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(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

Advanced Engineering Mathematics
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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How to solve part (d)?

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: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.
(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
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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