PART V : SECOND ORDER LINEAR ORDINARY DIFFERENTIAL EQUATIONS
5.1
INTRODUCTION
Second order linear O D E ’s have the general form :
d2 y
dy
a(x) 2 + b(x)
+ c(x) y = f (x) ,
dx
dx
where the coefficients a(x) , b(x) , c(x) and the inhomogeneous or ” forcing ”
term f (x)
5.2
are all given functions of
SPECIAL CASE
:
(53)
x.
c(x) = 0
d2 y
dy
ie
a(x) 2 + b(x)
= f (x) ,
(54)
dx
dx
As this second order ODE does not contain y but only its derivatives , it is a first order
equation in the function
dy
.
dx
dy
= p(x) , say , then
dx
If we put
d2 y
dp
=
dx2
dx
and (54) becomes :
dp
a(x)
+ b(x) p = f (x) ,
dx
We can now solve (55) for p(x) , by using an integrating factor , and then integrate
p(x)
to obtain a solution of (54). Note that this is the general solution of
it contains two arbitrary constants, from integrating (55)
Example 34
Find the solution of
which satisfies the conditions
(i)
x
(ii)
dy
= 0 at x = 0 ;
dx
y=1 ,
(54) as
and then integrating p(x) .
dy
d2 y
+
=x
2
dx
dx
y=1 ,
(55)
dy
= 0 at x = 1.
dx
5.3
:
EQUATIONS WITH CONSTANT COEFFICIENTS
We here consider
a
d2 y
dy
+b
+ c y = f (x) ,
dx2
dx
(56)
where a , b , c are constants.
We first take
f (x) = 0
, so
a
dy
d2 y
+b
+ cy = 0 ,
2
dx
dx
This equation is said to be homogeneous in
λ is a constant , leaves
Equation
5.31
:
y
d2 y
dx2
ie replacing
y
by
λy
, where
(57) unchanged.
(56) is said to be inhomogeneous in
y
THE GENERAL SOLUTION OF
(57)
As the coefficients are constant , a solution of
and
(57)
or just inhomogeneous.
(57) must be such that
all have the same dependence on x
y(x)
,
dy
dx
. The only function whose derivative
is proportional to itself is the exponential function . We therefore look for a solution of
(57) in the form
where
A
Hence ,
and
λ
y(x) = A eλ x ,
are constants to be determined .
dy
= A λ eλ x
dx
and
d2 y
= A λ 2 eλ x .
dx2
Substitution in (57) gives :
A eλ x ( a λ2 + b λ + c ) ≡ 0.
so
∴
a λ2 + b λ + c = 0 .
This is called the auxiliary equation of the O D E (57).
(58)
In general , this quadratic equation has two distinct roots ,
Note that
λ 1 + λ2 = −
b
a
It follows from (58) that both
and
e λ1 x
λ 1 λ2 =
and
e λ2 x
λ1
and λ2
say , where
c
.
a
are solutions of
(57).
What is the general solution of
(I)
repeated
(I)
REPEATED ROOTS OF THE AUXILIARY EQUATION
Let
2
λ =
or
(II)
(57) if the roots of the auxiliary equation are either
λ = λ 1 = λ2
c
.
a
complex ?
be the repeated root of
It follows that eλ x
(58) . So,
is a solution of
2λ = −
b
a
and
(57) but what is the other ?
To find a second solution , we make use of the solution that we know :
We look for a solution of
(57) in the form :
y(x) = eλ x u(x)
where
u(x)
is to be determined .
du
dy
= λ eλ x u + eλ x
dx
dx
From (60) :
du
d2 u
d2 y
2
= λ eλ x u + 2 λ eλ x
+ eλ x 2 .
2
dx
dx
dx
and
Substitution in
ie
(60)
(57) gives :
e
λx
On substituting for λ ,
d2 u
du
2
a 2 + ( 2λa +b)
+ (aλ + bλ + c)u = 0
dx
dx
this reduces to
d2 u
= 0.
dx2
y(x) = ( c1 + c2 x ) eλ x
and so
− the general solution of
(57) when the roots of the auxiliary equation are repeated.
Note that the second solution has the form
Examples 36 :
(i)
(II)
x ( First Solution ) .
Find the general solution of
y 00 + 2 y 0 + y = 0
,
(ii)
9 y 00 − 12 y 0 + 4 y = 0.
COMPLEX ROOTS OF THE AUXILIARY EQUATION
λ1 = λ∗2 = α + i β , say , where
Here
The general solution of
(57) is formally
α = −
∴
where k1
and k2
and β =
( 4 a c − b2 )1/2
.
2a
y(x) = c1 e( α + i β ) x + c2 e( α − i β ) x
y(x) = eα x ( c1 e i β
ie
b
2a
x
+ c2 e − i β x )
y(x) = eα x ( k1 cos βx + k2 sin βx )
(61)
are real constants.
NOTE :
(i)
In (61) ,
α = Re ( λ ) and β = Im ( λ ) , where λ is a complex root of (58).
(ii)
ie
If we take
k1 = A sin γ
replace the constants k1
and
and k2
oscillations with amplitude
Examples 37 :
5.32
:
Let Y (x)
by the constants A and γ
,
y(x) = A eα x sin( βx + γ ) ,
(61) reduces to
-
k2 = A cos γ ,
A eα x
, period
2π
β
and phase γ .
Find the general solution of
THE GENERAL SOLUTION OF
denote any solution of
(56).
(56)
ie
a
d2 Y
dY
+b
+ c Y = f (x) .
2
dx
dx
Such a solution is called a Particular Integral of (56)
-
it contains no arbitrary
constants.
y(x) = y(x) + Y (x)
Now consider
(i)
(62) is a solution of
is a solution of
(ii)
(56) as
y(x)
is a solution of
(62)
(57) and
Y (x)
(56) ;
(62) contains two arbitrary constants , within
y(x)
, and so is the general
solution of a second order O D E ;
(iii)
of
(56) is a second order O D E so it follows that
(56) .
(62) is the general solution
So, the general solution of
(56)
ie
GS =CF + PI
We can always find the general solution of equation (57) . How can we find a P I of
(56) for a given f (x) ?
ADDENDUM III
To find the general solution of the differential equation :
a
d2 y
dy
+b
+ c y = a f (x) ,
dx2
dx
where a , b , c are constants and f (x)
is a given function of
(1)
x , we first consider
the corresponding homogeneous equation :
a
dy
d2 y
+b
+ cy = 0 ,
2
dx
dx
If the auxiliary equation of (2) has roots λ1
c
= λ 1 λ2
a
and λ2
(2)
then
b
= − λ 1 − λ2
a
and
so (2) may be written as :
d2 y
dy
− ( λ 1 + λ2 )
+ λ 1 λ2 y = 0 ,
2
dx
dx
Differentiation may be considered as an operation carried out on y(x)
in
dy
.
dx
(3)
which results
We can therefore consider a differential equation as a differential operator
acting on y(x).
So , (3) may be written as
d2
d
− ( λ 1 + λ2 )
+ λ 1 λ2
dx2
dx
y=0 ,
and the differential operator may be factorised to give
d
− λ1
dx
If we introduce a new function
then (4)
y=0 ,
z(x) such that
d
− λ2
dx
(4)
y = z.
can be replaced by two first order O D E ’s :
dz
− λ1 z = 0
dx
and
We can find the general solution of (5i)
(5i)
and (5ii)
by separation of variables.
dy
− λ 2 y = k 1 e λ1 x .
dx
(5ii) therefore becomes
This equation has an integrating factor
On integrating and dividing by
and k3
dy
− λ2 y = z .
dx
z(x) = k1 eλ1 x .
We find that
where k2
d
− λ2
dx
e − λ2 x
e − λ2 x
and so may be written as
, we obtain the general solution of (3) :
are arbitrary constants , provided that λ1 6= λ2 .
SPECIAL CASE :
λ1 = λ2 , = λ , say .
Now, having applied the integrating factor to (5ii) , we find that
is the general solution of
(3) when the auxiliary equation has repeated roots.
Note : When the roots of the auxiliary equation are repeated , the second solution has
the form
x ( First solution ).
THE GENERAL SOLUTION OF THE INHOMOGENEOUS EQUATION
With
f (x) 6= 0 , (4) becomes
d
− λ1
dx
dz
− λ1 z = f (x)
dx
and
d
− λ2
dx
y = f (x) ,
(6)
and equations (5) become
Using the integrating factor in
Hence ,
dy
− λ2 y = z .
dx
(7i) , we find that
z(x) = e λ1 x
Z
f (x) e− λ1 x dx + k4 e λ1 x
(7i) and (7ii)
dy
− λ2 y = e λ1 x F (x) + k4 e λ1 x .
dx
∴ From (7ii) :
Using an integrating factor again :
provided
λ1 6= λ2
d
( e− λ2 x y ) = e( λ1 −λ2 ) x F (x) + k4 e (λ1 −λ2 ) x
dx
.
So, the general solution of the inhomogeneous equation (6) , for
where the first term is a P I of (6) as it contains f (x)
λ1 6= λ2
, is
( and no arbitrary constants )
and the last two terms , which contain two arbitrary constants , form the general solution
of the corresponding homogeneous equation ie the C F of (6) .
VERY SPECIAL CASE : λ1 = λ2 , = λ , say , and f (x) = eλx
Integrating (7i) , we now get :
and so
z(x) = x e λ x + k6 e λ x
∴ Using the integrating factor in (7ii) , we find that :
ie
F (x) = x .
∴
The general solution of the equation :
is
ie
y(x) =
a P I of
1 2 λx
1
+ k6 x eλ x + k7 eλ x = ( x2 + k6 x + k7 ) eλ x ,
x e
2
2
2
y 00 − 2 λ y 0 + λ y = eλx
Note : it follows that , if
λ ,
then a P I is
f (x) = xn eλx
is
1 2 λx
x e .
2
and the auxiliary equation has repeated roots
xn+2 eλx
.
(n + 1)(n + 2)
How can we find P I ’s other than from the general expression given above ?
We use
5.33
:
THE METHOD OF UNDETERMINED COEFFICIENTS.
This method involves guessing , from
f (x)
, the form of a particular integral ,
containing a number of unspecified constants ( undetermined coefficients ). This
guess is then substituted in the O D E and the constants chosen to make it satisfy
the equation ie to make it a Particular Integral .
The method is applied as follows :
(A)
f (x) = a polynomial of degree n
:
We look for a P I in the form of a polynomial of degree n with unknown coefficients.
Example 38 :
Find the general solution of the differential equation
f (x) = ( a polynomial of degree n ) eα x
(B)
We look for a P I in the form
h(x) eα x
, where
:
h (x)
is a polynomial of
degree n with unknown coefficients.
Example 39 :
(C)
f (x) = a cos x + b sin x
Example 40 :
NOTE :
is
Find a P I of the equation :
:
we look for a P I in the form
Find a P I of the equation :
A P I of
y 00 + y 0 − 2y = x2 ex .
A cos x + B sin x.
y 00 + y 0 − 2y = 3 cos x
y 00 + y 0 − 2y = 1 + 2x + x2 + x2 ex + 3 cos x
( P I )38 + ( P I )40 + (P I )12 .
5.34
THE METHOD OF UNDETERMINED COEFFICIENTS : SPECIAL CASES :
(D)
If λ = α , ie
eαx occurs in the C F , and
αx
(i)
f (x) = a e
, we look for a P I in the form
(ii)
f (x) = a x eαx ,
(iii)
f (x) = a x2 eαx , we look for a P I in the form
we look for a P I in the form
A x eαx ;
( A1 x + A2 x2 ) eαx
;
( A1 x + A2 x2 + A3 x3 ) eαx ...
(E)
If λ = α
is a repeated root , ie
(i) f (x) = a eαx
both
eαx
and x eαx
B x2 eαx ;
, we look for a P I in the form
(ii) f (x) = a x eαx , we look for a P I in the form
If λ = ± i β
(F )
If
B x3 eαx
λ = α ± iβ
, we look for a P I in the form
(i)
(iii)
...
x ( A sin βx + B cos βx ).
, ie eα x sin β x and eα x cos β x both occur in the C F , and
f (x) = a eα x sin βx + b eα x cos βx , we look for a P I in the form
Examples 41 :
,
, ie sin β x and cos β x both occur in the C F , and
f (x) = a sin βx + b cos βx
(G)
occur in the C F , and
x eα x ( A sin βx + B cos βx ).
Find P I ’s for the following equations :
y 00 − 3 y 0 + 2 y = 5 e2x ,
y 00 + 4 y = 5 sin 2 x + 3 cos 2x,
(ii)
(iv)
y 00 + 2 y 0 + y =
x2 e−x ,
y 00 + 4 y 0 + 5 y = e−2 x sin x .