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 problemafare continuous functions in some rectangle R = {(x,y):a0, satisfies the initial value problem =y with y(0) = 0.dydx(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.=dydx13dydx(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)=afду=Answer parts (a) through (d).for x ≥ 0., and yC113dyThus, 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 anddx

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Answered: In applications, most initial value…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

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

Expert Solution

Step 1: Part a)

$Use space the space method space of space seperation space of space variables space to space find space the space solution space of space dy over dx equals straight y to the power of 1 third end exponent straight a right parenthesis We space can space write dy equals straight y to the power of 1 third end exponent dx box enclose bold y to the power of bold minus bold 1 over bold 3 end exponent end enclose bold dy bold equals box enclose bold 1 bold space bold dx Using space the space integration space rules fraction numerator straight y to the power of negative 1 third plus 1 end exponent over denominator negative 1 third plus 1 end fraction equals straight x plus straight C fraction numerator straight y to the power of 2 over 3 end exponent over denominator 2 over 3 end fraction minus straight x equals straight C box enclose fraction numerator bold 3 bold y to the power of bold 2 over bold 3 end exponent over denominator bold 2 end fraction bold minus bold x bold equals bold C end enclose$