MA 108-ODE- D3
Lecture 14
Debanjana Mitra
Department of Mathematics
Indian Institute of Technology Bombay
Powai, Mumbai - 76
May 29, 2023
nth order linear DE: Constant coefficient
Laplace transform
Constant Differential Operators
Aim. Give a formulation for the basis of ker L, where
L = D n + a1 D n−1 + . . . + an I ,
with constant coefficients a1 , a2 , · · · an .
Constant Differential Operators
Constant differential operator: L = D n + a1 D n−1 + . . . + an I , with
constant coefficients a1 , a2 , · · · an .
Case I: PL has distinct real roots:
Theorem
Let L be a constant coefficient linear differential operator of order n such
that
PL (x) = (x − r1 ) . . . (x − rn )
where r1 , r2 , . . . , rn are distinct real numbers. Then the general solution
of L(y ) = 0 is given by
y = c1 e r1 x + c2 e r2 x + . . . + cn e rn x
where c1 , c2 , . . . , cn ∈ R.
Proof: We have
L = (D − r1 ) . . . (D − rn )
and the Null-space
N(D − ri ) = {ce ri x : c ∈ R}.
Constant Differential Operators
It follows that
e r1 x , . . . , e rn x ∈ N(L).
Check! {e r1 x , . . . , e rn x } are linearly independent over R as r1 , r2 , . . . , rn
are distinct real numbers.
As dimension of N(L) is n by the Dimension Theorem, we get
N(L) = {c1 e r1 x + . . . + cn e rn x : c1 , . . . , cn ∈ R}.
Constant Differential Operators
Case II: PL (x) has some repeated real roots: What happened in the
n = 2 case? m1 = m2 = m gave us only one solution - f (x) = e mx . The
other solution was obtained using the method of looking for a solution of
the form vf . This method yielded xe mx as the other solution.
Proposition
For any real number r , the functions,
u1 (x) = e rx , u2 (x) = xe rx , . . . , um (x) = x m−1 e rx
are linearly independent and
u1 (x), . . . , um (x) ∈ Ker((D − r )m ).
Constant Differential Operators
Proof. Since {1, x, x 2 , . . . , x m } is linearly independent, it follows that
{e rx , xe rx , . . . , x m−1 e rx } is also linearly independent. We need to show
that these functions are in Ker (D − r )m . Let’s first verify this when
m = 1. We need to show that
u1 (x) = e rx ∈ Ker((D − r )),
which is true, since
(D − r )(e rx ) = re rx − re rx = 0.
Suppose m = 2. Since u1 is in Ker of (D − r ), it’s in Ker of (D − r )2
(why?). What about u2 ?
(D − r )2 (xe rx ) =
(D − r )(D − r )(xe rx )
=
(D − r )(xre rx + e rx − rxe rx )
=
(D − r )(e rx ) = 0.
Constant Differential Operators
So how do we prove this in general? Induction. Assume
u1 , u2 , . . . , um−1 ∈ Ker((D − r )m−1 ),
and we need to show that
u1 , u2 , . . . , um ∈ Ker((D − r )m ).
That
u1 , u2 , . . . , um−1 ∈ Ker((D − r )m )
is easy since
Ker((D − r )m−1 ) ⊆ Ker((D − r )m ).
To show that um is also in Ker((D − r )m ), consider
(D − r )m (um (x)) = (D − r )m (x m−1 e rx )
(D − r )m−1 (D − r )(x m−1 e rx )
+ (m − 1)x m−2 e rx − rx m−1 e rx
=
= (D − r )m−1 x m−1 re rx
=
(D − r )m−1 ((m − 1)x m−2 e rx )
=
0.
Constant Differential Operators
Thus, if
PL (x) = (x − r1 )m1 (x − r2 )m2 . . . (x − r` )m` ,
where
`
X
mi = n, a basis of Ker L is given by
i=1
e r1 x , . . . , x m1 −1 e r1 x , e r2 x , . . . , x m2 −1 e r2 x , . . . , e r` x , . . . , x m` −1 e r` x .
Note that that the above functions are linearly independent and since
dim Ker L = n, these form a basis.
Exercise: Check that the above functions are linearly independent.
Constant Differential Operators
Example: Find the general solution of the DE:
L(y ) = (D 3 − D 2 − 8D − 12)(y ) = 0.
We have
PL (x) = x 3 − x 2 − 8x − 12 = (x − 2)2 (x + 3),
and therefore,
L = (D − 2)2 (D + 3).
Hence, e 2x , xe 2x ∈ N (D − 2)2 and e −3x ∈ N (D + 3) . As dimension
of N(L) = 3 and e 2x , xe 2x and e −3x are linearly independent, they form a
basis of the solution space of L. Thus the general solution is
y = c1 e 2x + c2 xe 2x + c3 e −3x ,
where c1 , c2 , c3 ∈ R.
Constant Differential Operators
Example: Find the general solution of the DE:
L(y ) = (D 6 + 2D 5 − 2D 3 − D 2 )(y ) = 0.
Now,
PL (x) = x 6 + 2x 5 − 2x 3 − x 2 = x 2 (x − 1)(x + 1)3 .
Hence,
L = D 2 (D − 1)(D + 1)3 .
Ker D 2 is h1, xi. Ker D − 1 is he x i. Ker (D + 1)3 is he −x , xe −x , x 2 e −x i.
Thus, the general solution is
c1 + c2 x + c3 e x + c4 e −x + c5 xe −x + c6 x 2 e −x ,
with ci ∈ R.
Proposition (Linear independency of a set of products of
polynomial and exponential functions)
Let Q1 , Q2 , . . . , Qn be non-zero polynomials and let r1 , r2 , . . . , rn be
distinct real numbers. Then Q1 (x)e r1 x , Q2 (x)e r2 x , . . . , Qn (x)e rn x are
linearly independent functions.
Proof: Induction on n.
Proposition
Consider the functions
upq (x) = x q−1 e rp x ,
where p = 1, 2, . . . , k, q = 1, 2, . . . , mp , r1 , . . . , rk are distinct real
numbers and m1 , m2 , . . . , mk are positive integers such that
m1 + . . . + mk = n.
Then the n functions upq are linearly independent.
Proof: Suppose that
mp
k X
X
p=1 q=1
cpq upq (x) = 0
Proof: Suppose that
mp
k X
X
cpq upq (x) = 0
p=1 q=1
i.e.,
mp
k
X
X
p=1
Denoting Qp (x) =
Pmp
!
cpq x
q−1
e rp x = 0
q=1
q=1 cpq x
q−1
k
X
for each p = 1, · · · k, we have
Qpq (x)e rp x = 0.
p=1
From the above Proposition [Linear independency of a set of products of
polynomial and exponential functions], it follows that
mp
X
cpq x q−1 = 0.
q=1
This implies that cpq = 0 for every p, q as 1, x, . . . , x mp −1 are linearly
independent.
Constant Differential Operators
III. PL has some complex roots: In the second order case, if
m1 = a + ıb, m2 = a − ıb, we got a basis:
e m1 x − e m2 x
e m1 x + e m2 x
= e ax sin bx.
= e ax cos bx,
2i
2
If
PL (x) = x n + p1 x n−1 + . . . + pn
has a complex root a + ıb, it also has a − ıb as a root. Thus,
(x − (a + ıb))(x − (a − ıb)) = x 2 − 2ax + (a2 + b 2 )
is a factor of PL (x), i.e., D 2 − 2aD + (a2 + b 2 ) is a factor of L. Ker
(D 2 − 2aD + (a2 + b 2 )) has a basis
{e ax cos bx, e ax sin bx}.
Constant Differential Operators
If a ± ıb is a root of multiplicity m, we need to consider
m Ker D 2 − 2aD + (a2 + b 2 )
.
Can you guess a basis for this space?
e ax cos bx, xe ax cos bx, . . . , x m−1 e ax cos bx,
e ax sin bx, xe ax sin bx, . . . , x m−1 e ax sin bx.
Exercise: Check this.
Constant Differential Operators
Example: Find the general solution of
y (5) − 9y (4) + 34y (3) − 66y (2) + 65y 0 − 25y = 0.
The characteristic polynomial is
(x − 1)(x 2 − 4x + 5)2 .
The roots are
1, 2 ± ı, 2 ± ı.
Hence, the general solution is
y = c1 e x + c2 e 2x cos x + c3 xe 2x cos x + c4 e 2x sin x + c5 xe 2x sin x,
where ci ∈ R.
Summary: n-th order linear ODE with constant coefficients
L = D n + a1 D n−1 + . . . + an I , with constant coefficients a1 , a2 , · · · an .
Characteristic polynomial: PL (x) = x n + a1 x n−1 + · · · + an x.
Case I If PL (x) = (x − r1 ) · · · (x − rn ), where r1 , r2 , . . . , rn are distinct real
numbers, then the general solution of L(y ) = 0 is given by
y (x) = c1 e r1 x + c2 e r2 x + . . . + cn e rn x
where c1 , c2 , . . . , cn ∈ R.
m1
Case II If PL (x) = (x − r1 ) (x − r2 )
m2
m`
. . . (x − r` ) , where
`
X
mi = n,
i=1
then the general solution of L(y ) = 0 is given by
y (x) =
mp
` X
X
cpq x q−1 e rp x ,
p=1 q=1
where cpq , for p = 1, · · · , ` and q = 1, · · · mp are real constants.
Summary: Contd....
m
Case 3 If PL (x) has a factor D 2 − 2aD + (a2 + b 2 ) , then the general
solution of L(y ) = 0 will have terms involving
e ax cos bx, xe ax cos bx, . . . , x m−1 e ax cos bx,
e ax sin bx, xe ax sin bx, . . . , x m−1 e ax sin bx.
Non-homogeneous 2nd order Linear ODE’s - RECALL
Consider
y 00 + p(x)y 0 + q(x)y = r (x)
(1)
where p, q, r are continuous functions on an interval I .
Let yp be any solution of (1) and y1 , y2 be a basis of the solution space of the
corresponding homogeneous DE.
Then the set of solutions of (1) is
{c1 y1 + c2 y2 + yp | c1 , c2 ∈ R}.
Summary: In order to find the general solution of a non-homogeneous DE, we
need to
I
get the general solution of the corresponding homogeneous DE.
I
get one particular solution of the non-homogeneous DE.
Annihilator Operator
If A is a linear differential operator with constant coefficients and r (·) is a
sufficiently smooth differentiable function such that
Ar (x) = 0,
then A is said to be the annihilator of the function r (x).
Examples :
1. D + 3 annihilates e −3x .
2. (D − 2)2 annihilates 4e 2x − 10xe 2x .
3. D 2 + 16 annihilates cos 4x, sin 4x or any of their linear combinations.
4. D 2 + 2D + 5 annihilates 5e −x cos 2x − 9e −x sin 2x.
5. D 4 annihilates 1 − 5x 2 + 8x 3 .
6. D n annihilates 1, x, x 2 , · · · , x n−1 .
7. (D − α)n annihilates e αx , xe αx , · · · , x n−1 e αx .
8. (D 2 − 2αD + (α2 + β 2 ))n annihilates e αx cos βx, xe αx cos βx, · · ·
x n−1 e αx cos βx, e αx sin βx, xe αx sin βx, · · · x n−1 e αx sin βx.
Annihilator Method - Formalising the method of
undetermined coefficients
I
Let L(y ) = y (n) + a1 y (n−1) + . . . + an−1 y 0 + an y = r (·) be a
non-homogeneous DE with constant coefficients a1 , a2 , . . . , an ∈ R
and r (·) is a sum or product of exponential, polynomial, or
sinusoidal terms.
I
In the method of annihilators, we assume that the function r is in
Null-space of A, where A is a constant coefficient linear differential
operator, i.e., Ar (x) = 0.
I
Since L(y ) = r (·), we get (AL)(y ) = A(r (·)) = 0.
I
If y is a solution of the non-homogeneous DE L(y ) = r (·), then y is
a solution of the homogeneous DE with constant coefficients:
AL(y ) = 0.
I
To find solutions of the above homogeneous DE, we look for roots
of its charactersitic polynomial:
PAL (x) = PA (x)PL (x).
Example
Example: Find a particular solution of DE:
L(y ) = (D 2 − 5D + 6)(y ) = xe x .
Note that for r (x) = xe x ,
(D − 1)2 (xe x ) = 0.
Thus, for r (x) = xe x , an annihilator operator is A = (D − 1)2 .
If Ly = xe x ,
AL(y ) = A(xe x ) = 0,
i.e.,
(D − 1)2 (D 2 − 5D + 6)y = 0.
We know that the characteristic polynomial of the above DE is
PAL (x) = PA (x)PL (x) = (x − 1)2 (x 2 − 5x + 6) = (x − 1)2 (x − 2)(x − 3).
The roots of this polynomial are
1, 1, 2, 3.
Hence, the linearly independent solutions of AL(y ) = 0 are
e x , xe x , e 2x , e 3x ,
Annihilator Method
The general solution of AL(y ) = 0 is given by
ae x + bxe x + ce 2x + de 3x ,
for a, b, c, d ∈ R.
We need to find a, b, c, d ∈ R such that
L(ae x + bxe x + ce 2x + de 3x ) = xe x .
Since for Ly = (D 2 − 5D + 6)y = 0, the characteristic polynomial is
m2 − 5m + 6 = (m − 2)(m − 3).
Thus, N(L) = he 2x , e 3x i.
As L(ce 2x + de 3x ) = 0, we may choose
c = d = 0.
Annihilator Method
So we need to find a, b such that
L(ae x + bxe x ) = xe x .
i.e.,
(D 2 − 5D + 6)(ae x + bxe x ) = xe x .
To find a, b satisfying the above equation, set
y1 = ae x + bxe x ,
Dy1 = ae x + be x + bxe x ,
D 2 y1 = ae x + 2be x + bxe x .
Annihilator Method
Hence,
L(y1 ) =(D 2 − 5D + 6)(y1 ),
=D 2 (y1 ) − 5D(y1 ) + 6y1 ,
=(a + 2b)e x + bxe x − 5(a + b)e x − 5bxe x + 6ae x + 6bxe x ,
=xe x .
This gives (2a − 3b)e x + 2bxe x = xe x . Thus a = 3/4 and b = 1/2 and
hence,
3
1
y = e x + xe x
4
2
is a particular solution of the non-homgeneous DE.
Example
Find a particular solution of
y (4) − 16y = x 4 + x + 1.
Here, the linear differential operator associated to the above DE is
L = D 4 − 16,
and note that r (x) = x 4 + x + 1 is in Null-space of the operator D 5 . Let
us take
A = D 5.
Then Ar (x) = D 5 (x 4 + x + 1) = 0. Hence a solution y of L(y ) = r (x) is
also a solution of
AL(y ) = D 5 (D 4 − 16)y = 0.
To find solutions of AL(y ) = 0, consider the characteristic equation of AL
PAL (x) = m5 (m4 − 16) = m5 (m − 2)(m + 2)(m2 + 4).
Roots are 0, ±2, ±2ι.
Annihilator Method
Thus, a general solution of (AL)(y ) = 0 is of the form
c1 + c2 x + c3 x 2 + c4 x 3 + c5 x 4 + c6 e 2x + c7 e −2x + c8 cos 2x + c9 sin 2x.
So in order to solve
L(y ) = x 4 + x + 1,
we should look for a solution of the form
c1 + c2 x + c3 x 2 + c4 x 3 + c5 x 4 .
Why? First note that L annihilates the last four terms. Hence, we may
choose c6 , c7 , c8 , c9 arbitrarily. We choose
c6 = c7 = c8 = c9 = 0.
We need to find c1 , c2 , c3 , c4 , c5 so that
L c1 + c2 x + c3 x 2 + c4 x 3 + c5 x 4 = r (x).
i.e.,
(D 4 − 16)(c1 + c2 x + c3 x 2 + c4 x 3 + c5 x 4 ) = x 4 + x + 1.
Annihilator Method
(D 4 − 16)(c1 + c2 x + c3 x 2 + c4 x 3 + c5 x 4 ) = x 4 + x + 1
gives
24c5 − 16(c1 + c2 x + c3 x 2 + c4 x 3 + c5 x 4 ) = x 4 + x + 1.
Comparing coefficients in the above equation, we get
24c5 − 16c1 = 1, −16c2 = 1, −16c3 = 0, −16c4 = 0 and − 16c5 = 1,
and hence
c5 = −
1
1
5
, c4 = 0, c3 = 0, c2 = − , c1 = − .
16
16
32
Hence, a particular solution of the given non-homogeneous equation is
yp = −
1 4
1
5
x − x− .
16
16
32
Annihilator Method
r (x)
r (x) = x m−1
Annihilator A such that A r (x) =
0.
A = Dm
e ax
(D − a)
x m−1 e ax
(D − a)m
cos bx or sin bx
(D 2 + b 2 )
e ax cos bx or e ax sin bx
(D 2 − 2aD + a2 + b 2 )
x m−1 e ax cos bx
(D 2 − 2aD + a2 + b 2 )m
x m−1 e ax sin bx
(D 2 − 2aD + a2 + b 2 )m .
Annihilator Method
Exercise: Get candidate solutions by the annihilator method:
1. y (4) + 2y 00 + y = 3 sin x − 5 cos x.
2. y (4) − y (3) − y 00 + y 0 = x 2 + 4 + x sin x.
3. y (4) − 2y 00 + y = x 2 e x + e 2x .
Undetermined coefficient method and variation of
parameters
Remark As described for the 2nd order non-homogeneous linear
differential equation, the undetermined coefficient method and variation
of parameters method are as well applicable to the case of n-order
non-homogeneous linear differential equation with a suitable
modifications.
Variation of Parameters- finding yp
Let
L(y ) = y (n) + p1 (x)y (n−1) + . . . + pn (x)y = r (x).
Here, p1 , p2 , . . . , pn , r are continuous on an open interval I .
Recall the case n = 2: We replace c1 , c2 in the general solution of the
associated homogeneous equation by functions v1 (x), v2 (x), so that
y (x) = v1 (x)y1 (x) + v2 (x)y2 (x)
is a solution of
y 00 + p(x)y 0 + q(x)y = r (x).
Now,
y 0 = v1 y10 + v2 y20 + v10 y1 + v20 y2 .
We also demanded
v10 y1 + v20 y2 = 0.
Variation of Parameters
Then we substituted y , y 0 , y 00 in the given non-homogeneous ODE, and
rearranged terms, to obtain:
v10 y10 + v20 y20 = r (x).
Together with,
v10 y1 + v20 y2 = 0,
this yielded :
v10 =
0
y2
r (x) y20
, v20 =
W (y1 , y2 )
y1
0
y10 r (x)
,
W (y1 , y2 )
y2 r (x)
dx, v2 =
W (y1 , y2 )
Z
and hence
Z
v1 = −
y1 r (x)
dx.
W (y1 , y2 )
Thus,
Z
y = v1 y1 + v2 y2 = y2
y1 r (x)
dx − y1
W (y1 , y2 )
Z
y2 r (x)
dx.
W (y1 , y2 )
Variation of Parameters
For general n, we proceed exactly the same way.
The method of variation of parameters assumes that a basis of solutions
y1 , . . . , yn of the associated homogeneous equation
y (n) + p1 (x)y (n−1) + . . . + pn (x)y = 0,
is known, i.e., the general solution of the above homogeneous equation is
y (x) = c1 y1 (x) + c2 y2 (x) + · · · + cn yn (x)
where c1 , · · · , cn are arbitrary constants. The idea is to replace ci in the
above equation by functions vi , which are continuous on I and such that
yp = v1 (x)y1 (x) + v2 (x)y2 (x) + · · · + vn (x)yn (x)
is a solution of the non-homogeneous equation.
Variation of Parameters
c 1 y1 + c 2 y2
y = v1 y1 + v2 y2
→ c 1 y1 + c 2 y2 + . . . + c n yn
→ y = v1 y1 + v2 y2 + . . . + vn yn
y 0 = v1 y10 + v2 y20 + v10 y1 + v20 y2
→
y 0 = v1 y10 + . . . + vn yn0 + v10 y1 + . . . + vn0 yn
Variation of Parameters
Demand
v10 y1 + v20 y2 = 0
→
Demand
v10 y1 + . . . + vn0 yn = 0.
Take
y 00 = v1 y100 + . . . + vn yn00 + v10 y10 + . . . + vn0 yn0
and demand
v10 y10 + . . . + vn0 yn0 = 0.
......
Variation of Parameters
Take
(n−1)
y (n−1) = v1 y1
(n−2)
+ . . . + vn yn(n−1) + v10 y1
+ . . . + vn0 yn(n−2)
and demand
(n−2)
v10 y1
+ . . . + vn0 yn(n−2) = 0.
Variation of Parameters
Substituting y , y 0 , y 00 ,
rearranging
→ Substituting y , y 0 , . . . , y (n) ,
→ rearranging
and using L(y1 ) = L(y2 ) = 0 → and using L(y1 ) = . . . = L(yn ) = 0
get v10 y10 + v20 y20 = r (x) →
get
(n−1)
v10 y1
+ . . . + vn0 yn(n−1) = r (x).
Variation of Parameters
Thus,
takes the shape

y1
 y10


·
(n−1)
y1
y1
y10
y2
y20
·
(n−1)
y2
y2
y20
v10
v20
=
0
r (x)
 0
·
yn
v1
0
 v20
·
yn 

 ·
·
·
(n−1)
vn0
· yn


0



 =  0 .

 · 
r (x)
Use Cramer’s rule to solve for
v10 , v20 , . . . , vn0 ,
and thus get
v1 , v2 , . . . , vn ,
and form
y = v1 y1 + v2 y2 + . . . + vn yn .

Variation of Parameters
Let w (x) = det W (y1 , y2 , . . . , yn )(x) be the determinant of the
Wronskian matrix of the functions y1 , y2 , . . . , yn . By Cramer’s rule,
det
(n−1)
vk0 (x) =
···
···
..
.
y1 (x)
y10 (x)
..
.
y1
(x) · · ·
0
0
..
.
···
···
..
.
r (x) · · ·
w (x)
yn (x)
yn0 (x)
..
.
(n−1)
yn
(x)
=:
for each k = 1, 2, . . . , n. Hence,
Z
x
vk (x) =
x0
wk (t)
dt
w (t)
and therefore,
Z
x
y (x) = y1 (x)
x0
wk (t)
dt + . . . + yn (x)
w (t)
Z
x
x0
wk (t)
dt.
w (t)
wk (x)
w (x)
Variation of Parameters
Example: Solve
y (3) − y (2) − y (1) + y = r (x).
Characteristic polynomial for the homogeneous equation is
x 3 − x 2 − x + 1 = (x − 1)2 (x + 1).
Hence, a basis of solutions is
{e x , xe x , e −x }.
We need to calculate W (t). Use Abel’s formula:
W (x) = W (0)e −
Rx
0
p1 (t)dt
= W (0) · e x .
Variation of Parameters
Now,
W (x) =
ex
ex
ex
xe x
e + xe x
2e x + xe x
x
e −x
−e −x
e −x
Thus,
W (0) =
1
1
1
0
1
2
1
−1
1
Hence,
W (x) = 4e x .
= 4.
.
Variation of Parameters
Let
W1 (x) =
0
0
r (x)
xe x
e + xe x
2e x + xe x
x
e −x
−e −x
e −x
= −r (x)(2x + 1).
Similarly,
W2 (x) = 2r (x), W3 (x) = r (x)e 2x .
Therefore,
y (x) = e x
Z
0
x
−r (t)(2t + 1)
dt + xe x
4e t
Z
0
x
2r (t)
dt + e −x
4e t
Z
0
x
r (t)e 2t
dt.
4e t
Laplace Transforms
Let f : (0, ∞) → R be a function. The Laplace transform L(f ) of f is
the function defined by
Z ∞
Z a
−st
L(f )(s) =
e f (t)dt = lim
e −st f (t)dt,
0
a→∞
0
for all values of s for which the integral exists. Sometimes we denote
F (s) = L(f )(s).
The integral above may not converge for every s.
We may impose suitable restrictions on f later under which the intergral
exists. What is the meaning of the improper integral?
The Improper Integral of the first kind
Definition
Let a, b be two real numbers such that 0 < a < b < ∞. A function
f : [a, b] → R is said to be piecewise continuous on [a, b] if there is a
partition
a = t0 < t1 < . . . < tn−1 < tn = b
such that
(i) f is continuous on (ti−1 , ti ) for i = 1, 2, . . . , n.
(ii) lim+ f (t) and lim− f (t) both exist for i = 1, 2, . . . , n − 1
t→ti
t→ti
and lim+ f (t) and lim f (t) both exist.
t→t0
t→tn−
A piecewise continuous function on an interval [a, b] is continuous except
possibly for finitely many jump discontinuties.