Page 168 - Numerical Analysis Using MATLAB and Excel

Solutions of Ordinary Differential Equations (ODE)

Proof:

Let us assume that y t() is a solution of (5.13); then by substitution,
1

n
d y d n – 1 y dy
1
1
1
a ----------- + a n – 1 ------------------ + … + a -------- + a y = 0 (5.15)
0
n
1
1
dt n dt n – 1 dt
A solution of the form k y t() will also satisfy (5.13) since
1 1
n d n – 1 d
d
a ------- k y( n 1 1 ) + a n – 1 ------------- k y( 1 1 ) + … a ---- k y ) ( + 1 1 a k y ) ( + 1 1
-
1
0
dt n dt n – 1 dt
(5.16)
n
⎛ d y d n – 1 y dy ⎞
1
1
1
⎜
= k a ----------- + a n – 1 ------------------ + … + a -------- + a y ⎟ 0 1 = 0
n
1
1
⎝ dt n dt n – 1 dt ⎠
If y = y t() and y = y t() are any two solutions, then y = y t() + y t() will also be a solution
2
1
1
2
since
n n – 1
d y d y dy
1
1
1
a ----------- + a n – 1 ------------------ + … + a -------- + a y = 0
1
1
0
n
dt n dt n – 1 dt
and
n
d y d n – 1 y dy
2
2
2
a ----------- + a n – 1 ------------------ + … + a ---------- + a y = 0
1
n
2
0
dt n dt n – 1 dt
Therefore,
n n – 1
d d d
(
a ------- y +( n 1 y ) 2 + a n – 1 ------------- y +( 1 y ) 2 + … a ----- y + y ) 2 + a y + y ) 2 (5.17)
( +
1
0
1
1
dt n dt n – 1 dt
d n d n – 1 d
= a ------- y + a n – 1 ------------- y + … + a ----- y + a y 1
n
0
1
1
1
1
dt n dt n – 1 dt
d n d n – 1 d
-
+ a ------- y + a n – 1 ------------- y + … + a ---- y + a y = 0
2
1
2
2
2
0
n
dt n dt n – 1 dt
In general, if
y= k y t() k y t() k y t() … k y t(),,, 2 1 3 3 , n n
1 1
are the solutions of the homogeneous ODE of (5.13), the linear combination
n
y= k y t() + k y t() + k y t() + … + k y t()
n n
3 3
1 1
2 1
is also a solution.
In our subsequent discussion, the solution of the homogeneous ODE, i.e., the complementary
Numerical Analysis Using MATLAB® and Excel®, Third Edition 5−7
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