1550078481-Ordinary_Differential_Equations__Roberts_

544 Ordinary Differential Equations

Exercises 2.4.4 Pitfalls of Numerical Methods

1. The given differential equation is linear with a( x) = 0 and b( x) =

l/(x -1). These functions a re both continuous on t he intervals (-oo, 1)

and (1, oo). Since 0 E (- oo, 1) and the differential equation is linear,

the solution exists and is unique on the interval (- oo, 1 ). [Note: The

unique, explicit solution on (-oo, 1) is y = ln Ix - ll + 1. ]

3. The differential equation is linear wit h a(x) = l/x and b(x) = 0.

Since a(x) and b(x) are both continuous on (- oo,O) and (O,oo) and

since -1 E (- oo, 0), there exists a unique solution to both initial value

problems on the interval (-oo, 0).

a. The unique, explicit solution on (- oo, 0) is y = -x.

b. The unique, explicit solution on (- oo, 0) is y = x.

5. The given differential equation is nonlinear. The functions f(x, y) = y^2

and fy(x, y) = 2y are continuous on the entire plane, so there exists a

unique solution until x ---+ -oo, x ---+ +oo, y ---+ - oo, or y ---+ +oo.

a. The unique, explicit solution on (- oo,O) is y = -1/x.

b. The unique, explicit solut ion on ( - oo, oo) is y = 0.

c. The unique, explicit solution on (- oo, 3) is y = -1/(x - 3).

7. The differential equation is nonlinear and f(x, y) = - 3x^2 /(2y) and
   fy(x,y) = 3x^2 /(2y^2 ) are both continuous for y-=f-0.

a. and b. Thus, there exists a unique solution until x ---+ -oo,

x ---7 +oo, y ---7 +oo, or y ---7 o+.

d. Hence, there exists a unique solution until x ---+ - oo, x ---+ +oo,

y ---+ o-, or y ---+ - oo.

a. The unique, explicit solution on ( - oo, 0) is y = V-XJ.

b. The unique, explicit solution on (-oo, ij=3,/4) is y = J-x3 -3/4.

c. There is no solut ion, because the differential equation is undefined

for y = 0.

d. The unique, explicit solution on ( - oo, 0) is y = -V-XJ.