = -1 and r3 = -3 are the roots.
=> y1(t) = et, y2(t) = e-t and y3(t) = e-3t are solutions.
So, y(t) = c1et + c2e-t + c3e-3t is a general solution.
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4.2 Homogeneous Linear Equations
 Exercise 4.1: Find a general solution to each given DE.
 Exercise 4.2: Solve the given initial value problems.
 Exercise 4.3: Determine whether the functions y1 and
y2 are linearly dependent on the interval (0, 1).
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To
Be
Continued……
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End
Of
Chapter IV
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