General solutions of linear equations of higher order
Higher order linear equations
â The general n-th order linear differential equation is one of the
form
Math 216
Differential Equations
P0 (x)y (n) + P1 (x)y (n−1) + . . . + Pn−1 (x)y 0 + Pn (x)y = F (x).
â The associated homogeneous equation is the equation with the
right-side set to 0:
Kenneth Harris
kaharri@umich.edu
P0 (x)y (n) + P1 (x)y (n−1) + . . . + Pn−1 (x)y 0 + Pn (x)y = 0.
Department of Mathematics
University of Michigan
â We will assume that the coefficients Pi (x) and F (x) are
continuous on some open interval I, and that P0 (x) is never zero
on I. Under these assumptions, we write the general form of the
n-order linear equation as
September 29, 2008
y (n) + p1 (x)y (n−1) + . . . + pn−1 (x)y 0 + pn (x)y = f (x).
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General solutions of linear equations of higher order
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General solutions of linear equations of higher order
Principle of Superposition
ConcepTest
Question. What interval can we expect to find a unique solution to the
second order initial value problem
Theorem
Let y1 , y2 , . . . , yk be k solutions to the homogeneous linear equation
27y 00 + 14y + 3 = 0,
y (0) = 1, y 0 (0) = −1?
y (n) + p1 (x)y (n−1) + . . . + pn−1 (x)y 0 + pn (x)y = 0
(a) (−∞, ∞).
on the interval I. Then for any constants c1 , c2 , . . . , ck , the linear
combination
c1 y1 + c2 y2 + . . . + ck yk
(b) A small interval around x = 0.
(c) Impossible to tell from the given data.
is also a solution to the homogeneous equation on I.
Kenneth Harris (Math 216)
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Answer. (a) – A linear equation always has a unique solution on the
entire interval on which its coefficients are continuous. The constant
coefficients are continuous on the entire real line.
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General solutions of linear equations of higher order
General solutions of linear equations of higher order
Existence and Uniqueness Theorem
Example
Theorem
Supose the functions p1 , p2 , . . . , pn and f are continuous on the open
interval I containing the point a. Then given any n numbers
b0 , b1 , . . . , bn−1 , the nth-order linear equation
Let p1 , p2 , . . . , pn be continuous on some interval I containing a.
What is the unique solution on I of the nth-order initial value problem
y (n) + p1 (x)y (n−1) + . . . + pn−1 (x)y 0 + pn (x)y = 0
y (n) + p1 (x)y (n−1) + . . . + pn−1 (x)y 0 + pn (x)y = f (x).
y (a) = y 0 (a) = . . . = y (n−1) (a) = 0
has a unique solution on the entire interval I that satisfies the n initial
conditions
0
y (a) = b0 , y (a) = b1 , . . . , y
(n−1)
Answer. The trivial solution y ≡ 0 is always a solution to a
homogeneous equation, and satisfies the initial conditions. So, y ≡ 0
is the unique solution satisfying the nth-order initial value problem.
(a) = bn−1 .
An nth-order equation with n initial conditions is called an nth-order
initial value problem.
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Kenneth Harris (Math 216)
General solutions of linear equations of higher order
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Linear Independence
Example
ConcepTest
Solve the homogeneous third-order equation
x2
x3
The functions y1 =
and y2 =
and y3 ≡ 0 are three different
solutions to the second-order initial value problem
x 2 y 00 − 4xy + 6y = 0,
y (3) = 0
y (0) = y 0 (0) = 0
Answer. The general solution is
Why doesn’t this contradict the Uniqueness Theorem?
y (t) = c1 + c2 x + c3 x 2
Answer. The uniqueness theorem requires the leading coefficient of
y 00 to be nonzero in the interval containing 0.
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Goal. An n-order equation has “n parameters" and “n distinct
solutions". By distinct solutions we mean linearly independent
solutions.
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Linear Independence
Linear Independence
Linear dependence
Equivalent formulation
Definition
We say that n functions f1 , f2 , . . . , fn are linearly dependent on an
interval I when there are constants c1 , c2 , . . . , cn , not all zero, such that
Lemma
The function f1 , f2 , . . . , fn are linearly dependent if and only if one of the
functions can be written as a linear combination of the other functions.
c1 f1 (x) + c2 f2 (x) + . . . + cn fn (x) = 0
That is, there is some fi and constants c1 , c2 , . . . , ci−1 , ci+1 , . . . , cn such
that
X
fi (x) =
cj fj (x)
for all x in I.
for all x in I.
The functions are linearly independent if they are not linearly
dependent.
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j6=i
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Kenneth Harris (Math 216)
Linear Independence
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Linear Independence
Example
ConcepTest
Show the following functions are linearly dependent
cosh x, sinh x, ex .
Show the following functions are linearly dependent on (−∞, ∞)
Recall the definition of the hyperbolic functions
cosh x =
ex
+
2
e−x
sinh x =
ex
−
2
0, sin x, ex .
e−x
Answer. Let c1 = 27, c2 = c3 = 0
Answer. Let c1 = c2 = 1 and c3 = −1
0 = (27)0 + 0 sin x + 0ex .
0 = cosh x + sinh x − ex
In general, 0, f1 , f2 , . . . , fn are linearly dependent for any n functions.
or equivalently,
ex = cosh x + sinh x
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Kenneth Harris (Math 216)
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Linear Independence
The Wronskian for two functions
ConcepTest
Linear independence
Show the following functions are linearly dependent on (−∞, ∞)
This is a rephrasing of our earlier definition in terms of linear
dependence.
Definition
We say that n functions f1 , f2 , . . . , fn are linearly independent on an
interval I when there are no nontrivial choice of constants
c1 , c2 , . . . , cn , such that
20x, 5x sin2 x, 10x cos2 x.
Answer. Let c1 = −1, c2 = 4 and c3 = 2
−20x + 20x sin2 x + 20x cos2 x = −20x + 20x sin2 x + cos2 x = 0.
c1 f1 (x) + c2 f2 (x) + . . . + cn fn (x) = 0
for all x in I.
(The trivial choice, c1 = c2 = . . . = cn = 0, is always possible.)
Question. How can we test for linear independence?
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Kenneth Harris (Math 216)
The Wronskian for two functions
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The Wronskian for two functions
A test for linear independence
Proof of Lemma
Suppose f and g are linearly dependent on I. We will show that
Lemma
Let f and g be any once differentiable functions on an interval I. If
there is no x in I such that the following equation holds,
f (x)g 0 (x) − g(x)f 0 (x) = 0.
for all x in I.
f (x)g 0 (x) − g(x)f 0 (x) = 0,
Let c1 and c2 not both zero such that
then f and g are linearly independent on the interval I.
0 = c1 f (x) + c2 g(x)
Equivalently, if f and g are linearly dependent on an interval I, then
there is an a in I satisfying the equation
Suppose c1 6= 0. Then f (x) =
2
Let k = −c
c1 ; so, for any x in I
f (a)g 0 (a) − g(a)f 0 (a) = 0.
−c2
c1 g(x)
for all x in I
for all x in I.
f (x)g 0 (x) − g(x)f 0 (x) = kg(x)g 0 (x) − g(x)kg 0 (x)
(In fact, if f and g are linearly dependent on I then every a in I satisfies
this equation.)
= 0.
If c1 = 0 then g ≡ 0; but f , 0 are always linearly dependent.
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Kenneth Harris (Math 216)
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The Wronskian for two functions
The Wronskian for two functions
A test for linear independence
ConcepTest
We define the Wronskian of f and g to be the 2 × 2 determinant
f (x) g(x) = f (x)g 0 (x) − g(x)f 0 (x)
W (f , g)(x) = 0
f (x) g 0 (x)
Show that sin(ax) and cos(ax) are linearly independent when a 6= 0.
Answer.
Note that W (f , g) is a function of x.
cos(ax)
sin(ax) W (cos(ax), sin(ax))(x) = −a sin(ax) a cos(ax)
The following restates our condition for linear independence.
= a cos2 (ax) + a sin2 (ax)
Lemma
Let f and g be once differentiable functions. If the Wronskian is
nonzero for every x in I,
W (f , g)(x) = f (x)g 0 (x) − g(x)f 0 (x) 6= 0
= a 6= 0
for all x. By the Lemma, cos(ax) and sin(ax) are linearly independent.
for every x in I,
then f and g are linearly independent on I.
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Kenneth Harris (Math 216)
The Wronskian for two functions
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The Wronskian for two functions
ConcepTest
ConcepTest
Show that eax and ebx are linearly independent when a 6= b.
Compute W (cos2 x, sin2 x).
Answer. Let x be any number.
Answer.
ax
e
ax
bx
W (e , e )(x) = ax
ae
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W (cos2 x, sin2 x)(x)
ebx bebx =
sin2 x 2 sin x cos x cos2 x 2 sin x cos x) − sin2 x − 2 cos x sin x)
2 cos x sin x cos2 x + sin2 x
=
2 cos x sin x.
=
=
= be(a+b)x − ae(a+b)x
= (b − a)e(a+b)x
cos2 x
−2 sin x cos x
6= 0
By the Lemma,
eax
Kenneth Harris (Math 216)
and
e−ax
The Wronskian W (cos2 x, sin2 x)(0) = 0 but cos2 x and sin2 x are linearly
independent. So, we do not have a test for linear dependence.
are linearly independent.
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The Wronskian for two functions
The Wronskian for two functions
A test for linear dependence
A test for linear independence
Lemma
Let y1 and y2 are solutions on an interval I to the second-order
homogeneous equation
Here is an equivalent formulation in terms of linear independence.
Lemma
Let y1 and y2 are solutions on an interval I to the second-order
homogeneous equation
y 00 + p0 (x)y 0 + p1 (x)y = 0.
Then y1 and y2 are linearly dependent if and only if there is a point a in
I such that the Wronskian is zero: W (y1 , y2 )(a) = 0.
Under these conditions, iff the Wronskian is zero on one point a in I,
then it is zero on all x in I.
y 00 + p0 (x)y 0 + p1 (x)y = 0.
Then y1 and y2 are linearly independent if and only if the Wronskian is
never zero on I: W (y1 , y2 )(x) 6= 0 for any x in I.
Recall,
Recall,
W (y1 , y2 )(a) = y1 (a)y20 (a) − y2 (a)y10 (a)
W (y1 , y2 )(a) = y1 (a)y20 (a) − y2 (a)y10 (a)
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Kenneth Harris (Math 216)
The Wronskian for two functions
We are supposing y1 and y2 are solutions on an interval I to a
second-order linear equation and for some a in I
−
y2 (a)y10 (a)
y1 (a)y20 (a) − y2 (a)y10 (a) = 0.
= 0.
Case 2. y1 (a) = 0 but y10 (a) 6= 0. Note that our hypothesis implies
y 0 (a)
Case 1. y1 (a) 6= 0. Let k = yy21 (a)
(a) and consider the solution to the
equation y (x) = ky1 (x). Then
y2 (a)
y1 (a) = y2 (a);
y1 (a)
y 0 (a) =
y2 (a) = 0. Let k = y20 (a) and consider the solution to the equation
1
y (x) = ky1 (x). Then
y2 (a) 0
y1 (a) = y20 (a);
y1 (a)
y (a) =
Since y2 is the unique solution determined by the values y2 (a) and
y20 (a), we have y = y2 on I. So, ky1 = y2 on I, and y1 and y2 are
linearly dependent on I.
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y20 (a)
y1 (a) = 0 = y2 (a);
y10 (a)
y 0 (a) =
y20 (a) 0
y (a) = y20 (a);
y10 (a) 1
where the first equality is from our hypothesis that y1 (a) = 0 and so,
y2 (a) = 0.
where the last equality is from our hypothesis about a.
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Proof of Lemma: (⇒)
Suppose y1 and y2 are solutions on an interval I to a second-order
linear equation and for some a in I
y (a) =
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The Wronskian for two functions
Proof of Lemma: (⇒)
y1 (a)y20 (a)
Math 216 Differential Equations
Since y2 is the unique solution determined by the values y2 (a) and
y20 (a), we have y = y2 on I. So, ky1 = y2 on I, and y1 and y2 are
linearly dependent on I.
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The Wronskian for two functions
The Wronskian for two functions
Proof of Lemma: (⇒)
Caveat
The test for linear dependence of f and g on an interval I given by
W (f , g)(a) = f (a)g 0 (a) − g(a)f 0 (a) = 0
We are supposing y1 and y2 are solutions on an interval I to a
second-order linear equation and for some a in I
for some a in I, is only appropriate when f and g are both solutions on
I to a homogeneous linear equation
y1 (a)y20 (a) − y2 (a)y10 (a) = 0.
y 00 + p0 (x)y 0 + p1 (x)y = 0.
Case 3. y1 (a) = y10 (a) = 0. Then y1 ≡ 0 on I, since the constant zero
function is the unique solution y with y (0) = y 0 (0) = 0. So, y1 and y2
are linearly dependent (since y1 ≡ 0 on I).
Example. sin2 x and cos2 x are linearly independent on the interval
(−∞, ∞), however
W (cos2 x, sin2 x)(x) = 2 cos x sin x cos2 x − sin2 x
and W (cos2 x, sin2 x)(0) = 0.
It follows that sin2 x and cos2 x are not both solutions to the same
homogeneous second-order linear equation.
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The Wronskian for three functions
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Example
Definition
We define the Wronskian for three functions f , g and h which are twice
differentiable using the 3 × 3 determinant.
f (x) g(x) h(x) 0
W (f , g, h)(x) = f (x) g 0 (x) h0 (x) f 00 (x) g 00 (x) h00 (x)
0
g (x) h0 (x) = f (x) 00
g (x) h00 (x)
g(x) h(x) 0
−f (x) 00
g (x) h00 (x)
g(x) h(x) 00
+f (x) 0
g (x) h0 (x)
Math 216 Differential Equations
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The Wronskian for three functions
The Wronskian for three functions
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Compute the Wronskian for ex , cos x, sin x.
Answer.
x
e
cos x
sin x cos x W (ex , cos x, sin x)(x) = ex − sin x
ex − cos x − sin x − sin x
cos x = ex − cos x − sin x cos x
sin x −ex − cos x − sin x − sin x
cos x +ex − cos x − sin x = 2ex
Note that W (ex , cos x, sin x)(x) 6= 0 for all x.
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The Wronskian for three functions
The Wronskian for three functions
A test for linear independence
A test for linear dependence
We get a test for linear dependence if the functions are solutions to a
third-order homogeneous equation.
Analogous to two functions, we have a test of linear independence
using the Wronskian for three functions.
Lemma
Let y1 , y2 and y3 be solutions on an interval I to the third-order
homogeneous equation
Lemma
Let f , g and h be twice differentiable functions. If the Wronskian is
nonzero for every x in I,
W (f , g, h)(x) 6= 0
y (3) + p0 (x)y 00 + p1 (x)y 0 + p2 (x)y = 0.
for every x in I,
Then y1 , y2 and y3 are linearly dependent if and only if there is a point
a in I such that the Wronskian vanishes
then f , g and h are linearly independent on I.
W (y1 , y2 , y3 )(a) = 0.
Example. ex , cos x, sin x are linearly independent.
In fact, if the Wronskian is zero on one point a in I then it is zero on all
x in I.
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Kenneth Harris (Math 216)
The Wronskian for three functions
y
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ConcepTest
Lemma
Let y1 , y2 and y3 be solutions on an interval I to the third-order
homogeneous equation
00
September 29, 2008
The Wronskian for three functions
A test for linear independence
(3)
Math 216 Differential Equations
Show that ex , cos x, sin x are linearly independent on (−∞, ∞).
Answer. We have seen that W (ex , cos x, sin x)(x) = 2ex 6= 0 for all x.
It follows that ex , cos x, sin x are linearly independent.
0
+ p0 (x)y + p1 (x)y + p2 (x)y = 0.
Then y1 , y2 and y3 are linearly independent if and only if the
Wronskian W (y1 , y2 , y3 ) is never zero on I.
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Linear independence for n functions
Linear independence for n functions
General criterion for linear independence
Application of linear independence
â There is a generalization of the Wronskian to n functions which
are n-times differentiable: it is an n × n determinant,
W (f1 , f2 , . . . , fn ).
â Let f1 , f2 , . . . , fn are n-times differentiable on an interval I. If
W (f1 , f2 , . . . , fn )(x) 6= 0 for all x in I, then the functions are linearly
independent.
â In addition, if y1 , y2 , . . . , yn be n solutions on an interval I to the
homogeneous equation
The following sets of functions are linearly independent.
â 1, x, x 2 , x 3 , . . . , x n .
â 1, sin x, cos x, sin(2x), cos(2x), . . . , sin(nx), cos(nx).
â ea1 x , ea2 x , . . . , ean x where the ai are distinct constants.
y (n) + p1 (x)y (n−1) + . . . + pn−1 (x)y 0 + pn (x)y = 0,
then, y1 , y2 , . . . , yn are linearly independent on I if and only if
W (f1 , f2 , . . . , fn )(x) 6= 0 for all x in I.
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General solutions to homogeneous equations
Kenneth Harris (Math 216)
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General solutions to homogeneous equations
General solutions
General solutions
Theorem
Let y1 , y2 , . . . , yn be n linearly independent solutions to the
homogeneous equation
y (n) + p1 (x)y (n−1) + . . . + pn−1 (x)y 0 + pn (x)y = 0
on an open interval I in which each of p1 , p2 , . . . , pn−1 are continuous.
Then, for any solution to the equation, there are constants
c1 , c2 , . . . , cn such that
Definition
Given n linearly independent solutions y1 , y2 , . . . , yn on an interval I,
and parameters c1 , c2 , . . . , cn , the general solution to the
homogeneous equation is
y = c1 y1 + c2 y2 + . . . + cn yn
y (x) = c1 y1 (x) + c2 y2 (x) + . . . + cn yn (x)
for all x in I.
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Kenneth Harris (Math 216)
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General solutions to homogeneous equations
General solutions to homogeneous equations
Proof of Theorem
Proof of Theorem
Let y be a solution to the same equation with y (a) = Y0 and
y 0 (a) = Y1 . We want solutions c1 and c2 for the equations:
We prove the theorem in the case of n = 2. The general case is
essentially the same.
Y0 = c1 y1 (a) + c2 y2 (a)
Y1 = c1 y10 (a) + c2 y20 (a)
Let y1 and y2 be linearly independent solutions to the homogeneous
equation
y 00 + p1 (x)y 0 + p2 (x)y = 0
Multiply the first equation by y20 (a) and the second by y2 (a):
Y0 y20 (a) = c1 y20 (a)y1 (a) + c2 y20 (a)y2 (a)
on an open interval I in which each of p1 and p2 are continuous.
Y1 y2 (a) = c1 y2 (a)y10 (a) + c2 y2 (a)y20 (a)
We know that the Wronskian is never zero on I:
W (y1 , y2 )(x) 6= 0
Subtract the second equation from the first and solve for c1 :
for all x in I
c1 =
Y0 y20 (a) − Y1 y2 (a)
W (y1 , y2 )(a)
This is OK since W (y1 , y2 )(a) 6= 0.
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Kenneth Harris (Math 216)
General solutions to homogeneous equations
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General solutions to homogeneous equations
Proof of Theorem
Proof of Theorem
So, the choice of constants
Y0 = c1 y1 (a) + c2 y2 (a)
c1 =
Y1 = c1 y10 (a) + c2 y20 (a)
+
guarantee
c2 y1 (a)y20 (a)
y (a) = c1 y1 (a) + c2 y2 (a)
y 0 (a) = c1 y10 (a) + c2 y20 (a)
Subtract the second equation from the first and solve for c2 :
c2 =
Y0 y10 (a) − Y1 (a)y1 (a)
W (y1 , y2 )(a)
Then, by the uniqueness of the solution on I satisfying y (a) and y 0 (a),
y (x) = c1 y1 (x) + c2 y2 (x).
This is OK since W (y1 , y2 )(a) 6= 0.
Kenneth Harris (Math 216)
Math 216 Differential Equations
Y0 y10 (a) − Y1 (a)y1 (a)
W (y1 , y2 )(a)
c1 y1 + c2 y2
Y0 y10 (a) = c1 y10 (a)y1 (a) + c2 y10 (a)y2 (a)
Y1 y1 (a) =
c2 =
for
Multiply the second equation by y10 (a) and the second by y1 (a):
c1 y1 (a)y10 (a)
Y0 y20 (a) − Y1 y2 (a)
;
W (y1 , y2 )(a)
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