In applications, most initial value problems will have a unique solution. In fact, the existence of unique solutions is so important that there is a theorem about the existence and uniqueness of a solution. Consider the initial value problem =f(x,y). y(x) =yo. Iff and dx af dy are continuous functions in some rectangle R={(x,y):a
In applications, most initial value problems will have a unique solution. In fact, the existence of unique solutions is so important that there is a theorem about the existence and uniqueness of a solution. Consider the initial value problem =f(x,y). y(x) =yo. Iff and dx af dy are continuous functions in some rectangle R={(x,y):a
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 17EQ
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Transcribed Image Text:In applications, most initial value problems will have a unique solution. In fact, the existence of unique solutions is so important that there is a theorem about the existence and uniqueness of a solution. Consider the initial value problem
af
are continuous functions in some rectangle R = {(x,y):a<x<b, c<y<d} that contains the point (Xo Yo), then the initial value problem has a unique solution (x) in some interval xo -8<x<x+8, where is a positive number. The method for separable equations
ду
dy 3
can give a solution, but it may not give all the solutions. To illustrate this, consider the equation =y
dx
1
dy
3
(a) Use the method of separation of variables to find the solution to =y
Begin by separating the variables.
dx
dy=dx
Solve the differential equation, ignoring lost solutions, if any.
dy
nie by
(b) Show that the initial value problemy with y(0)=0 is satisfied for C =0 by y=
For y = 0, y(0)=
Now, determine
of
ду
Using the general solution, substituting for x and for y(x) implies that C=. So, the solution
x>0, satisfies the initial value problem =y with y(0) = 0.
dy
dx
(c) Now show that the constant function y = 0 also satisfies the initial value problem given in part (b). Hence, this initial value problem does not have a unique solution.
=
dy
dx
1
3
dy
dx
(d) Finally, show that the conditions for the theorem are not satisfied for the initial value problem in part (b). Begin by identifying f(x,y) in the initial value problem in (b).
f(x,y)=
af
ду
=
Answer parts (a) through (d).
for x ≥ 0.
, and y
C
1
1
3
dy
Thus, it holds that the constant function y = 0 satisfies the initial value problem -= y³ with y(0) = 0.
dx
=f(x,y). y (xo) yo. Iff and
dx
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