1550078481-Ordinary_Differential_Equations__Roberts_

14 Ordinary Differential Equations

For x < 0, xy' - 2y = x(-2x) - 2(-x^2 ) = -2x^2 + 2x^2 = 0, so y(x)
satisfies the given differential equation on the interval (-oo, 0). For x > 0,

xy' - 2y = x(2x) - 2(x^2 ) = 2x^2 - 2x^2 = 0, so y(x) satisfies the differential

equation on the interval (0, oo). At x = 0, xy' -2y = 0(0)-2(0) = 0, so y(x)
satisfies the differential equation at x = 0. Therefore, the piecewise defined,
differentiable function y( x ) satisfies the differential equation xy' - 2y = 0 on
the interva l (-oo, oo ).

EXERCISES 1.2

In exercises 1-14 determine the order of the given ordinary differ-

ential equation and state whether the equation is linear or nonlinear.

1. y^1 + x^2 y = 3 COSX 2. y' + a(x)y = b(x)


2. y' - 2exy = y2 4. y' + a(x)yn = b(x) (where n =/. 0 and n =/. 1)

5. (y')^3 - xy^2 = sinx 6. 3x^2 dx - 4ydy = 0 (Hint: Divide by dx.)

7. 
   

   ydx -xdy = 0 8. y(l + (y')^2 ) = 5

   
   


8. 
   

   y" - 3y'y = 4 10. y" + x^2 y' - x^3 y = tan x

   
   


9. 
   

   y" + ysinx = 3 12. y" + x sin y = 3

   
   


10. 
    

    (y(3))2 _ 4 (yC2))4 + x2y = 0

    
    


11. 
    

    yC^4 l + x^2 yC^3 l - (sinx)yC^2 l = y

    
    


12. 
    

    Is y(x) = l/x a solution of the differentia l equation x^2 y" + xy' - y = O

    
    

a. on the interval ( -1, 1)? Why?

b. on the interval (0, oo)? Why?

c. on the interval (- oo, 0)? Why?

16. Is y = lx l a solution of the differential equation xy' - y = 0

a. on the interval (-1, 1)? Why?

b. on the interval (O,oo)? Why?

c. on the interval (-oo, O)? Why?

17. Is y = Vx a solution of the differential equation 2x^2 y" + 3xy^1 - y = 0
    a. on the interval ( -1, 1)? Why?
    b. on the interval (O,oo)? Why?
    c. on the interval (- oo, 0)? Why?