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58
Linear Ordinary Differential
Equations of Second and Higher Order
2.1. DEFINITIONS
A linear differential equation of n th order is that in which the dependent variable and its derivatives occur
only in the first degree and are not multiplied together. Thus, the general linear differential equation of the n th
order is of the form
1 2
P 1 1 P 2 2 P 1 P
n n n
n n n n^ n
d y d y d y dy y dx dx dx dx
- (^) - + (^) - + L + (^) - + = X, where P 1 , P 2 , ……, P n – 1 , P n and
X are functions of x only.
A linear differential equation with constant coefficients is of the form
1 2
1 1 2 2 1 X
n n n
n n n n^ n
d y d y d y dy a a a a y dx dx dx dx
- (^) - + (^) - + L + (^) - + = …(1)
where a 1 , a 2 , …… , a (^) n – 1, a (^) n are constants and X is either a constant or a function of x only.
First of all we discuss solution of linear differential equation with constant coefficients.
2.2. THE OPERATOR D
The part
d
dx
of the symbol
dy
dx
may be regarded as an operator such that when it operates on y , the result
is the derivative of y.
Similarly,
2 3
2 3
n
n
d d d
dx dx dx
L may be regarded as operators.
For brevity, we write
2 2 2
D, D , , D
n n n
d d d
dx (^) dx dx
º º L º
Thus, the symbol D is a differential operator or simply an operator.
Written in symbolic form, equation (1) becomes (D
n
n – 1
n – 2
- …… + a (^) n – 1 D + a (^) n ) y = X
or f (D) y = X
where f (D) =
1 2 D 1 D 1 D ....... 1 D
n n n a a an an
i.e., f (D) is a polynomial in D.
The operator D can be treated as an algebraic quantity.
i.e., D ( u + v ) = D u + D v
D (l u ) = l D u
D
p D
q u = D
p + q u
D
p D
q u = D
q D
p u
The polynomial f (D) can be factorised by ordinary rules of algebra and the factors may be written in any
order.
LINEAR ORDINARY DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER 59
2.3. THEOREMS
Theorem 1. If y = y 1 , y = y 2 , ……… , y = yn are n linearly independent solutions of the differential
equation
( D
_n
_n – 1
_n – 2
then u = c 1 y 1 + c 2 y 2 + ……… + cn yn is also its solution, where c 1 , c 2 , ……… , c (^) n are arbitrary constants.
Proof. Since y = y 1 , y = y 2 , ……, y = y (^) n are solution of equation ( i ).
1 2 1 1 1 2 1 1 1 2 2 1 2 2 2 2
1 2 1 2
D D D 0
D D D 0
D D D 0
n n n n n n n n
n n n n n n n n
y a y a y a y
y a y a y a y
y a y a y a y
\ + + + + = ü ï
L
L
L L L L L
L L L L L
L
…( ii )
Now, D
n u + a 1 D
n – 1 u + a 2 D
n – 2 u + …… + a (^) n u
= D
n ( c 1 y 1 + c 2 y 2 + …… + c (^) n y (^) n ) + a 1 D
n – 1 ( c 1 y 1 + c 2 y 2 + …… + cn yn )
n – 2 ( c 1 y 1 + c 2 y 2 + …… + c (^) n yn ) + … … … … + a (^) n ( c 1 y 1 + c 2 y 2 + … + cn yn )
= c 1 (D
n y 1 + a 1 D
n – 1 y 1 + a 2 D
n – 2 y 1 + … + an y 1 ) + c 2 (D
n y 2 + a 1 D
n – 1 y 2 + a 2 D
n – 2 yn + … + an y 2 )
n y (^) n + a 1 D
n – 1 y (^) n + a 2 D
n – 2 yn + … + an yn )
= c 1 (0) + c 2 (0) + … + cn (0) [ Q of ( ii )]
= 0
which shows that u = c 1 y 1 + c 2 y 2 + …… + cn yn is also the solution of equation ( i ).
Since this solution contains n arbitrary constants, it is the general or complete solution of equation ( i ).
Theorem 2. If y = u is the complete solution of the equation f ( D ) y = 0 and y = v is a particular solution
( containing no arbitrary constants ) of the equation f ( D ) y = X, then the complete solution of the equation
f ( D ) y = X is y = u + v.
Proof. Since y = u is the complete solution of the equation f (D) y = 0 …( i )
\ f (D) u = 0 …( ii )
Also, y = v is a particular solution of the equation f (D) y = X …( iii )
\ f (D) v = X …( iv )
Adding ( ii ) and ( iv ), we have f (D) ( u + v ) = X
Thus y = u + v satisfies the equation ( iii ), hence it is the complete solution (C.S.) because it contains n
arbitrary constants.
The part y = u is called the complementary function (C.F.) and the part y = v is called the particular
integral (P.I.) of the equation ( iii ). (P.T.U., Jan. 2010)
\ The complete solution of equation ( iii ), is y = C.F. + P.I.
Thus in order to solve the equation ( iii ), we first find the C.F. i.e., the C.S. of equation ( i ) and then the P.I.
i.e., a particular solution of equation ( iii ).
2.4. AUXILIARY EQUATION (A.E.)
Consider the differential equation (D
n
n – 1
n – 2
- ……… + a (^) n ) y = 0 …( i )
Let y = e
mx be a solution of ( i ), then D y = m e
mx , D
2 y = m
2 e
mx , …… , D
n – 2 y = m
n – 2 e
mx
D
n – 1 y = m
n – 1 e
mx , D
n y = m
n e
mx
LINEAR ORDINARY DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER 61
\ (D – m 1 ) y (^) = 1 1 or 1 1 1
m x dy m x c e m y c e dx
which is a linear equation and I.F. = e - m x^1
\ Its solution is y. 1 1 1.^12 1
m x m x m x e c e e dx c c x c
or y = ( c 1 x + c 2 ) 1
m x e
Thus, the complete solution of equation ( i ) is
y = ( c 1 x + c 2 ) 1 3 3
m x m x m x e + c e + L+ c en n
If, however, three roots of the A.E. are equal, say m 1 = m 2 = m 3 , then proceeding as above, the solution
becomes
y = (^) ( ) 1 4
2 1 2 3 4
m x m x m (^) nx c x + c x + c e + c e + L+ cn e
Case III. If two roots of the A.E. are imaginary,
Let m 1 = a + i b and m 2 = a – i b (Q in a real equation imaginary roots occur in conjugate pair)
The solution obtained in equation ( iv ) becomes
y =
1 2 3
a + b a - b
i x i x (^) m x m x c e c e c e cn e
= (^) ( 1 2 ) 3 3
a b - b
x i x i x m x mn x e c e c e c e cn e
= 1 ( cos sin ) 2 ( cos sin )
a é b + b + b - b ù ë û
x e c x i x c x i x + 3 3 n
m x m x c e + L+ cn e
By Euler's Theorem, cos sin é q ù = q + q ë û
Q
i e i
= (^) ( 1 2 ) cos (^) ( 1 2 ) sin 3 3
a é (^) + b + - b ù+ + + ë û
L n
x m x^ m^ x e c c x i c c x c e cn e
= (^) ( C cos 1 b + C sin 2 b (^) ) + 3 3 + L+ n
ax m x m x e x x c e c en
[Taking c 1 + c 2 = C 1 , i ( c 1 – c 2 ) = C 2 ]
Case IV. If two pairs of imaginary roots be equal
Let m 1 = m 2 = a + i b and m 3 = m 4 = a – i b
Then by case II, the complete solution is
y = (^) ( 1 2 ) cos^ ( 3 4 ) sin^55 n
x m x m x e c x c x c x c x c e cn e
a é + b + + b ù+ + + ë û
L.
ILLUSTRATIVE EXAMPLES
Example 1. Solve: 9y¢¢¢ + 3y¢¢ – 5y¢ + y = 0. (P.T.U., May 2008)
Sol. Symbolic form of given equation is
(9D
3
2
A.E. is 9D
3
2
or (D + 1) (3D – 1)
2 = 0
or D = – 1,
\ C.S is y = (^) ( )
1 3 1 2 3
x^ x c e c c x e
Example 2. Solve :.
4
4
d x 4x 0 dt
Sol. Given equation in symbolic form is (D
4
d
dt
62 A TEXTBOOK OF ENGINEERING MATHEMATICS
Its A.E. is D
4
4
2
2 = 0
or (D
2
2
2 = 0 or (D
2
2
whence D =
and 2 2
i.e ., D = – 1 ± i and 1 ± i
Hence the C.S. is x = e
- t ( c 1 cos t + c 2 sin t ) + e
t ( c 3 cos t + c 4 sin t ).
Example 3. If ( )
2
2
d x g x a 0 dt b
- = ; ( a > 0 , b > 0 , g > 0 ) and x = a,
dx 0 dt
= when t = 0 , show that
x = a + (a – a ) cos
g t b
. (P.T.U., May 2002)
Sol.
2
2
d x g x a dt b
Put x – a = y \
2
2
d x
dt
2
2
d y
dt
\
2
2
d y g y dt b
2 0
g m b
g
b
(–ve)
\ m =
g i b
± ; \ y = 1 cos^2 sin
g g c t c t b b
x – a = 1 cos 2 sin
g g c t c t b b
when x = a, t = 0; a – a = c 1
\ x – a = (a – a ) cos 2 sin
g g t c t b b
\
dx
dt
= ( ) sin 2 cos
g g g g a t c t b b b b
t = 0, 0
dx
dt
= \ 0 = 2
g c b
\ c 2 = 0
\ x – a = (a – a ) cos^
g t b
Hence, x = a + (a – a ) cos
g t b
TEST YOUR KNOWLEDGE
Solve the following differential equations :
1.
2
2
3 4 0
d y dy y dx dx
2
2
( ) 0
d y dy a b aby dx dx
3.
2
2
4 0
d y dy y dx dx
2
2
6 9 0
d x dx x dt dt
5.
3 2
3 2
3 3 0
d y d y dy y dx dx dx
3
3
0
d y y dx
64 A TEXTBOOK OF ENGINEERING MATHEMATICS
Theorem 2. Prove that : (^) ò
X = X dx D
. (P.T.U., May 2002)
Proof. Let
X
D
= y
Operating both sides by D, we have
D X D or X D
æ ö ç ÷=^ = è ø
dy y dx Integrating both sides w.r.t. x
y = (^) òX dx ,
no arbitrary constant being added since y =
X
D
contains no arbitrary constant.
\
X
D
= X dx ò
Theorem 3. Prove that :
=
ò
(^1) ax ax X e X e dx D a
Proof. Let
X
D
y a
Operating on both sides by (D – a ), (D – a )
X (D )
D
æ ö ç ÷=^ - è (^) - ø
a y a
or X = -. ., - =X
dy dy ay i e ay dx dx
which is a linear equation and I.F. =
a dx (^) ax e e
\ Its solution is ye
ax e dx , no constant being added
or y = X^
ò
ax ax e e dx
Hence,
X X
D
=
ax ax e e dx a
2.7. RULES FOR FINDING THE PARTICULAR INTEGRAL (P.T.U., Dec. 2004)
Consider the differential equation, (D
n
n – 1
n – 2
- … + a (^) n – 1 D + an ) y = X
It can be written as f (D) y = X, where f (D) = D
n
n – 1
n – 2
- … + a (^) n – 1 D + a (^) n
\ P.I. =
X
f (D)
Case I. When X = e
ax
Since D e
ax = a e
ax
D
2 e
ax = a
2 e
ax
D
n – 1 e
ax = a
n – 1 e
ax
D
n e
ax = a
n e
ax
\ f^ (D) e
ax = (D
n
n – 1
n – 2
- L + a (^) n – 1 D + a (^) n ) e
ax
= ( a
n
n – 1
n – 2
- L + a (^) n – 1 a + an ) e
ax
= f ( a ) e
ax .
Operating on both sides by
f (D)
LINEAR ORDINARY DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER 65
( ) ( )
(D) ( ) or ( ) (D) (D) (D)
ax ax ax ax f e f a e e f a e f f f
Dividing both sides by f ( a ),
( ) (D)
ax ax e e f a f
= , provided f ( a ) ¹ 0
Hence,
(D) ( )
ax ax e e f f a
= , provided f ( a ) ¹ 0.
Case of failure. If f ( a ) = 0, the above method fails.
Since f ( a ) = 0, D = a is a root of A.E. f (D) = 0
\ D – a is a factor of f (D).
Let f (D) = (D – a ) f (D), where f ( a ) π 0 …( i )
Then
(D) (D ) (D) D (D) D ( )
= = = ×
ax ax ax ax e e e e f a a a a
( ) D – ( )
ax ax ax ax e e e e dx a a a
× = × f f ò^
[By Theorem 3]
= ×
f ò f
ax ax e dx x e a a
…( ii )
Differentiating both sides of ( i ) w.r.t. D, we have f ¢(D) = (D – a ) f¢ (D) + f (D)
Þ f ¢( a ) = f ( a )
\ From ( ii ), we have
(D) ( )
= ×
ax ax e x e f f a
, provide f ¢( a ) ¹ 0
If f¢ ( a ) = 0, then
(D) ( )
ax ax e x e f f a
= ×
, provided f ¢¢ ( a ) ¹ 0
and so on.
Example 1. Find the P.I. of ( D^3 – 3D^2 + 4 ) y = e 2x.
Sol. P.I. =
2 3 2
D - 3D + 4
x e (^).
The denominator vanishes when D is replaced by 2. It is a case of failure.
We multiply the numerator by x and differentiate the denominator w.r.t. D.
\ P.I.=
2 2
3 D 6D
×
x x e
It is again a case of failure. We multiply the numerator by x and differentiate the denominator w.r.t. D.
\ P.I. =
2 2 1 2 2 1 2 2
6D 6 6(2) 6 6
× = × =
x x x x x e x e e (^).
Case II. When X = sin ( ax + b ) or cos ( ax + b ) (P.T.U., Dec. 2005)
D sin ( ax + b ) = a cos ( ax + b )
D^2 sin ( ax + b ) = (– a^2 ) sin ( ax + b )
D^3 sin ( ax + b ) = – a^3 cos ( ax + b )
D
4 sin ( ax + b ) = a
4 sin ( ax + b )
or (D
2 )
2 sin ( ax + b ) = (– a
2 )
2 sin ( ax + b )
.....................................................................
.....................................................................
In general, (D
2 )
n sin ( ax + b ) = (– a
2 )
n sin ( ax + b )
LINEAR ORDINARY DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER 67
P.I. =
3 2
sin 3 D – D + 4D – 4
x
sin 3
x [Q Put D
2 = – 9]
sin 3
x = ( )
sin 3 5 1 - D
x
( )
2
1 D
sin 3 5 1 – D
x = ( )
1 D
sin 3 5 1 + 9
x [Q Putting D^2 = – 9]
= (^) ( )
sin 3 D sin 3 50
é + ù ë û
x x = (^) [ ]
sin 3 3 cos 3 50
x + x
Case III. When X = x m , m being a positive integer
Here P.I. =
(D)
m x f
Take out the lowest degree term from f (D) to make the first term unity (so that Binomial Theorem for a
negative index is applicable).
The remaining factor will be of the form 1 + f (D) or 1 – f (D)
Take this factor in the numerator. It takes the form [1 + f (D)]–1^ or [1 – f (D)] –
Expand it in ascending powers of D as far as the term containing D m , since D m^ + 1^ ( xm ) = 0, D m^ + 2^ ( xm ) = 0
and so on.
Operate on x m^ term by term.
Example 3. Find the P.I. of ( D^2 + 5D + 4 ) y = x^2 + 7x + 9.
Sol. P.I. = (^) ( ) ( )
2 2 (^2 )
D 5D 4 5D D
x x x x
= (^) ( ) ( )
(^1 ) 2 2 2 1 5D D 2 1 5D D 5D D 2 1 7 9 1 7 9 4 4 4 4 4 4 4 4
é ù é^ ù æ ö æ ö æ ö ê + + ú + + = ê^ - + + + - ú + + ç ÷ ç ÷ ç ÷ ê è^ ø^ ú ê^ è^ ø^ è^ ø ú ë û (^) ë û
x x L x x
= (^) ( ) ( )
2 2 2 1 5D D 25 D 2 1 5D 21D 2 1 7 9 1 7 9 4 4 4 16 4 4 16
æ ö æ ö
ç -^ -^ +^ ÷ +^ +^ =^ ç -^ +^ ÷ +^ + è ø è ø
L x x L x x
= (^) ( ) ( ) ( )
7 9 D 7 9 D 7 9
é ù
ê ú ë û
x x x x x x
= (^) ( ) ( )
é ù æ ö
- = (^) ç + + ÷ ê ú (^) è ø ë û
x x x x x.
Case IV. When X = eax^ V, where V is a function of x
Let u be a function of x , then by successive differentiation, we have
D ( e
ax u ) = e
ax D u + a e
ax u = e
ax (D + a ) u
D^2 ( e ax^ u ) = D [ e ax^ (D + a ) u ] = e ax^ (D 2 + a D) u + aeax^ (D + a ) u
= e
ax (D
2
2 ) u = e
ax (D + a )
2 u
Similarly, D^3 ( e ax^ u ) = e ax^ (D + a ) 3 u
68 A TEXTBOOK OF ENGINEERING MATHEMATICS
In general, D
n ( e
ax u ) = e
ax (D + a )
n u
\ f (D) ( e ax^ u ) = eax^ f (D + a ) u
Operating on both sides by
f (D)
( )
(D)
(D)
ax f e u f
é ù ë û
(D )
(D)
ax e f a u f
é (^) + ù ë û
Þ e
ax u =
(D )
(D)
ax e f a u f
é (^) + ù ë û
Now, let f (D + a ) u = V, i.e ., u =
V
f (D + a )
…( i )
\ From ( i ), we have eax^ ( )
V V
(D ) (D)
ax e f a f
or (^) ( )
V V
(D) (D )
ax ax e e f f a
Thus, e
ax which is on the right of
f (D)
may be taken out to the left provided D is replaced by D + a.
Example 4. Find the P.I. of ( D^2 – 4D + 3 ) y = ex^ cos 2x.
Sol. P.I. =
( )
2 2
cos 2 cos 2 D 4D (^3) D 1 4 (D 1) 3
x x e x e x
2 2
cos 2 cos 2 D 2D 2 2D
x x e x e x [Putting D
2 = – 2
2 ]
1 1 1 2 D
cos 2 cos 2 2 2 D 2 (2 D) (2 D)
x x e x e x
( )
(^2 )
1 2 D 1 2 D
cos 2 cos 2 (^2 4) D (^24 )
x x e x e x
= ( ) ( ) ( )
2 cos 2 D cos 2 2 cos 2 2 sin 2 cos 2 sin 2. 16 16 8
x x x e x x e x x e x x
Case V. When X is any other function of x****.
Resolve f (D) into linear factors.
Let f (D) = (D – m 1
) (D – m 2
) ……… (D – m n
) X
Then P.I. = ( 1 )( 2 ) ( )
X X
f (D)^ D^ m D^ m D^ mn
- - L -
1 2
1 2 2
A A A
X
D D D
æ ö
ç ÷ è -^ -^ - ø
L
n
m m m
(By Partial fractions)
1 2
A X A X A X
D D D
L (^) n m m mn
= A 1 1 X^1 A^22 X^2 A^ X^
ò ò L^ ò
m x m x m x m x m xn m xn e e dx e e dx (^) ne e dx
See solved example 11 (art. 2.8)
X X
D
é (^) - ù ê = ú ë -^ ò û
Q
mx mx e e dx m
70 A TEXTBOOK OF ENGINEERING MATHEMATICS
Step 3. Write the complementary function with the help of following table.
Roots of the A.E. C.F.
- If roots are real and distinct say 1 1 + 2 2 + 3 3 +
m x m x m x c e c e c e ……
m 1 , m 2 , m 3
- If two real roots are equal say C.F. = ( c 1 + c 2 x ) e mx^ + 3 3
m x c e + ……
m 1
= m 2
= m
- If three roots are equal m 1
= m 2
= m 3
= m C.F. = ( c 1
x + c 3
x^2 ) emx^ + 4 4
m x c e + ……
- If roots are a pair of imaginary C.F. = e
a x ( c 1 cos b x + c 2 sin b x )
(non-repeated) numbers (say) a ± i b.
- If pair of imaginary roots is repeated C.F. = e a x^ {( c 1 + c 2 x ) cos b x + ( c 3 + c 4 x ) sin b x }
twice, i.e ., a ± i b, a ± i b.
Step 4. Find the particular integral i.e ., P.I. = (^1 )
1 2
D D D
n n n a a an
X with the help of
following rules.
Functions Particular Integrals
- When X = e ax^ then P.I. = (D)
ax e
f
Put D = a
ax e
f a
, provided f ( a ) ¹ 0.
In case f ( a ) = 0 then multiply by x and differentiate the denominator w.r.t. D and continue this process
untill denominator ceases to be zero on putting D = a.
- When X = sin ( ax + b ) P.I. =
( )
2
fD
sin ( ax + b ) or cos ( ax + b ) Put D 2 = –a^2
or cos ( ax + b )
2
sin ( ) ( )
f -
ax b a
or cos ( ax + b ) provided
2 f( - a ) ¹ 0
In case of failure apply to above mentioned rule of (1) case.
- When X = x m^ then P.I. = [ f (D)] – 1^ xm^ expand f (D) By binomial theorem up to D m^ and
then operate on x m.
- When X = e ax^ V, P.I. =
V
(D + )
ax e f a
- If X is any other function of x , then P.I. =
X
f (D)
. Resolve
f (D)
into partial fractions and operate
each partial fraction on X.
- Remember
X X and X X D D
= = ò (^) - ò
ax ax dx e e dx a
Step 5. Then write the C.S. which is C.S. = C.F. + P.I.
or cos ( ax + b )
LINEAR ORDINARY DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER 71
ILLUSTRATIVE EXAMPLES
Example 1. Solve : (D^2 + D + 1)y = (1 + sin x)^2_._ (P.T.U., May 2007)
Sol. (D
2
2
A.E. is D^2 + D + 1 = 0 \ D =
= - ± i
C.F. = 2 1
cos sin 2 2
x
e c x c x
P.I. = (^) ( )
2 2
1 sin D D 1
x = (^) { }
2 2
1 2 sin sin D D 1
x x
2
1 1 cos 2 1 2sin D D 1 2
ì - ü í +^ + ý
x x (^) = 2
2sin cos 2 D D 1 2 2
ì ü í +^ - ý
x x
2 2 2
. 2 sin cos 2 (^2) D D 1 D D 1 2 D D 1
x × (^) e + (^) x - x
(Put D = 0) (Put D^2 = - 1) (Put D 2 = - 4)
1 2 sin cos 2 2 D 2 D 3
× + × x - x
= (^) ( ) 2
3 1 D 3
2 cos cos 2 2 2 D 9
x x
3 1 D 3
2 cos cos 2 2 2 13
x x = (^) [ ]
2 cos 2 sin 2 3 cos 2 2 26
2 cos sin 2 cos 2 2 13 26
C.S. is y = (^2 1 )
cos sin 2 2
x
e c x c x
2 cos 2
sin 2 cos 2 13 26
Example 2. Solve : ( D – 2 )^2 y = 8 ( e 2x^ + sin 2x + x 2 ). (P.T.U., May 2009)
Sol. A.E. is (D – 2)^2 = 0 \ D = 2, 2
C.F. = ( c 1 + c 2 x ) e
2 x
P.I. = (^) ( )
2 2 2
8 sin 2 (D 2)
x é (^) e + x + x ù
2 2 2 2 2
8 sin 2 (D 2) (D 2) (D 2)
x e x x
é ù ê +^ + ú êë -^ -^ - úû
Now,
2 2
(D 2)
x e
Put D = 2; case of failure
2(D 2)
×
x x e | Put D = 2. Case of failure
x x ×^ e =
2 2 2
x (^) x e
2
sin 2 (D 2)
x
2 2
sin 2 sin 2 D 4D 4 2 4D 4
x = x
[Q Putting D 2 = – 2 2 ]
LINEAR ORDINARY DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER 73
\ P.I. =
2
9 6 4
x (^) x x x x e e e
Hence the C.S. is y = c 1 e
x
2
9 6 4
x (^) x x x x e e e.
Example 4. Solve :
4
4
d y
dx
- y = cos x cosh x. (P.T.U., May 2007, 2011)
Sol. Given equation in symbolic form is (D^4 – 1) y = cos x cosh x
A.E. is D^4 – 1 = 0 or (D 2 – 1) (D 2 + 1) = 0 \ D = ± 1, ± i
C.F. = c 1
ex^ + c 2
e –^ x^ + e^0 x^ ( c 3
cos x + c 4
sin x )
= c 1
e x^ + c 2
e –^ x^ + c 3
cos x + c 4
sin x
P.I. =
4 4
cos cosh cos D 1 D 1 2
x x e e x x x
æ - ö
= (^) ç ÷
4 4
cos cos (^2) D 1 D 1
x x e x e x
é (^) - ù ê + ú ë -^ - û
4 4
cos cos (^2) (D 1) 1 (D 1) 1
x x e x e x
é (^) - ù ê + ú êë +^ -^ -^ - úû
4 3 2 4 3 2
cos cos (^2) D 4D 6D 4D D 4D 6D 4D
x x e x e x
é (^) - ù ê + ú ë +^ +^ +^ -^ +^ - û
Put
2 D =– 1
( ) ( ) ( )
2 2
cos cos (^2 1) 4D 1 6 1 4 D ( 1) 4D( 1) 6( 1) 4D
x x e x e x
é ù ê (^) + ú ê (^) - + - + - + - - - + - - ú ë û
cos cos cos cosh cos 2 5 5 5 2 5
x x x x e^ e e x e x x x x
é ù æ^ + ö ê +^ ú = -^ ç ÷ = - ë -^ -^ û è^ ø
Hence the C.S. is y = c 1
ex^ + c 2
e –^ x^ + c 3
cos x + c 4
sin x –
cos cosh 5
x x.
Example 5. Solve:
2
2
d y 4 y dx
- = x sin 2x. (P.T.U., Dec. 2002)
Sol. S.F. of given equation is
(D^2 + 4) y = x sin 2 x
A.E. is D
2
C.F. is e^0 .x^ (cos 2 x + i sin 2 x ) = cos 2 x + i sin 2 x
P.I. =
2
sin 2 D 4
x x
= Imaginary part of
2 2
D 4
i x x e
= Imaginary part of
( )
2 2
D + 2 4
i x e x i +
= Imaginary part of
2 2
D 4 D – 4 + 4
i x e x
74 A TEXTBOOK OF ENGINEERING MATHEMATICS
= Imaginary part of
2 2
D
4 D 1 +
4 D
i x e x
i i
é ù ê ú êë úû
= Imaginary part of
2 1 D 1 4 D 4
i x e i x i
é ù
ê ú ë û
= Imaginary part of
2 1 D 1 4 D 4
i x i e i x
= Imaginary part of
2 1
4 D 4
i x i e i x
= Imaginary part of
( )
2 cos 2 sin 2
4 2 4
- i x + i x æ^ x i x ö × (^) ç + ÷ è ø
2
cos 2 sin 2 8 16
x x x + x
C.S. is y = C.F. + P.I.
2
1 cos 2^2 sin 2^ cos 2^ sin 2 8 16
x x c x + c x - x + x.
Example 6. Solve :
2 x 2
d y dy
- 2 y x e sin x dx dx
(P.T.U., Dec. 2003, Jan. 2010, Dec. 2010, May 2011, Dec. 2012)
Sol. Given equation in symbolic form is (D 2 – 2D + 1) y = x ex^ sin x
A.E. is D
2
2 = 0 \ D = 1, 1
C.F. = ( c 1 + c 2 x ) e x
P.I. =
2 2
. sin. sin (D 1) (D 1 1)
x x e x x = e x x
2
sin sin D D
x x e x x = e x x dx ò
Integrating by parts (^) = ( )
( cos ) 1( cos ) cos sin D D
x x e x x x dx e x x x
é ù
êë ò úû
= ( cos sin ) (^) { sin 1 .sin (^) } cos
x x e - x x + x dx = e é^ - x x - x dx - x ù ò (^) ë ò û
= (^) [ – sin – cos – cos (^) ] = – ( sin +2 cos )
x x e x x x x e x x x
Hence the C.S. is y = ( c 1 + c 2 x ) e
x
x ( x sin x + 2 cos x ).
Example 7. Solve :
2
2
d y dy 2 + y dx dx
x sin x. (P.T.U., May 2006, Dec. 2011)
Sol.
2
2
d y dy y dx dx
S.F. is (D 2 – 2D + 1) y = e x^ sin x
A.E. is m^2 – 2 m + 1 = 0 i.e., m = 1, 1.
C.F. is ( c 1
x ) e x
76 A TEXTBOOK OF ENGINEERING MATHEMATICS
\ Four values of m are
± i - ± i
C.F. =
1 1 2 2 1 2 3 4
cos sin cos sin 2 2 2 2
æ ö - æ ö
ç +^ ÷ +^ ç + ÷ è ø è ø
x x e c x c x e c x c x
P.I. = 2
4 2
cos D D 1 2
ç ÷
x
e x
4 2
cos 1 1 2 D D 1 2 2
ç ÷ æ ö æ ö è^ ø ç -^ ÷ +^ ç -^ ÷ + è ø è ø
x
e x [Using art. 2.7 Case IV]
4 3 2
cos (^5 3 21 ) D 2D D D 2 2 16
ç ÷
x
e x
Put D^2 =
D D
x
e
cos 2
x
æ ö
ç ÷ è ø
x
e
cos 2
x
æ ö
ç ÷ è ø
i.e., case of failure
3 2
cos (^3 ) 4D 6D 5D 2
x x e x
ç ÷ è ø
Put D^2 =
cos (^9 3 ) 3D 5D 2 2
ç ÷ è ø
x x e x
2
1 3 2D 3 3
cos cos 2D (^3 2) 4D 9 2
ç ÷
x x
x e x x e x
2 2
(2D 3) 3 3 3 3
cos. 2. sin 3cos 12 2 12 2 2 2
- (^) - - ìï üï = (^) í- - ý
- (^) ï ï î þ
x x x e x e x x x
LINEAR ORDINARY DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER 77
3 sin 3cos 12 2 2
x
x e x x
C.S. is y = C.F. + P.I.
y = (^2 1 2 23 )
cos sin cos sin 2 2 2 2
æ ö - æ ö
ç +^ ÷ +^ ç + ÷ è ø è ø
x x
e c x c x e c x c x
. 3 sin 3cos 12 2 2
é ù
x
x e x x.
Example 9. Solve : (D
2
- 6D + 13) y = 8e
3x sin 4x + 2
x
. (P.T.U., Dec. 2005)
Sol. A.E. is D
2
D =
= 3 ± 2 i
C.F. = e
3 x ( c 1 cos 2 x + c 2 sin 2 x )
P.I. =
D 6D 13
2
(8 e
3 x sin 4 x + 2
x )
D 6D 13
2
e
3 x sin 4 x +
D 6D 13
2
x
= 8 e
3 x^1
D 3 6 D 3 13
2 b +^ g -^ b +^ g +
sin 4 x +
D 6D 13
2
e
log 2 x
= 8 e
3 x^1
D 4
2
sin 4 x +
D 6D 13
2
e
x log 2
(Put D
2 = – 16) (Put D = log 2)
= 8 e
3 x .
sin 4 x + ( )
2
log 2 - 6 log 2 + 13
e
x log 2
3 e
x
sin 4 x +
2 blog g -^ log +
x
C.S. is y = e
3 x ( c 1 cos 2 x + c 2 sin 2 x ) –
3 2
e x
x x sin log log
b g -^ +
Example 10. Solve : (D
_2
–x sec x. (P.T.U., Dec. 2002)
Sol. A.E. is D
2
D =
i = – 1 ± i
C.F. = e
- x [ c 1 cos x + c 2 sin x ]
