(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). 2 f(x,y) = -y
(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). 2 f(x,y) = -y
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter4: Calculating The Derivative
Section4.4: Derivatives Of Exponential Functions
Problem 23E: Find derivatives of the functions defined as follows. fz=2z+e-z22
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How to solve part (d)?
Transcribed Image Text:of
If f and
dy
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 o(x) in some interval x, - 6<x< X, + 8, where ô is a positive number. The method for separable equations can give
dy
= f(x,y), y (Xo) = Yo-
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
dx
1
dy
= y
dx
3
a solution, but it may not give all the solutions. To illustrate this, consider the equation
Answer parts (a) through (d).
- .-
dy
3
(a) Use the method of separation of variables to find the solution to
=y
dx
Begin by separating the variables.
1
3
dy = 1 dx
У
Solve the differential equation, ignoring lost solutions, if any.
2
3 3
1
dy
=y° with y(0) = 0 is satisfied for C = 0 by y =
2x
(b) Show that the initial value problem
for x20.
dx
3
3
1
2x
dy
3
Using the general solution, substituting 0 for x and 0 for y(x) implies that C = 0. So, the solution y=
3
, x20, satisfies the initial value problem
= y
dx
with y(0) = 0.
Transcribed Image Text:(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).
2
2x
f(x,y) =
- y
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