Solution and
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Consider the second-order, linear, homogeneous differential equation with
initial conditions,
x²y" +xy - 4y = 0
y(0) = 0
y'(0) = 0
This initial value problem has many solutions, for instance y1 = x2 and y2 = 0. Does
this violate the existence and uniqueness theorem for linear differential equations, from
section 4.1? Explain.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0bda9c1d-08ea-450e-a130-838d16131279%2F31dd115e-e3a7-4817-98e6-93058f95e9d3%2Fujx6dcg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider the second-order, linear, homogeneous differential equation with
initial conditions,
x²y" +xy - 4y = 0
y(0) = 0
y'(0) = 0
This initial value problem has many solutions, for instance y1 = x2 and y2 = 0. Does
this violate the existence and uniqueness theorem for linear differential equations, from
section 4.1? Explain.
![THEOREM4.1
Existence of a Unique Solution
Let a,(x), a-1(x),.., a (x), a(x), and g(x) be continuous on an interval /
and let a,(x) + 0 for every x in this interval. If x = xo is any point in this in-
terval, then a solution y(x) of the initial-value problem (1) exists on the in-
terval and is unique.
EXAMPLE 1 Solution of an IVP
You should verify that the function y = 3e + e 2
initial-value problem
3x is a solution of the
SECTION 4.1 Preliminary Theory
113
y" - 4y = 12x, y(0) = 4, y'(0) = 1.
Now the differential equation is linear, the coefficients as well as g(x) = 12x are
continuous, and a(x) = 1+ 0 on any interval containing x = 0. We conclude
from Theorem 4.1 that the given function is the unique solution.
EXAMPLE 2
Trivial Solution of an IVP
The initial-value problem
3y" + 5y" - y' + 7y = 0, y(1) -0, y'(1) = 0, y"(1) = 0
possesses the trivial solution y = 0. Since the third-order equation is linear with
constant coefficients, it follows that all the conditions of Theorem 4.1 are ful-
filled. Hence y = 0 is the only solution on any interval containing x = 1.
EXAMPLE 3 Solution of an IVP
The function y = sin 4x is a solution of the initial-value problem
y" + 16y = 0, y(0) = 0, y'(0) = 1.
It follows from Theorem 4.1 that on any interval containing x = 0 the solution
is unique.
The requirements in Theorem 4.1 that a,(x), i = 0, 1, 2, ..., n be continu-
ous and a,(x) + O for every x in / are both important. Specifically, if a,(x) = 0
for some x in the interval, then the solution of a linear initial-value problem may
not be unique or even exist.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0bda9c1d-08ea-450e-a130-838d16131279%2F31dd115e-e3a7-4817-98e6-93058f95e9d3%2Fs8fixmc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:THEOREM4.1
Existence of a Unique Solution
Let a,(x), a-1(x),.., a (x), a(x), and g(x) be continuous on an interval /
and let a,(x) + 0 for every x in this interval. If x = xo is any point in this in-
terval, then a solution y(x) of the initial-value problem (1) exists on the in-
terval and is unique.
EXAMPLE 1 Solution of an IVP
You should verify that the function y = 3e + e 2
initial-value problem
3x is a solution of the
SECTION 4.1 Preliminary Theory
113
y" - 4y = 12x, y(0) = 4, y'(0) = 1.
Now the differential equation is linear, the coefficients as well as g(x) = 12x are
continuous, and a(x) = 1+ 0 on any interval containing x = 0. We conclude
from Theorem 4.1 that the given function is the unique solution.
EXAMPLE 2
Trivial Solution of an IVP
The initial-value problem
3y" + 5y" - y' + 7y = 0, y(1) -0, y'(1) = 0, y"(1) = 0
possesses the trivial solution y = 0. Since the third-order equation is linear with
constant coefficients, it follows that all the conditions of Theorem 4.1 are ful-
filled. Hence y = 0 is the only solution on any interval containing x = 1.
EXAMPLE 3 Solution of an IVP
The function y = sin 4x is a solution of the initial-value problem
y" + 16y = 0, y(0) = 0, y'(0) = 1.
It follows from Theorem 4.1 that on any interval containing x = 0 the solution
is unique.
The requirements in Theorem 4.1 that a,(x), i = 0, 1, 2, ..., n be continu-
ous and a,(x) + O for every x in / are both important. Specifically, if a,(x) = 0
for some x in the interval, then the solution of a linear initial-value problem may
not be unique or even exist.
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