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Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Differential Equations HW problem. Solution and work needed, please. I have attached a picture of the section 4.1 theorem from my textbook as well. However, you can refer to the textbook if needed. It is called "A first course in differential equations the classic 5th edition" by Dennis Zill.
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.
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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