Ordinary Differential Equations
Fundamental Solution, Wronskians, and Linearly Independence Abadi
Universitas Negeri Surabaya
Ch 3.2: Fundamental Solutions of Linear Homogeneous Equations
Let p, q be continuous functions on an interval I
= (, ), which could be infinite. For any function y that is twice differentiable on I, define the
differential operator L by
Note that L[y] is a function on I, with output value
For example,
y y p y q yL
y (t) y (t) p(t) y (t) q(t) y(t)L
( ) sin( ) cos( ) 2 sin( )2 , 0 ),
sin(
) ( , )
( , )
(
2 2
2 2
t e
t t
t t
y L
I t
t y e
t q t
t p
t t
Differential Operator Notation
In this section we will discuss the second order linear homogeneous equation L[y](t) = 0, along with initial conditions as indicated below:
We would like to know if there are solutions to this initial value problem, and if so, are they unique.
Also, we would like to know what can be said about the form and structure of solutions that might be helpful in finding solutions to particular problems.
These questions are addressed in the theorems of this section.
1 0
0
0) , ( )
(
0 )
( )
(
y t
y y
t y
y t q y
t p y
y L
Theorem 3.2.1
Consider the initial value problem
where p, q, and g are continuous on an open interval I that contains t0. Then there exists a unique solution y
= (t) on I.
Note: While this theorem says that a solution to the initial value problem above exists, it is often not
possible to write down a useful expression for the
solution. This is a major difference between first and second order linear equations.
0 0
0
0) , ( )
(
) ( )
( )
(
y t
y y
t y
t g y
t q y
t p y
Example 1
Consider the second order linear initial value problem
In Section 3.1, we showed that this initial value problem had the following solution:
Note that p(t) = 0, q(t) = -1, g(t) = 0 are each continuous on (-, ), and the solution y is
defined and twice differentiable on (-, ).
t
t e
e t
y( ) 2
0 3,
0 1,
0
y y y
y
Example 2
Consider the second order linear initial value problem
where p, q are continuous on an open interval I containing t0.
In light of the initial conditions, note that y = 0 is a solution to this homogeneous initial value problem.
Since the hypotheses of Theorem 3.2.1 are
satisfied, it follows that y = 0 is the only solution of this problem.
0 0,
0 0, 0 )
( )
(
p t y q t y y y y
Example 3
Determine the longest interval on which the given initial value problem is certain to have a unique twice
differentiable solution. Do not attempt to find the solution.
First put differential equation into standard form:
The longest interval containing the point t = 0 on which the coefficient functions are continuous is (-1, ).
It follows from Theorem 3.2.1 that the longest interval on which this initial value problem is certain to have a twice differentiable solution is also (-1, ).
t 1
y (cost)y 3y 1, y
0 1, y
0 0
0 1,
0 01, 1 1
3 1
cos
y y
y t y t
t y t
Theorem 3.2.2 (Principle of Superposition)
If y1and y2 are solutions to the equation
then the linear combination c1y1 + y2c2 is also a solution, for all constants c1 and c2.
To prove this theorem, substitute c1y1 + y2c2 in for y in the equation above, and use the fact that y1 and y2 are
solutions.
Thus for any two solutions y1 and y2, we can construct an infinite family of solutions, each of the form y = c1y1 + c2 y2.
Can all solutions can be written this way, or do some
solutions have a different form altogether? To answer this question, we use the Wronskian determinant.
0 )
( )
( ]
[y y p t y q t y L
The Wronskian Determinant
(1 of 3) Suppose y1 and y2 are solutions to the equation
From Theorem 3.2.2, we know that y = c1y1 + c2 y2 is a solution to this equation.
Next, find coefficients such that y = c1y1 + c2 y2 satisfies the initial conditions
To do so, we need to solve the following equations:
0 )
( )
( ]
[y y p t y q t y L
0 0
0
0) , ( )
(t y y t y
y
0 0
2 2 0
1 1
0 0
2 2 0
1 1
) ( )
(
) ( )
(
y t
y c t
y c
y t
y c t
y c
The Wronskian Determinant
(2 of 3) Solving the equations, we obtain
In terms of determinants:
) ( ) ( )
( ) (
) ( )
(
) ( ) ( )
( ) (
) ( )
(
0 2 0 1 0
2 0 1
0 1 0 0
1 2 0
0 2 0 1 0
2 0 1
0 2 0 0
2 1 0
t y t y t
y t y
t y y t
y c y
t y t y t
y t y
t y y t
y c y
) ( )
(
) ( )
(
) (
) ( ,
) ( )
(
) ( )
(
) (
) (
0 2 0
1
0 2 0
1
0 0
1
0 0
1
2
0 2 0
1
0 2 0
1
0 2 0
0 2 0
1
t y t
y
t y t
y
y t
y
y t
y c
t y t
y
t y t
y
t y y
t y y
c
The Wronskian Determinant
(3 of 3) In order for these formulas to be valid, the
determinant W in the denominator cannot be zero:
W is called the Wronskian determinant, or more simply, the Wronskian of the solutions y1and y2. We will sometimes use the notation
) ( ) ( )
( ) ) (
( )
(
) ( )
(
0 2 0 1 0
2 0 1 0
2 0
1
0 2 0
1 y t y t y t y t
t y t
y
t y t
W y
W y t
y
y t
y W c
t y y
t y y
c 1 0 0
0 0
1
2 0
2 0
0 2 0
1
) (
) ( ) ,
( ) (
y1, y2
t0W
Theorem 3.2.3
Suppose y1 and y2 are solutions to the equation
and that the Wronskian
is not zero at the point t0 where the initial conditions
are assigned. Then there is a choice of
constants c1, c2 for which y = c1y1 + c2 y2 is a solution to the differential equation (1) and initial conditions (2).
) 1 ( 0 )
( )
( ]
[y y p t y q t y L
2 1 2
1y y y
y
W
) 2 ( )
( ,
)
(t0 y0 y t0 y0
y
Example 4
Recall the following initial value problem and its solution:
Note that the two functions below are solutions to the differential equation:
The Wronskian of y1 and y2 is
Since W 0 for all t, linear combinations of y1 and y2 can be used to construct solutions of the IVP for any initial value t0.
t
t y e
e
y1 , 2
2 2 0
2 1 2
1 2
1
2
1
y y y y e e e e e
y y
y
W y t t t t
y
y t et e ty y
y 0, 0 3, 0 1 ( ) 2
2 2 1
1y c y
c
y
Theorem 3.2.4 (Fundamental Solutions)
Suppose y1 and y2 are solutions to the equation
If there is a point t0 such that W(y1,y2)(t0) 0,
then the family of solutions y = c1y1 + c2 y2 with arbitrary coefficients c1, c2 includes every
solution to the differential equation.
The expression y = c1y1 + c2 y2 is called the general solution of the differential equation above, and in this case y1 and y2 are said to form a fundamental set of solutions to the differential equation.
. 0 )
( )
( ]
[y y p t y q t y L
Example 5
Recall the equation below, with the two solutions indicated:
The Wronskian of y1 and y2 is
Thus y1 and y2 form a fundamental set of
solutions to the differential equation above, and can be used to construct all of its solutions.
The general solution is
t
t y e
e y
y
y 0, 1 , 2
. all for 0 2
2 0
2 1
2
1 e e e e e t
y y
y
W y t t t t
t
t c e
e c
y 1 2
Example 6
Consider the general second order linear equation below, with the two solutions indicated:
Suppose the functions below are solutions to this equation:
The Wronskian of y1and y2 is
Thus y1and y2 form a fundamental set of solutions to the equation, and can be used to construct all of its
solutions.
The general solution is
2 1
2
1 e 1 , y e 2 , r r
y rt r t
1 2 0 for all .2 1
2 1
1 2
2 2 1
1
2
1 r r e t
e r e
r
e e
y y
y
W y rt r t r r t
t r t
r
0 )
( )
(
p t y q t y y
t r t
r c e
e c
y 1 1 2 2
Example 7: Solutions
(1 of 2) Consider the following differential equation:
Show that the functions below are fundamental solutions:
To show this, first substitute y1 into the equation:
Thus y1 is a indeed a solution of the differential equation.
Similarly, y2 is also a solution:
1 2
2 / 1
1 t , y t y
0 ,
0 3
2t2y t y y t
0 2 1
3 2 1 3 2
2 4 1/2 1/2
2 / 1 2
/ 3
2
t t t
t t t
2 3
4 3 1
02t2 t3 t t2 t1 t1
Example 7: Fundamental Solutions
(2 of 2) Recall that
To show that y1 and y2 form a fundamental set of solutions, we evaluate the Wronskian of y1 and y2:
Since W 0 for t > 0, y1, y2 form a fundamental set of solutions for the differential equation
3 2
/ 3 2
/ 3 2
/ 2 3
2 / 1
1 2
/ 1
2 1
2 1
2 3 2
3 2
1 2
1 t t t t
t t
t t
y y
y
W y
1 2
2 / 1
1 t , y t y
0 ,
0 3
2t2y t y y t
Theorem 3.2.5: Existence of Fundamental Set of Solutions
Consider the differential equation below, whose coefficients p and q are continuous on some open interval I:
Let t0 be a point in I, and y1 and y2 solutions of the equation with y1 satisfying initial conditions
and y2 satisfying initial conditions
Then y1, y2 form a fundamental set of solutions to the given differential equation.
0 )
( )
( ]
[y y p t y q t y L
0 )
( ,
1 )
( 0 1 0
1 t y t
y
1 )
( ,
0 )
( 0 2 0
2 t y t
y
Example 7: Theorem 3.2.5
(1 of 3) Find the fundamental set specified by Theorem 3.2.5 for the differential equation and initial point
We showed previously that
were fundamental solutions, since W(y1, y2)(t0) = -2 0.
But these two solutions don’t satisfy the initial
conditions stated in Theorem 3.2.5, and thus they do not form the fundamental set of solutions mentioned in that theorem.
Let y3 and y4 be the fundamental solutions of Thm 3.2.5.
t
t y e
e
y1 , 2 0 ,
0 0
y t y
1 ) 0 ( ,
0 )
0 (
; 0 )
0 ( ,
1 ) 0
( 3 4 4
3 y y y
y
Example 7: General Solution
(2 of 3) Since y1 and y2 form a fundamental set of solutions,
Solving each equation, we obtain
The Wronskian of y3 and y4 is
Thus y3, y4 forms the fundamental set of solutions indicated in Theorem 3.2.5, with general solution in this case
1 ) 0 ( , 0 )
0 ( ,
0 )
0 ( , 1 ) 0 ( ,
4 4
2 1
4
3 3
2 1
3
y y
e d e
d y
y y
e c e
c y
t t
t t
) sinh(
2 1 2
) 1 ( ),
cosh(
2 1 2
) 1
( 4
3 t e e t y t e e t
y t t t t
0 1 sinh
cosh cosh sinh
sinh
cosh 2 2
2 1
2
1
t t
t t
t t
y y
y W y
) sinh(
) cosh(
)
(t k1 t k2 t
y
Example 7:
Many Fundamental Solution Sets
(3 of 3) Thus
both form fundamental solution sets to the differential equation and initial point
In general, a differential equation will have
infinitely many different fundamental solution sets. Typically, we pick the one that is most convenient or useful.
e e
S
t t
S1 t, t , 2 cosh ,sinh
0 ,
0 0
y t y
Summary
To find a general solution of the differential equation
we first find two solutions y1 and y2.
Then make sure there is a point t0 in the interval such that W(y1, y2)(t0) 0.
It follows that y1 and y2 form a fundamental set of solutions to the equation, with general solution y = c1y1 + c2 y2.
If initial conditions are prescribed at a point t0 in the interval where W 0, then c1 and c2 can be chosen to satisfy those conditions.
p t y q t y t
y ( ) ( ) 0,
Ch 3.3:
Linear Independence and the Wronskian
Two functions f and g are linearly dependent if there exist constants c1 and c2, not both zero, such that
for all t in I. Note that this reduces to determining whether f and g are multiples of each other.
If the only solution to this equation is c1 = c2 = 0, then f and g are linearly independent.
For example, let f(x) = sin2x and g(x) = sinxcosx, and consider the linear combination
This equation is satisfied if we choose c1 = 1, c2 = -2, and hence f and g are linearly dependent.
0 )
( )
( 2
1 f t c g t c
0 cos
sin 2
sin 2
1 x c x x
c
Solutions of 2 x 2 Systems of Equations
When solving
for c1 and c2, it can be shown that
Note that if a = b = 0, then the only solution to this system of equations is c1 = c2 = 0, provided D 0.
b y
c y
c
a x
c x
c
2 2 1
1
2 2 1
1
2 1
2 1 1
1 2
1 2
1
1 1
2
2 2
2 1 2
1
2 1 2
where ,
,
y y
x D x
D bx ay
x y y
x
bx c ay
D bx ay
x y y
x
bx c ay
Example 1: Linear Independence
(1 of 2) Show that the following two functions are linearly independent on any interval:
Let c1 and c2 be scalars, and suppose
for all t in an arbitrary interval (, ).
We want to show c1 = c2 = 0. Since the
equation holds for all t in (, ), choose t0 and t1 in (, ), where t0 t1. Then
t
t g t e
e t
f ( ) , ( )
0 )
( )
( 2
1 f t c g t c
0 0
1 1
0 0
2 1
2 1
t t
t t
e c e
c
e c e
c
Example 1: Linear Independence
(2 of 2) The solution to our system of equations
will be c1 = c2 = 0, provided the determinant D is nonzero:
Then
Since t0 t1, it follows that D 0, and therefore f and g are linearly independent.
0 1 1
1 0 1 0
0 1
1
0
0 t t t t t t t t
t t
t t
e e
e e e
e e e
e
D e
1 0
2
1
1 1
0
1 0
1 0 1
0 1
0 0
1 1
0
t t
e
e e e
e e
D
t t
t t t
t t
t t
t t
t
0 0
1 1
0 0
2 1
2 1
t t
t t
e c e
c
e c e
c
Theorem 3.3.1
If f and g are differentiable functions on an open interval I and if W(f, g)(t0) 0 for some point t0 in I, then f and g are linearly independent on I. Moreover, if f and g are linearly dependent on I, then W(f, g)(t) = 0 for all t in I.
Proof (outline): Let c1 and c2 be scalars, and suppose
for all t in I. In particular, when t = t0 we have
Since W(f, g)(t0) 0, it follows that c1 = c2 = 0, and hence f and g are linearly independent.
0 ) ( )
( 2
1 f t c g t c
0 ) ( )
(
0 )
( )
(
0 2
0 1
0 2
0 1
t g c t
f c
t g c t
f c
Theorem 3.3.2 (Abel’s Theorem)
Suppose y1 and y2 are solutions to the equation
where p and q are continuous on some open interval I. Then W(y1,y2)(t) is given by
where c is a constant that depends on y1 and y2 but not on t.
Note that W(y1,y2)(t) is either zero for all t in I (if c = 0) or else is never zero in I (if c ≠ 0).
0 )
( )
( ]
[y y p t y q t y L
cep t dt t
y y
W ( 1, 2)( ) ( )
Example 2: Wronskian and Abel’s Theorem
Recall the following equation and two of its solutions:
The Wronskian of y1and y2 is
Thus y1 and y2 are linearly independent on any
interval I, by Theorem 3.3.1. Now compare W with Abel’s Theorem:
Choosing c = -2, we get the same W as above.
t
t y e
e y
y
y 0, 1 , 2
. all for 0 2
2 0
2 1
2
1 e e e e e t
y y
y
W y t t t t
c ce
ce t
y y
W( 1, 2)( ) p(t)dt 0dt
Theorem 3.3.3
Suppose y1 and y2 are solutions to equation
below, whose coefficients p and q are continuous on some open interval I:
Then y1 and y2 are linearly dependent on I iff W(y1, y2)(t) = 0 for all t in I. Also, y1 and y2 are
linearly independent on I iff W(y1, y2)(t) 0 for all t in I.
0 )
( )
( ]
[y y p t y q t y L
Summary
Let y1 and y2 be solutions of
where p and q are continuous on an open interval I.
Then the following statements are equivalent:
The functions y1 and y2 form a fundamental set of solutions on I.
The functions y1 and y2 are linearly independent on I.
W(y1,y2)(t0) 0 for some t0 in I.
W(y1,y2)(t) 0 for all t in I.
0 )
( )
(
p t y q t y y
Linear Algebra Note
Let V be the set
Then V is a vector space of dimension two, whose bases are given by any fundamental set of
solutions y1 and y2.
For example, the solution space V to the differential equation
has bases with
e e
S
t t
S1 t, t , 2 cosh ,sinh
: ( ) ( ) 0, ,
y y p t y q t y t V
0
y y
2
1 Span
Span S S
V
