MA-108 Differential Equations I

Ronnie Sebastian

Department of Mathematics
Indian Institute of Technology Bombay
Powai, Mumbai - 76

26th March, 2018

D2 - Lecture 7

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Constant Coefficient Differential Operators

Definition
Consider

L = a0 Dn + a1 Dn−1 + . . . + an−1 D + an ,

where a0 , a1 , . . . , an ∈ R. Thus for any function f ∈ C n (I),

L(f ) = a0 f (n) + a1 f (n−1) + . . . + an−1 f 0 + an f.

Such an L is called a constant coefficient differential operator.

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Constant Coefficient Differential Operators

Example
D2 − 5D + 6 = (D − 3)(D − 2) as a linear transformation from
C 2 (I) → C(I), i.e. for any y ∈ C 2 (I),

(D − 3)(D − 2)y = (D − 3)(y 0 − 2y)

= (y 00 − 5y 0 + 6y)
(D − 2)(D − 3)y = (D − 2)(y 0 − 3y)
= y 00 − 5y 0 + 6y

Hence (D − 2)(D − 3) = (D − 3)(D − 2).

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Constant Coefficient Differential Operators

Example
Check that (D + 1)(D2 + D + 1) = (D2 + D + 1)(D + 1) as a
function from C 3 (I) → R. i.e. for each y ∈ C 3 (I),

(D + 1)(D2 + D + 1)y = (D2 + D + 1)(D + 1)y

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Constant Coefficient Differential Operators

• The main point is that

Dr ◦ Ds = Ds ◦ Dr = Dr+s

Hence if L = ni=0 an−i Di and M = m i

P P
i=0 bm−i D are differential
operators with constant coefficient, then

L(M (f )) = M (L(f )), ∀f ∈ C m+n =⇒ LM = M L

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Constant Coefficient Differential Operators
Note that it is important that L and M are constant coefficient
equations.
Example
If L = D + x and M = D + 1, then

LM (f ) = (D + x)(D + 1)(f )
= (D + x)(f 0 + f )
= (f 00 + f 0 ) + x(f 0 + f )

M L(f ) = (D + 1)(D + x)(f )

= (D + 1)(f 0 + xf )
= (f 00 + f + xf 0 ) + (f 0 + xf )
= LM (f ) + f 6= LM (f )

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Constant Coefficient Differential Operators

Example
Solve y 000 − 7y 0 − 6y = 0.
We notice that this is same as the solution space of

Ly = (D3 − 7D − 6I)y = 0

But L = (D3 − 7D − 6I) = (D − 3)(D + 1)(D + 2)

Now, note that if y is such that (D + 2)y = 0, then Ly = 0.

L = (D − 3)(D + 1)(D + 2)
= (D + 2)(D + 1)(D − 3) = (D − 3)(D + 2)(D + 1)
If (D + 1)y = 0 or (D − 3)y = 0, then Ly = 0.
This gives us that f (x) = e−x , g(x) = e−2x and h(x) = e3x are all
solutions to Ly = 0.

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Constant Coefficient Differential Operators

By dimension theorem, if they are linearly independent, then they

give basis for all solutions.
We can check that these are linearly independent by checking that
the Wronskian is nonzero at any one point of our choice .
 
1 1 1

Then W (f, g, h; 0) = −1 −2 3 = −30 + 12 − 2 = −20 6= 0.
1 4 9
=⇒ f , g and h are linearly independent solutions of the ODE
y 000 − 7y 0 − 6y = 0 on R.

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Constant Coefficient Differential Operators
Consider

L = a0 Dn + a1 Dn−1 + . . . + an−1 D + an , ai ∈ R

We define the characteristic polynomial to be

PL (x) = a0 xn + a1 xn−1 + . . . + an−1 x + an .

Theorem
Let L and M be two constant coefficient linear differential
operators. Then,
1 L = M if and only if PL (x) = PM (x).
2 PL+M (x) = PL (x) + PM (x).
3 PLM (x) = PL (x) · PM (x).

All the above statements are obvious.

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Constant Coefficient Differential Operators

Definition
Given a constant coefficient differential operator L we define
Ker(L) to be the space of functions y such that Ly = 0.

By the dimension theorem, this is a vector space of dimension n,

where n is the order of the differential operator.
We shall next investigate how to find Ker(L) for any L.
Let us begin with the simplest case, when

PL (x) = a0 xn + a1 xn−1 + . . . + an ai ∈ R

Assume that the roots of PL (x) are distinct and are all real. Then

PL (x) = a0 (x − r1 )(x − r2 ) . . . (x − rn )

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Constant Coefficient Differential Operators

Let us assume that

r1 < r2 < · · · < rn
Since D − ri commute with each other, and since eri x is a solution
to (D − ri )y = 0, we get that

{er1 x , er2 x , . . . , ern x } ⊂ Ker(L)

We claim that these functions are linearly independent. If not,

suppose there is a relation of the type

λ1 er1 x + λ2 er2 x + · · · + λn ern x = 0 λi ∈ R

Not all the λi are 0. Let j be the largest such that λj 6= 0.

Multiply the equation with −1 if necessary and assume that
λj > 0. Also multiply the equation with e−rj x to get

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Constant Coefficient Differential Operators

λ1 e(r1 −rj )x + λ2 e(r2 −rj )x + · · · + λj = 0

Since we have
r1 − rj < r2 − rj < . . . < 0
and λj > 0, taking x >> 0 we get a contradiction.
Alternatively, we could have used Abel’s theorem, but then we
would have to show that the determinant of the matrix
 
1 1 ··· 1
 r1
 r2 ··· rn 
 r2 r 2 · · · r 2 
 1 2 n  6= 0
 .. .. .. .. 
 . . . . 
r1n−1 r2n−1 · · · rnn−1

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Constant Coefficient Differential Operators

Example
Solve y (3) − 7y 0 + 6y = 0.

L = D3 − 7D + 6,

PL (x) = x3 − 7x + 6 = (x − 1)(x − 2)(x + 3).

Therefore,
L = (D − 1)(D − 2)(D + 3).
Thus, ex , e2x , e−3x ∈ Ker(L). Hence, {ex , e2x , e−3x } is a basis of
Ker L.
Thus, the general solution is of the form

c1 ex + c2 e2x + c3 e−3x

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Constant Coefficient Differential Operators

Q. What if PL (x) has some repeated real roots?

Let us begin with the following result.
Proposition
For any real number r, the functions

u1 (x) = erx , u2 (x) = xerx , . . . , um (x) = xm−1 erx

are linearly independent and

u1 (x), . . . , um (x) ∈ Ker((D − r)m ).

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Constant Coefficient Differential Operators
Proof. That these functions are linearly independent is obvious,
since {1, x, x2 , . . . , xm } is linearly independent (erx is non-zero).
We need to show that these functions are in Ker (D − r)m .
When m = 1, we need to show
u1 (x) = erx ∈ Ker((D − r)),
which is true since
(D − r)(erx ) = rerx − rerx = 0.
Suppose m = 2. Since u1 is in Ker of (D − r), it is also in Ker of
(D − r)2 .
What about u2 = xerx ?
(D − r)2 (xerx ) = (D − r)(D − r)(xerx )
= (D − r)(xrerx + erx − rxerx )
= (D − r)(erx ) = 0.
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Constant Coefficient Differential Operators
Let us use induction to prove the general case. Assume

u1 , u2 , . . . , um−1 ∈ Ker((D − r)m−1 ),

and we need to show that

u1 , u2 , . . . , um ∈ Ker((D − r)m ).

Clearly

u1 , u2 , . . . , um−1 ∈ Ker((D − r)m−1 ) ⊂ Ker((D − r)m ).

To show that um is also in Ker((D − r)m ), compute

(D − r)m (xm−1 erx )

= (D − r)m−1 (D − r)(xm−1 erx )

= (D − r)m−1 (xm−1 rerx + (m − 1)xm−2 erx − rxm−1 erx )
= (D − r)m−1 ((m − 1)xm−2 erx ) = 0.
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Constant Coefficient Differential Operators
Therefore, a basis for solution space of (D − r)m is

erx , xerx , . . . , xm−1 erx

Thus, if

PL (x) = (x − r1 )e1 (x − r2 )e2 . . . (x − r` )e` ,

`
X
where ei = n, then a basis of Ker L is given by
i=1

{er1 x , . . . , xe1 −1 er1 x }∪{er2 x , . . . , xe2 −1 er2 x }∪. . .∪{er` x , . . . , xe` −1 er` x }

The above functions are linearly independent and since dim Ker
L = n, these form a basis.
Ex: Check that the above functions are linearly independent by
evaluating the Wronskian at 0.
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Constant Coefficient Differential Operators

Example
Find the general solution of the ODE:

L(y) = (D3 − D2 − 8D − 12)(y) = 0.

We have

PL (x) = x3 − x2 − 8x − 12 = (x − 2)2 (x + 3),

and therefore,
L = (D − 2)2 (D + 3).
Thus the general solution is

y = c1 e2x + c2 xe2x + c3 e−3x ,

where c1 , c2 , c3 ∈ R.

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Constant Coefficient Differential Operators
Example
Find the general solution of the ODE:

L(y) = (D6 + 2D5 − 2D3 − D2 )(y) = 0.

L = D2 (D − 1)(D + 1)3
Ker(D2 ) = {1, x}
Ker(D − 1) = {ex }
Ker(D + 1)3 = {e−x , xe−x , x2 e−x }

Thus, the general solution is

c1 + c2 x + c3 ex + c4 e−x + c5 xe−x + c6 x2 e−x ,

with ci ∈ R.
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Constant Coefficient Differential Operators: complex roots

Assume PL (x) has some complex roots. In the 2nd order case, if
m1 = a + ıb, m2 = a − ıb, then y1 = eax cos bx and y2 = eax sin bx
were the basis for Ker(L).
If PL (x) has a complex root a + ıb, then it also has a − ıb as a
root. Thus,

(x − (a + ıb))(x − (a − ıb)) = (x − a)2 + b2

is a factor of PL (x).
Null space of (D − a)2 + b2 has a basis

{eax cos bx, eax sin bx} ⊂ Ker(L)

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Constant Coefficient Differential Operators: complex roots

If a ± ıb is a root of PL (x) of multiplicity m, then

((D − a)2 + b2 )m is a factor of PL (x).
Can we find the null space of ((D − a)2 + b2 )m ?
Ex. Check that

eax cos bx, xeax cos bx, . . . , xm−1 eax cos bx,
eax sin bx, xeax sin bx, . . . , xm−1 eax sin bx.

are a basis for null space of ((D − a)2 + b2 )m .

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Constant Coefficient Differential Operators: complex roots

Example
Find the general solution of

y (5) − 9y (4) + 34y (3) − 66y (2) + 65y 0 − 25y = 0.

The characteristic polynomial is

(x − 1)(x2 − 4x + 5)2 .

The roots are

1, 2 ± ı, 2 ± ı.
Hence, the general solution is

y = c1 ex + e2x [c2 cos x + c3 sin x + c4 x cos x + c5 x sin x],

where ci ∈ R.

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Examples

Example
Ex: Find the fundamental set of solutions to

D3 (D − 2I)2 (D2 + 4I)2 y = 0

The fundamental set will be given by

{1, x, x2 , e2x , xe2x , cos 2x, sin 2x, x cos 2x, x sin 2x}

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Annihilator or Undetermined Coefficient Method
The Annihilator method or method of undetermined
coefficients helps us in finding a particular solution of a
non-homogeneous equation.

Example
Find a particular solution of

y (4) − 16y = x4 + x + 1 = r(x).

Here, L = D4 − 16,
and let us take A = D5 . Then Ar(x) = 0.
We say A annihilates or kills r(x).
A solution y of L(y) = r(x) is also a solution of

D5 (D4 − 16) = 0.

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Annihilator or Undetermined Coefficient Method

Example (continued ...)

AL = D5 (D4 − 16) has characteristic equation

x5 (x4 − 16) = x5 (x − 2)(x + 2)(x2 + 4).

A general solution of (AL)(y) = 0 is of the form

c1 +c2 x+c3 x2 +c4 x3 +c5 x4 +c6 e2x +c7 e−2x +c8 cos 2x+c9 sin 2x.

Here c6 e2x + c7 e−2x + c8 cos 2x + c9 sin 2x is a solution of the

homogeneous part (D4 − 16)y = 0.
We want a particular solution yp for y (4) − 16y = x4 + x + 1
This implies that we can take yp = c1 + c2 x + c3 x2 + c4 x3 + c5 x4 ,
since all the other terms are solutions to the corresponding
homogenous ODE.

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Annihilator or Undetermined Coefficient Method

Example
To find ci ’s in yp = c1 + c2 x + c3 x2 + c4 x3 + c5 x4 , solve
(4)
yp − 16yp = x4 + x + 1.
Then 24c5 − 16(c1 + c2 x + c3 x2 + c4 x3 + c5 x4 ) = x4 + x + 1.
Equating the coefficients, we get

24c5 − 16c1 = 1
−16c2 = 1
−16c3 = 0
−16c 4 = 0
−16c5 = 1

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Annihilator or Undetermined Coefficient Method

Example (continued ...)

This gives

c3 = c4 = 0,
c5 = c2 = −1/16,
c1 = −5/32.

5 1 1
Therefore yp = − − x − x4 .
32 16 16

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Annihilator or Undetermined Coefficient Method

Example
Solve y (4) − 4y 00 = ex + x2 .
Let L = D4 − 4D2 = D2 (D − 2)(D + 2).
Let z(x) and w(x) be such that Lz = ex and Lw = x2 . Then
L(z + w) = ex + x2 .
Let us first solve Lz = ex . We know that ex is a solution of
M y = (D − I)y = 0.
Now, M Lz = (D − I)D2 (D − 2)(D + 2)z = 0.
Clearly z satisfies this equation. Hence z will be of the form
z = c1 + c2 x + c3 e2x + c4 e−2x + c5 ex .

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Annihilator or Undetermined Coefficient Method

Example (continued ...)

But {1, x, e2x , e−2x } are all solution to Ly = 0 and therefore,
z = c5 ex for some c5 ∈ R.
Plugging z = c5 ex into the equation

y (4) − 4y 00 = ex

we have
c5 − 4c5 = 1 =⇒ c5 = −1/3
Thus
z = (−1/3)ex .

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Annihilator or Undetermined Coefficient Method

Example (continued ...)

Next we have to solve Lw = x2 , where L = (D4 − 4D2 ).
Let N = D3 , then x2 is a solution to D3 y = 0.
Then N Lw = D3 D2 (D − 2)(D + 2)w = D5 (D − 2)(D + 2)w = 0.

Clearly w = c1 + c2 x + c3 x2 + c4 x3 + c5 x4 + c6 e2x + c7 e−2x .

But c1 + c2 x + +c6 e2x + c7 e−2x are solutions to Ly = 0.
Therefore, w = c3 x2 + c4 x3 + c5 x4 .
Substituting in Lw = (D4 − D2 )w = x2 , we get

24c5 − 2c3 + 6c4 x + 12c5 x2 = x2

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Annihilator or Undetermined Coefficient Method

Example (continued ...)

24c5 − 2c3 = 0,
c4 = 0,
c5 = 1/12
c3 = 1

1 4
Therefore, w = x2 + x .
12
Hence a particular solution to Ly = ex + x2 is given by
1 1
yp = z + w = − ex + x2 + x4
3 12

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Summary: Anhilator Method

Given a linear differential operator L with constant

coefficients, we want to solve Ly = r(x).
We find a particular solution as follows.
We first find linear a differential operators M which have the
property that M (r(x)) = 0.
Find a basis for the solution space of Ly = 0. Extend this to a
basis for the solution space of M Ly = 0.
Pick those elements in the basis which are not solutions to
Ly = 0.
Set yp to be a linear combination of these particular basis
elements and solve Lyp = r(x) for the constants.
A general solution to Ly = r is given by yp + z, where z is a
general solution to Ly = 0.

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Variation of Parameters
The variation of parameters method generalizes to nth order linear
ODE y (n) + p1 (x)y (n−1) + . . . + pn (x)y = r(x) where pi ’s and r
are continuous on I.
Let y1 , . . . , yn be a basis of solutions of homogeneous part.
Assume the particular solution yp is given by

yp = v1 (x)y1 + v2 (x)y2 + . . . + vn (x)yn

Assume
v10 y1 + . . . + vn0 yn =0 (1)
v10 y10 + . . . + v10 yn0 =0 (2)
.. ..
. .
(n−2) (n−2)
v10 y1 + . . . + vn0 yn = 0 (n − 1)
(n)
Compute yp0 , . . . , yp and put in the ODE, we get
(n−1)
v10 y1 + . . . + vn0 yn(n−1) = r(x) (n)
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Variation of Parameters
Thus,
 0
y1 y2 · yn
   
v1 0
 y10 y20 · yn0   v20   0 
=
  · .
  
 · · · ·  ·
(n−1) (n−1) (n−1)
y1 y2 · yn vn0 r(x)
Use Cramer’s rule to solve for v10 , v20 , . . . , vn0 , and thus get
v1 , v2 , . . . , vn , and form
y = v 1 y1 + v 2 y2 + . . . + v n yn
here vi0 is given by
· yi−1 ·
 
 y1 0 yi+1 yn 
 0 0 0 yn0 

 y1
 · yi−1 0 yi+1 ·
 · · · · · · · 
 (n−1) (n−1) (n−1) (n−1) 
y · yi−1 r(x) yi+1 · yn
vi = 1
0
W (y1 , . . . , yn ; x)
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Variation of Parameters

Example
Solve y (3) − y (2) − y (1) + y = r(x).
Here L = D3 − D2 − D + 1 = (D − 1)2 (D + 1).
Hence, a basis of solutions is {ex , xex , e−x }.
We need to calculate W (x). Use Abel’s formula:
Rx
W (x) = W (0) e− 0 p1 (t)dt
= W (0) · ex .

−x 
 x x

 e xe e
W (x) =  ex ex + xex −e−x  .
 
 ex 2ex + xex e−x 
 
 1 0 1 

=⇒ W (0) =  1 1 −1  = 4.
 1 2 1 
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Variation of Parameters
Example (continued ...)
Hence,
W (x) = 4ex .

xex e−x
 
 0 
ex + xex −e−x
 
W1 (x) =  0  = −r(x)(2x + 1).

 r(x) 2ex + xex e−x 

Similarly,

W2 (x) = 2r(x) W3 (x) = r(x)e2x

Therefore, a particular solution yp is given by yp =

x x x
−r(t)(2t + 1) r(t)e2t
Z Z Z
2r(t)
ex dt + xex dt + e−x dt.
0 4et 0 4et 0 4et
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