The Computable Differential Equation Lecture ... - Bruce E. Shapiro

CHAPTER 1. CLASSIFYING THE PROBLEM 11
for any values of the numbers a and b. The first boundary condition gives
Hence
From the second boundary condition,
1 = a cos 0 + b sin 0 = a (1.50)
y = cos t + b sin t (1.51)
1 = cos 2π + b sin 2π = 1 (1.52)
In other words, any value of b will satisfy the boundary value problem. There are
an infinite number of solutions.
Higher Order Equations
Any higher order equation y (n) = f(t, y, y ′ , y ′′ , ..., y n−1 ) can always be reduced to a
first order system by making the following change of variables:
We then have the n th order system
ξ 1 = y (1.53)
ξ 2 = y ′ (1.54)
. (1.55)
ξ n = y (n−1) (1.56)
ξ ′ 1ξ 2 (1.57)
ξ 2 ‘ = ξ 3 (1.58)
. (1.59)
ξ n = f(t, ξ 1 , ξ 2 , . . . , ξ n−1 ) (1.60)
Example 1.6. Rewrite the equation x ′′ + ω 2 x = 0 as a first order system.
Solution. We define u = x, v = x ′ . Then u ′ = x ′′ = v and v ′ = x ′′ = −ω 2 x = −ω 2 u.
We can write this as ⇐=
( ) ( ( )
u
′ 0 1 u
v ′ =
−ω 0) 2 ⇐= (1.61)
v
( u
or y ′ = Ay, where y = and A =
v)
( ) 0 1
−ω 2 .
0
Example 1.7. Repeat the previous example with boundary conditions
y(0) = 1 (1.62)
y(2) = 1 (1.63)
c○2007, B.E.Shapiro
Last revised: May 23, 2007
Math 582B, Spring 2007
California State University Northridge