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# Notes-HigherOrderLinEq.pdf

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Pennsylvania State University

Mathematics

MATH 251

Unknown

Spring

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Higher Order Linear Equations with Constant Coefficients
The solutions of linear differential equations with constant coefficients of
the third order or higher can be found in similar ways as the solutions of
second order linear equations. For an n▯th order homogeneous linear
equation with constant coefficients:
(n) (n−1)
any + a n−1 y + … + a y″ 2 a y′ + 1 y = 0,0 a n 0.
It has a general solution of the form
y = C y1 1C y + 2 2 C n−1 n−1 + C n n
where y 1 y 2 … , y n−1, n are any n linearly independent solutions of the
equation. (Thus, they form a set of fundamental solutions of the differential
equation.) The linear independence of those solutions can be determined by
their Wronskian, i.e., W(y ,1y 2 … , y n−1 yn)(t) ≠ 0.
Note 1: In order to determine the n unknown coefficients C i each n▯th order
equation requires a set of n initial condi(n−1) in an(n−1)ial value problem:
y(t0) = y0, y′0t ) = 0′ , y0(t ) =0y″ , and y (0 ) = y 0.
Note 2: The Wronskian W(y , y ,1… ,2y n−1 yn)(t) is defined to be the
determinant of the following n × n matrix
y1 y2 .. .. yn
y'1 y' 2 .. .. y'n
y" 1 y" 2 .. .. y" n
.
: : :
(n 1) (n−1) (n−1)
y1 y 2 .. .. yn
© 2008 Zachary S Tseng B▯4 ▯ 1 Such a set of linearly independent solutions, and therefore, a general solution
of the equation, can be found by first solving the differential equation’s
characteristic equation:
a r + a r n−1 + … + a r + a r + a = 0.
n n−1 2 1 0
This is a polynomial equation of degree n, therefore, it has n real and/or
complex roots (not necessarily distinct). Those necessary n linearly
independent solutions can then be found using the four rules below.
rt
(i). If r is a distinct real root, then y = e is a solution.
(ii). If r = λ ± ▯i are distinct complex conjugate roots, then
y = e cos ▯t and y = e sin ▯t are solutions.
(iii). If r is a real root appearing k times, then y = e rt, y = te rt,
2 rt k−1 rt
y = t e , … , and y = t e are all solutions.
(iv). If r = λ ± ▯i are complex conjugate roots each appears k times,
then
y = e cos ▯t , y = e sin ▯t ,
λ t λ t
y = te2coλ tt , y = te 2inλ t ,
y = t e cos ▯t , y = t e sin ▯t ,
: :
k−1 : λ t : k−1 λ t
y = t e cos ▯t , and y = t e sin ▯t ,
are all solutions.
© 2008 Zachary S Tseng B▯4 ▯ 2 (4)
Example: y − y = 0
4 2
The characteristic equation is r − 1 = (r + 1)(r + 1)(r − 1) = 0,
which has roots r = 1, −1, i, −i. Hence, the general solution is
y = C e + C e −t + C cost + C sint.
1 2 3 4
(5) (4) (3)
Example: y − 3y + 3y − y″ = 0
The characteristic equation is r − 3r + 3r − r = r (r − 1) = 0, 3
which has roots r = 0 (a double root), and 1 (a triple root). Hence, the
general solution is
y = C e + C te + C e + C te + C t e t 2 t
1 2 3 4 5
t t 2 t
= C +1C t +2C e + C3te + C t4e . 5
Example: y(4) + 4y (3)+ 8y″ + 8y′ + 4y = 0
4 3 2 2 2
The characteristic equation is r + 4r + 8r + 8r + 4 = (r + 2r + 2)
= 0, which has roots r = −1 ± i (repeated). Hence, the general solution
is
− t − t − t − t
y = C e1 cost + C e 2 sin t + C t3 cost + C te4 sin t.
© 2008 Zachary S Tseng B▯4 ▯ 3 Example: What is a 4th order homogeneous linear equation whose general
solution is
t 2t 3t 4t
y = C 1 + C e + 2 e + C e 3 4
The solution implies that r = 1, 2, 3, and 4 are the four roots of the
characteristic equation. Therefore, r − 1, r − 2, r − 3, and r − 4 are its
factors. Consequently, the characteristic equation is
(r − 1)(r − 2)(r − 3)(r − 4) = 0
4 3 2
r − 10r + 35r − 50r + 24 = 0
Hence, an equation is
(4) (3)
y − 10y + 35y″ − 50y′ + 24y = 0.
Note: The above answer is not unique. Every nonzero constant multiple of
the above equation also has the same general solution. However, the
indicated equation is the only equation in the standard form that has the
given general solution.
© 2008 Zachary S Tseng

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