1550078481-Ordinary_Differential_Equations__Roberts_

156 Ordinary Differential Equations

From calculus we know the arc length s of the path of the dog satisfies the
differential equation

ds dx

• =-vi+ (dy/dx)^2 -d = sd.
  dt t

The minus sign is due to the fact that s increases as x decreases. Solving for
dt/dx, we h ave

(4)

dt

dx

Jl + (y')2

Differentiating (3) with respect to x, we get xy" + y' = y' - srdt/ dx. Solving
this equation for dt/dx, we obtain

dt -xy"

(5)

dx Sr

Equating (4) and (5) results in

(6)

xy"

Sr

Jl + (y')2

Let p = y'. Then p' = y". When t = 0, (x, y) = (b, 0) and p = -a/b.

Substituting for p and p' in equation (6) and solving for p', we see that p
satisfies the initial value problem

1 srJl+p^2

p = ;

XSd

-a

p(b) = t;·

Separating variables and letting R = Sr/ sd, we find

dp Rdx

Jl + p2 x

Integrating, we get

ln IP+ Jl + p^2 I = Rln lxl + C

where C is the constant of integration. Exponentiating, we obtain

(7) p+Jl+p^2 =KxR.

Subtracting p and squaring, we find

1 + p2 = K2x2R - 2KpxR + p2.

Eliminating p^2 and solving for p, yields

(8)

' K2x2R -1

p = y = 2KxR