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:
an y(n) + an−1 y(n−1) + … + a2 y″ + a1 y′ + a0 y = 0,
an ≠ 0.
It has a general solution of the form
y = C1 y1 + C2 y2 + … + C n−1 yn−1 + Cn yn
where y1, y2, … , yn−1, yn 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(y1, y2, … , yn−1, yn)(t) ≠ 0.
Note 1: In order to determine the n unknown coefficients Ci, each n-th order
equation requires a set of n initial conditions in an initial value problem:
y(t0) = y0, y′(t0) = y′0, y″(t0) = y″0, and y(n−1)(t0) = y(n−1)0.
Note 2: The Wronskian W(y1, y2, … , yn−1, yn)(t) is defined to be the
determinant of the following n × n matrix
 y1
 y'
 1
 y"1

 :
 y1( n −1)
© 2008, 2014 Zachary S Tseng
y2
y'2
y"2
:
y 2( n −1)
.. ..
.. ..
.. ..
.. ..
yn 
y ' n 
y"n 
.
: 
y n( n −1) 
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:
an r n + an−1 r n−1 + … + a2 r 2 + a1 r + a0 = 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 ± µi are distinct complex conjugate roots, then
y = e λ t cos µt and y = e λ t sin µt are solutions.
rt
rt
(iii). If r is a real root appearing k times, then y = e , y = te ,
y = t2e r t, … , and y = t k−1 e r t are all solutions.
(iv). If r = λ ± µi are complex conjugate roots each appears k times,
then
y = e λ t cos µt, y = e λ t sin µt,
y = t e λ t cos µt, y = t e λ t sin µt,
y = t2 e λ t cos µt, y = t2 e λ t sin µt,
:
:
:
:
k−1 λ t
y = t e cos µt, and y = t k−1 e λ t sin µt,
are all solutions.
© 2008, 2014 Zachary S Tseng
B-4 - 2
Example:
y(4) − y = 0
The characteristic equation is r 4 − 1 = (r 2 + 1)(r + 1)(r − 1) = 0,
which has roots r = 1, −1, i, −i. Hence, the general solution is
y = C1 e t + C2 e −t + C3 cos t + C4 sin t.
Example:
y(5) − 3 y(4) + 3 y(3) − y″ = 0
The characteristic equation is r 5 − 3r 4 + 3r 3 − r 2 = r 2(r − 1)3 = 0,
which has roots r = 0 (a double root), and 1 (a triple root). Hence, the
general solution is
y = C 1e 0 t + C 2 t e 0 t + C 3 e t + C 4 t e t + C 5 t 2 e t
= C1 + C2 t + C3 e t + C4 t e t + C5 t 2 e t.
Example:
y(4) + 4 y(3) + 8 y″ + 8 y′ + 4 y = 0
The characteristic equation is r 4 + 4r 3 + 8r 2 + 8r + 4 = (r 2 + 2r + 2)2
= 0, which has roots r = −1 ± i (repeated). Hence, the general solution
is
y = C1 e − t cos t + C2 e − t sin t + C3 t e − t cos t + C4 t e − t sin t .
© 2008, 2014 Zachary S Tseng
B-4 - 3
Example: What is a 4th order homogeneous linear equation whose general
solution is
y = C1 e t + C2 e 2t + C3 e 3t + C4 e 4t ?
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
r 4 − 10r 3 + 35r 2 − 50r + 24 = 0
Hence, an equation is
y(4) − 10 y(3) + 35 y″ − 50 y′ + 24 y = 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, 2014 Zachary S Tseng
B-4 - 4
Exercises B-4.1:
1 – 8 Find the general solution of each equation.
1.
y(3) + 25y′ = 0
2.
y(3) + 27y = 0
3.
y(4) − 18 y″ + 81y = 0
4.
y(4) − 3 y″ − 4 y = 0
5.
y(4) + 32 y″ + 256 y = 0
6.
y(5) + 5y(4) + 10y(3) + 10y″ + 5y′ + y = 0
7.
y(5) + 2y(4) + 5y(3) = 0
8.
y(6) − y = 0
9 – 12 Solve each initial value problem.
9.
y(3) + 4 y″ − 5 y′ = 0,
y(0) = 4,
y′(0) = −7, y″(0) = 23
10.
y(3) + 3 y″ + 3 y′ + y = 0,
11.
y(4) − 10y″ + 9y = 0, y(0) = 5, y′(0) = −1, y″(0) = 21, y(3)(0) = −49
12.
y(4) + 13y″ + 36y = 0, y(0) = 0, y′(0) = −3, y″(0) = 5, y(3)(0) = −3
y(0) = 7,
y′(0) = −7, y″(0) = 11
13. Same as #10, except with initial conditions y(6561) = 7,
y′(6561) = −7,
y″(6561) = 11.
14. Same as #12, except with initial conditions y(247) = 0, y′(247) = −3,
y″(247) = 5, y(3)(247) = −3
15. Find a 3rd order homogeneous linear equation which has a particular
solution
y = e t − 2e −2t + 5t e −2t.
16. Find a 5th order homogeneous linear equation whose general solution is
y = C1 e −t + C2 t e −t + C3 t2 e −t + C4 cos t + C5 sin t.
© 2008, 2014 Zachary S Tseng
B-4 - 5
17. Find a 6th order homogeneous linear equation whose general solution is
y = C1 + C2 t + C3 e −2 t cos t + C4 e −2 t sin t + C5 t e −2 t cos t + C6 t e −2 t sin t.
[Hint: the polynomial, with leading coefficient 1, that has complex conjugate
roots λ ± µi has the form r 2 − 2λr + (λ2 + µ2).]
18. Find a 6th order homogeneous linear equation whose general solution is
y = C1 cos 2t + C2 sin 2t + C3 t cos 2t + C4 t sin 2t + C5 t2 cos 2t + C6 t2 sin 2t.
© 2008, 2014 Zachary S Tseng
B-4 - 6
Answers B-4.1:
1. y = C1 + C2 cos 5t + C3 sin 5t
2. y = C1e
3.
4.
5.
6.
7.
− 3t
3
t
2
3
t
3 3
3 3
+ C 2 e cos
t + C 3 e 2 sin
t
2
2
y = C1 e 3t + C2 e −3t + C3 t e 3t + C4 t e −3t
y = C1 e 2t + C2 e −2t + C3 cos t + C4 sin t
y = C1 cos 4t + C2 sin 4t + C3 t cos 4t + C4 t sin 4t
y = C1 e −t + C2 t e −t + C3 t 2 e −t + C4 t 3 e −t + C5 t 4 e −t
y = C1 + C2 t + C3 t2 + C4 e −t cos 2t + C5 e −t sin 2t.
1
t
2
1
−1
−1
t
t
t
3
3
3
3
+ C 6 e 2 sin
t
8. y = C1e + C 2 e + C 3 e cos t + C 4 e 2 sin t + C5 e 2 cos
2
2
2
2
t
−t
9. y = 5 − 2e t + e −5t
10. y = 7e −t + 2t2 e −t
11. y = 4e t − e −t + 2e −3t
12. y = cos 2t − 3sin 2t − cos 3t + sin 3t
13. y = 7e – t + 6561 + 2(t – 6561)2 e – t + 6561
14. y = cos 2(t – 247) − 3sin 2(t – 247) − cos 3(t – 247) + sin 3(t – 247)
15. y(3) + 3y″ − 4y = 0
16. y(5) + 3y(4) + 4y(3) + 4y″ + 3y′ + y = 0
17. y(6) + 8y(5) + 26y(4) + 40y(3) + 25y″ = 0
18. y(6) + 12y(4) + 48y″ + 64y = 0
© 2008, 2014 Zachary S Tseng
B-4 - 7
Lastly, here is an example of an application of a very simple 4th order
nonhomgeneous linear equation that might be familiar to many engineering
students.
The static deflection of a uniform beam
Consider a horizontal beam of length L, of uniform cross section and made
of homogeneous material, acted upon by a vertical force. The beam is
positioned along the x-axis, with its left end at the origin. The deflection of
the beam (its vertical displacement relative to the horizontal axis) at any
point x is given by, according to the Euler–Bernoulli beam equation*,
EI u(4) = W(x),
0 < x < L.
The positive constants E and I are, respectively, the Young’s modulus of
elasticity of the material of the beam, and the beam’s cross-sectional
moment of inertia about the horizontal axis (i.e., the second moment of the
cross-sectional area). The value of the product EI is a measurement of the
beam’s stiffness. The forcing function W describes the load/force that the
beam bears per unit length. If the beam is not bearing external load, then
W(x) = w, which is the weight-density of the beam itself (acting downwards,
which is the positive direction per our usual convention).
The equation then becomes
EI u(4) = w.
It is a (very) simple 4th order nonhomgeneous linear equation. It could be
solved simply by integrating both sides four times with respect to x.
However it is certainly more illustrative for our purpose to solve it using the
general procedure that we have learned, namely by characteristic equation
and undetermined coefficients methods to obtain both parts of the solution of
this nonhomogeneous linear equation, in the form of u = uc + U.
*
2
2
4
The equation is originally d ( EI d u ) = W ( x) . When EI is constant, it simplifies to EI d u = W ( x) .
dx2
dx 2
dx 4
© 2008, 2014 Zachary S Tseng
B-4 - 8
4
Its characteristic equation is EI r = 0, which has zero as a (quadruple) root.
2
3
The complementary solution is, therefore, uc = C1 + C2 x + C3 x + C4 x .
The particular solution can be found using the method of Undetermined
Coefficients that we have already learned. What form would the particular
solution of the equation take?
The general solution of this deflection equation is
u ( x) = C1 + C 2 x + C 3 x 2 + C 4 x 3 +
w
x4 .
24 E I
The graph of the deflection function is called the deflection curve or elastic
curve of the beam.
Another interesting aspect of this problem is that this equation does not
come with initial conditions. Instead, it comes paired with four boundary
conditions – describing the physical conditions at the two ends, at x = 0 and
x = L, of the beam. For example, if the beam is securely embedded into
walls on both ends, the deflection function above must satisfy the boundary
conditions:
u(0) = u′(0) = u(L) = u′(L) = 0.
In this case the four conditions tell us that the displacement, u(x), is zero at
both ends, and that the slope of the beam, u′(x), is also zero at the two ends –
hence the beam is securely fixed at both ends (by embedding into the walls).
Notice that there are four conditions, which are necessary to solve the four
unknown coefficients present in the general solution of this fourth order
equation.
Being a fourth order equation, the boundary conditions in a beam problem
frequently involve u″ and uʺʹ. Physically, u″ represents the bending moment
and uʺʹ represents the shear force experienced by the beam at a given point.
© 2008, 2014 Zachary S Tseng
B-4 - 9
For a simply supported beam, as another example, where a beam is either
pinned or hinged at both ends (having no displacement nor resistance to
rotation at the two ends), the required boundary conditions are, therefore,
u(0) = u″(0) = u(L) = u″(L) = 0.
We will study simple boundary value problems, in a quite different context,
later in the course.
© 2008, 2014 Zachary S Tseng
B-4 - 10
Exercises B-4.2:
1. Deflection curves of uniformly loaded beams Based on the earlier
calculation, a beam bearing a uniformly distributed load of density w has a
deflection curve in the form
w
u ( x) = C1 + C 2 x + C3 x 2 + C 4 x 3 +
x4 .
24 E I
Find the deflection curve of each set of boundary conditions below.
(i) Beam with both ends embedded: u(0) = u′(0) = u(L) = u′(L) = 0.
(ii) Simply supported beam: u(0) = u″(0) = u(L) = u″(L) = 0.
(iii) Cantilever beam (left end embedded, right end free): u(0) = u′(0)
= u″(L) = u′″(L) = 0.
2. Consider a beam of length L bearing a load that is proportional to the
distance from the left endpoint, i.e. W(x) = wx, for some positive constant w.
(i) Find the general solution of its deflection curve.
(ii) Find the deflection curve given u(0) = u′(0) = u(L) = u′(L) = 0.
(iii) Find the deflection curve given u(0) = u″(0) = u(L) = u″(L) = 0.
(iv) Find the deflection curve given u(0) = u′(0) = u″(L) = u′″(L) = 0.
3. Consider a cantilever beam of length L bearing a load that is proportional
to the distance from the right endpoint, i.e. W(x) = w(L – x), for some
positive constant w. Find its deflection curve given that u(0) = u′(0)
= u″(L) = u′″(L) = 0.
© 2008, 2014 Zachary S Tseng
B-4 - 11
Answers B-4.2:
(
)
w
L2 x 2 − 2 Lx 3 + x 4
24 E I
w
L3 x − 2 Lx 3 + x 4
(ii) u ( x) =
24 E I
w
6 L2 x 2 − 4 Lx 3 + x 4
(iii) u ( x) =
24 E I
1. (i) u ( x) =
(
)
(
)
2
3
2. (i) u ( x) = C1 + C2 x + C3 x + C4 x +
(
w
x5
120 EI
)
w
2 L3 x 2 − 3L2 x 3 + x 5
120 E I
w
7 L4 x − 10 L2 x 3 + 3x 5
(iii) u ( x) =
360 E I
w
20 L3 x 2 − 10 L2 x 3 + x 5
(iv) u ( x) =
120 E I
3. The general solution is
wL 4
w
u ( x) = C1 + C2 x + C3 x 2 + C4 x 3 +
x −
x5 .
24 E I
120 EI
The cantilever beam’s deflection curve is
w
u ( x) =
10 L3 x 2 − 10 L2 x 3 + 5Lx 4 − x 5 .
120 E I
(ii) u ( x) =
(
(
(
© 2008, 2014 Zachary S Tseng
)
)
)
B-4 - 12