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Page 1
Edurev123
4. Second and Higher Order Linear
Equation with Constant coefficients
4.1 Use the method of undetermined coefficients to find the particular solution of
?? ''
+?? =?????? ?? +(?? +?? ?? )?? ??
and hence find its general solution.
(2010 : 20 Marks)
Solution:
Given equation is: ?? ''
+?? =sin ?? +(1+?? 2
)?? ??
Complementary Function (C.F.) :
The auxiliary equation is
?? 2
+1 =0??? =±?? ?? ?? =?? 1
?? ????
+?? 2
?? -????
(?? 1
,?? 2
are constants )
=?? cos ?? +?? sin ?? (?? ,?? are constants )
Particular Solution :
For this we will use method of undetermined coefficients.
As multiplicity of roots is 1 .
Let
?? ?? =?? (?? cos ?? +?? sin ?? )+(?? ?? 2
+???? +?? )?? ?? (1)
?? ?? '
= (?? cos ?? +?? sin ?? )+?? (-?? sin ?? +?? cos ?? )+(?? ?? 2
+???? +?? )?? ?? +(2???? +?? )?? ?? ?? ?? ''
= (-?? sin ?? +?? cos ?? )+(-?? sin ?? +?? cos ?? )+?? (-?? cos ?? -?? sin
?? )+(?? ?? 2
+???? +?? )?? ?? +(2???? +?? )?? ?? +(2???? +?? )?? ?? +2?? ?? ?? …(2)
? ?? ?? ''
= (-?? sin ?? +?? cos ?? )+(-?? sin ?? +?? cos ?? )+?? (-?? cos ?? -?? sin
?? )+(?? ?? 2
+???? +?? )?? ?? +(2???? +?? )?? ?? +(2???? +?? )?? ?? +2?? ?? ?? …(2)
? ?? ?? ''
+?? ?? =sin ?? +(1+?? 2
)?? ??
Using values from ( 1 ) and (2), we get
Page 2
Edurev123
4. Second and Higher Order Linear
Equation with Constant coefficients
4.1 Use the method of undetermined coefficients to find the particular solution of
?? ''
+?? =?????? ?? +(?? +?? ?? )?? ??
and hence find its general solution.
(2010 : 20 Marks)
Solution:
Given equation is: ?? ''
+?? =sin ?? +(1+?? 2
)?? ??
Complementary Function (C.F.) :
The auxiliary equation is
?? 2
+1 =0??? =±?? ?? ?? =?? 1
?? ????
+?? 2
?? -????
(?? 1
,?? 2
are constants )
=?? cos ?? +?? sin ?? (?? ,?? are constants )
Particular Solution :
For this we will use method of undetermined coefficients.
As multiplicity of roots is 1 .
Let
?? ?? =?? (?? cos ?? +?? sin ?? )+(?? ?? 2
+???? +?? )?? ?? (1)
?? ?? '
= (?? cos ?? +?? sin ?? )+?? (-?? sin ?? +?? cos ?? )+(?? ?? 2
+???? +?? )?? ?? +(2???? +?? )?? ?? ?? ?? ''
= (-?? sin ?? +?? cos ?? )+(-?? sin ?? +?? cos ?? )+?? (-?? cos ?? -?? sin
?? )+(?? ?? 2
+???? +?? )?? ?? +(2???? +?? )?? ?? +(2???? +?? )?? ?? +2?? ?? ?? …(2)
? ?? ?? ''
= (-?? sin ?? +?? cos ?? )+(-?? sin ?? +?? cos ?? )+?? (-?? cos ?? -?? sin
?? )+(?? ?? 2
+???? +?? )?? ?? +(2???? +?? )?? ?? +(2???? +?? )?? ?? +2?? ?? ?? …(2)
? ?? ?? ''
+?? ?? =sin ?? +(1+?? 2
)?? ??
Using values from ( 1 ) and (2), we get
2(-?? sin ?? +?? cos ?? )+?? (-?? cos ?? -?? sin ?? +(?? ?? 2
+???? +?? )?? ?? +(4???? +2?? )?? ?? +2?? ?? ?? +
(?? ?? 2
+???? +?? )?? ?? + ?? (?? cos ?? +?? sin ?? )=sin ?? +(1+?? 2
)?? ??
?2(-?? sin ?? +?? cos ?? )+2(?? ?? 2
+???? +?? )?? ?? +(4???? +2?? )?? ?? +2?? ?? ?? =sin ?? +(1+?? 2
)?? ??
Comparing LHS & RHS, we get
?? =0,?? =-
1
2
2?? =1??? =
1
2
2?? +4?? =0?2?? +2=0??? =-1
2?? +2?? +2?? =1?2?? -2+1=1??? =1
?? ?? =
-?? 2
cos ?? +(
?? 2
2
-?? +1)?? ??
So, general solution of given equation is
?? =?? cos ?? +?? sin ?? -
?? 2
cos ?? +(
?? 2
2
-?? +1)?? ??
4.2 Obtain the general solution of the second order ordinary differential equation
?? ''
-?? ?? '
+?? ?? =?? +?? ?? ?????? ??
where dashes denote derivatives w.r.t. ?? .
(2011 : 15 Marks)
Solution:
The given differential equation is
?? ''
+2?? +2?? =?? +?? ?? cos ??
The auxiliary equation is
?? 2
+2?? +2=0
? ?? =-1±??
? The complementary function is
?? ?? =?? -?? (?? 1
cos ?? +?? 2
sin ?? ) , where ?? 1
and ?? 2
are arbitrary constants.
For particular integral,
Page 3
Edurev123
4. Second and Higher Order Linear
Equation with Constant coefficients
4.1 Use the method of undetermined coefficients to find the particular solution of
?? ''
+?? =?????? ?? +(?? +?? ?? )?? ??
and hence find its general solution.
(2010 : 20 Marks)
Solution:
Given equation is: ?? ''
+?? =sin ?? +(1+?? 2
)?? ??
Complementary Function (C.F.) :
The auxiliary equation is
?? 2
+1 =0??? =±?? ?? ?? =?? 1
?? ????
+?? 2
?? -????
(?? 1
,?? 2
are constants )
=?? cos ?? +?? sin ?? (?? ,?? are constants )
Particular Solution :
For this we will use method of undetermined coefficients.
As multiplicity of roots is 1 .
Let
?? ?? =?? (?? cos ?? +?? sin ?? )+(?? ?? 2
+???? +?? )?? ?? (1)
?? ?? '
= (?? cos ?? +?? sin ?? )+?? (-?? sin ?? +?? cos ?? )+(?? ?? 2
+???? +?? )?? ?? +(2???? +?? )?? ?? ?? ?? ''
= (-?? sin ?? +?? cos ?? )+(-?? sin ?? +?? cos ?? )+?? (-?? cos ?? -?? sin
?? )+(?? ?? 2
+???? +?? )?? ?? +(2???? +?? )?? ?? +(2???? +?? )?? ?? +2?? ?? ?? …(2)
? ?? ?? ''
= (-?? sin ?? +?? cos ?? )+(-?? sin ?? +?? cos ?? )+?? (-?? cos ?? -?? sin
?? )+(?? ?? 2
+???? +?? )?? ?? +(2???? +?? )?? ?? +(2???? +?? )?? ?? +2?? ?? ?? …(2)
? ?? ?? ''
+?? ?? =sin ?? +(1+?? 2
)?? ??
Using values from ( 1 ) and (2), we get
2(-?? sin ?? +?? cos ?? )+?? (-?? cos ?? -?? sin ?? +(?? ?? 2
+???? +?? )?? ?? +(4???? +2?? )?? ?? +2?? ?? ?? +
(?? ?? 2
+???? +?? )?? ?? + ?? (?? cos ?? +?? sin ?? )=sin ?? +(1+?? 2
)?? ??
?2(-?? sin ?? +?? cos ?? )+2(?? ?? 2
+???? +?? )?? ?? +(4???? +2?? )?? ?? +2?? ?? ?? =sin ?? +(1+?? 2
)?? ??
Comparing LHS & RHS, we get
?? =0,?? =-
1
2
2?? =1??? =
1
2
2?? +4?? =0?2?? +2=0??? =-1
2?? +2?? +2?? =1?2?? -2+1=1??? =1
?? ?? =
-?? 2
cos ?? +(
?? 2
2
-?? +1)?? ??
So, general solution of given equation is
?? =?? cos ?? +?? sin ?? -
?? 2
cos ?? +(
?? 2
2
-?? +1)?? ??
4.2 Obtain the general solution of the second order ordinary differential equation
?? ''
-?? ?? '
+?? ?? =?? +?? ?? ?????? ??
where dashes denote derivatives w.r.t. ?? .
(2011 : 15 Marks)
Solution:
The given differential equation is
?? ''
+2?? +2?? =?? +?? ?? cos ??
The auxiliary equation is
?? 2
+2?? +2=0
? ?? =-1±??
? The complementary function is
?? ?? =?? -?? (?? 1
cos ?? +?? 2
sin ?? ) , where ?? 1
and ?? 2
are arbitrary constants.
For particular integral,
?? ?? =
?? +?? ?? cos ?? ?? 2
+2?? +2
,?? =
?? ????
=
1
?? 2
+2?? +2
?? +
1
?? 2
+2?? +2
·?? ?? cos ?? =
1
2[1+
?? 2
+2?? 2
]
]
?? +?? ?? 1
(?? +1)
2
+2(?? +1)+2
·cos ?? =
1
2
[1+
?? 2
+2?? 2
]
-1
·?? +?? ?? ·
1
?? 2
+4?? +5
·cos ?? =
1
2
[1-
?? 2
+2?? 2
]·?? +?? ?? ·
1
-1+4?? +5
·cos ?? =
1
2
[?? -1]+
?? ?? 4
·
1
?? +1
·cos ??
1
2
(?? -1)+
?? ?? 4
(?? -1)
?? 2
-1
cos ?? =
?? -1
2
+
?? ?? 4
·
(-1)
2
(-sin ?? -cos ?? )
=
?? -1
2
+
?? ?? 8
(sin ?? +cos ?? )
?? =?? ?? +?? ?? ?? =?? -?? (?? 1
cos ?? +?? 2
sin ?? )+
?? -1
2
+
?? ?? 8
(sin ?? +cos ?? )
is the required solution.
4.3 Find the general solution of the equation ?? '''
-?? ''
=???? ?? ?? +?? ?? .
(2012 : 20 Marks)
Solution:
The given equation is
?? ''
-?? ''
=12?? 2
+6?? (??)
Denote
?? ????
as Detc., the given equation reduces to
(?? 3
-?? 2
)?? =12?? 2
+?? ?? (???? )
The auxiliary equation of (ii) is :
?? 3
-?? 2
=0??? =0,0,1
4.4 Find a particular integral of
Page 4
Edurev123
4. Second and Higher Order Linear
Equation with Constant coefficients
4.1 Use the method of undetermined coefficients to find the particular solution of
?? ''
+?? =?????? ?? +(?? +?? ?? )?? ??
and hence find its general solution.
(2010 : 20 Marks)
Solution:
Given equation is: ?? ''
+?? =sin ?? +(1+?? 2
)?? ??
Complementary Function (C.F.) :
The auxiliary equation is
?? 2
+1 =0??? =±?? ?? ?? =?? 1
?? ????
+?? 2
?? -????
(?? 1
,?? 2
are constants )
=?? cos ?? +?? sin ?? (?? ,?? are constants )
Particular Solution :
For this we will use method of undetermined coefficients.
As multiplicity of roots is 1 .
Let
?? ?? =?? (?? cos ?? +?? sin ?? )+(?? ?? 2
+???? +?? )?? ?? (1)
?? ?? '
= (?? cos ?? +?? sin ?? )+?? (-?? sin ?? +?? cos ?? )+(?? ?? 2
+???? +?? )?? ?? +(2???? +?? )?? ?? ?? ?? ''
= (-?? sin ?? +?? cos ?? )+(-?? sin ?? +?? cos ?? )+?? (-?? cos ?? -?? sin
?? )+(?? ?? 2
+???? +?? )?? ?? +(2???? +?? )?? ?? +(2???? +?? )?? ?? +2?? ?? ?? …(2)
? ?? ?? ''
= (-?? sin ?? +?? cos ?? )+(-?? sin ?? +?? cos ?? )+?? (-?? cos ?? -?? sin
?? )+(?? ?? 2
+???? +?? )?? ?? +(2???? +?? )?? ?? +(2???? +?? )?? ?? +2?? ?? ?? …(2)
? ?? ?? ''
+?? ?? =sin ?? +(1+?? 2
)?? ??
Using values from ( 1 ) and (2), we get
2(-?? sin ?? +?? cos ?? )+?? (-?? cos ?? -?? sin ?? +(?? ?? 2
+???? +?? )?? ?? +(4???? +2?? )?? ?? +2?? ?? ?? +
(?? ?? 2
+???? +?? )?? ?? + ?? (?? cos ?? +?? sin ?? )=sin ?? +(1+?? 2
)?? ??
?2(-?? sin ?? +?? cos ?? )+2(?? ?? 2
+???? +?? )?? ?? +(4???? +2?? )?? ?? +2?? ?? ?? =sin ?? +(1+?? 2
)?? ??
Comparing LHS & RHS, we get
?? =0,?? =-
1
2
2?? =1??? =
1
2
2?? +4?? =0?2?? +2=0??? =-1
2?? +2?? +2?? =1?2?? -2+1=1??? =1
?? ?? =
-?? 2
cos ?? +(
?? 2
2
-?? +1)?? ??
So, general solution of given equation is
?? =?? cos ?? +?? sin ?? -
?? 2
cos ?? +(
?? 2
2
-?? +1)?? ??
4.2 Obtain the general solution of the second order ordinary differential equation
?? ''
-?? ?? '
+?? ?? =?? +?? ?? ?????? ??
where dashes denote derivatives w.r.t. ?? .
(2011 : 15 Marks)
Solution:
The given differential equation is
?? ''
+2?? +2?? =?? +?? ?? cos ??
The auxiliary equation is
?? 2
+2?? +2=0
? ?? =-1±??
? The complementary function is
?? ?? =?? -?? (?? 1
cos ?? +?? 2
sin ?? ) , where ?? 1
and ?? 2
are arbitrary constants.
For particular integral,
?? ?? =
?? +?? ?? cos ?? ?? 2
+2?? +2
,?? =
?? ????
=
1
?? 2
+2?? +2
?? +
1
?? 2
+2?? +2
·?? ?? cos ?? =
1
2[1+
?? 2
+2?? 2
]
]
?? +?? ?? 1
(?? +1)
2
+2(?? +1)+2
·cos ?? =
1
2
[1+
?? 2
+2?? 2
]
-1
·?? +?? ?? ·
1
?? 2
+4?? +5
·cos ?? =
1
2
[1-
?? 2
+2?? 2
]·?? +?? ?? ·
1
-1+4?? +5
·cos ?? =
1
2
[?? -1]+
?? ?? 4
·
1
?? +1
·cos ??
1
2
(?? -1)+
?? ?? 4
(?? -1)
?? 2
-1
cos ?? =
?? -1
2
+
?? ?? 4
·
(-1)
2
(-sin ?? -cos ?? )
=
?? -1
2
+
?? ?? 8
(sin ?? +cos ?? )
?? =?? ?? +?? ?? ?? =?? -?? (?? 1
cos ?? +?? 2
sin ?? )+
?? -1
2
+
?? ?? 8
(sin ?? +cos ?? )
is the required solution.
4.3 Find the general solution of the equation ?? '''
-?? ''
=???? ?? ?? +?? ?? .
(2012 : 20 Marks)
Solution:
The given equation is
?? ''
-?? ''
=12?? 2
+6?? (??)
Denote
?? ????
as Detc., the given equation reduces to
(?? 3
-?? 2
)?? =12?? 2
+?? ?? (???? )
The auxiliary equation of (ii) is :
?? 3
-?? 2
=0??? =0,0,1
4.4 Find a particular integral of
?? ?? ?? ?? ?? ?? +?? =?? ?? /?? ·??????
?? v?? ??
(2016: 10 marks)
Solution:
We write the given ODE as
(?? 2
+1)?? =?? ?? 2
·sin (
v3
2
?? )
P.I. =
1
?? 2
+1
(?? ?? 2
·sin
v3?? 2
)
=?? ?? 2
·
1
(?? +
1
2
)
2
+1
sin
v3
2
?? [
1
?? (?? )
?? ????
·?? =?? ????
??? ·?? (?? +?? )]
=?? ?? 2
·
1
?? 2
+?? +
5
4
·sin
v3?? 2
[?
1
?? (?? )
sin (???? )=
1
?? (-?? 2
)
sin (???? )]
=?? ?? 2
·
1
?? +
1
2
·sin
v3?? 2
=?? ?? 2
·
?? -
1
2
?? 2
-
1
4
·sin
v3?? 2
=?? ?? 2
·
?? -
1
2
-
3
4
-
1
4
·sin
v3?? 2
=?? ?? 2
·(
1
2
-?? )·sin
v3?? 2
=?? ?? 2
[
1
2
sin
v3
2
?? -
v3
2
cos
v3
2
?? ]
=?? ?? 2
·sin (
v3?? 2
-
?? 3
)
4.5 Solve: ?? ''
-?? =?? ?? ?? ?? ??
(2018 : 10 Marks)
Solution:
?????????? ?????????? ?????? ???? ?? ''
-?? =?? 2
?? 2??
or (?? 2
-1)?? =?? 2
?? 2??
C.F. : Auxiliary equation is (?? 2
-1)?? =0
Page 5
Edurev123
4. Second and Higher Order Linear
Equation with Constant coefficients
4.1 Use the method of undetermined coefficients to find the particular solution of
?? ''
+?? =?????? ?? +(?? +?? ?? )?? ??
and hence find its general solution.
(2010 : 20 Marks)
Solution:
Given equation is: ?? ''
+?? =sin ?? +(1+?? 2
)?? ??
Complementary Function (C.F.) :
The auxiliary equation is
?? 2
+1 =0??? =±?? ?? ?? =?? 1
?? ????
+?? 2
?? -????
(?? 1
,?? 2
are constants )
=?? cos ?? +?? sin ?? (?? ,?? are constants )
Particular Solution :
For this we will use method of undetermined coefficients.
As multiplicity of roots is 1 .
Let
?? ?? =?? (?? cos ?? +?? sin ?? )+(?? ?? 2
+???? +?? )?? ?? (1)
?? ?? '
= (?? cos ?? +?? sin ?? )+?? (-?? sin ?? +?? cos ?? )+(?? ?? 2
+???? +?? )?? ?? +(2???? +?? )?? ?? ?? ?? ''
= (-?? sin ?? +?? cos ?? )+(-?? sin ?? +?? cos ?? )+?? (-?? cos ?? -?? sin
?? )+(?? ?? 2
+???? +?? )?? ?? +(2???? +?? )?? ?? +(2???? +?? )?? ?? +2?? ?? ?? …(2)
? ?? ?? ''
= (-?? sin ?? +?? cos ?? )+(-?? sin ?? +?? cos ?? )+?? (-?? cos ?? -?? sin
?? )+(?? ?? 2
+???? +?? )?? ?? +(2???? +?? )?? ?? +(2???? +?? )?? ?? +2?? ?? ?? …(2)
? ?? ?? ''
+?? ?? =sin ?? +(1+?? 2
)?? ??
Using values from ( 1 ) and (2), we get
2(-?? sin ?? +?? cos ?? )+?? (-?? cos ?? -?? sin ?? +(?? ?? 2
+???? +?? )?? ?? +(4???? +2?? )?? ?? +2?? ?? ?? +
(?? ?? 2
+???? +?? )?? ?? + ?? (?? cos ?? +?? sin ?? )=sin ?? +(1+?? 2
)?? ??
?2(-?? sin ?? +?? cos ?? )+2(?? ?? 2
+???? +?? )?? ?? +(4???? +2?? )?? ?? +2?? ?? ?? =sin ?? +(1+?? 2
)?? ??
Comparing LHS & RHS, we get
?? =0,?? =-
1
2
2?? =1??? =
1
2
2?? +4?? =0?2?? +2=0??? =-1
2?? +2?? +2?? =1?2?? -2+1=1??? =1
?? ?? =
-?? 2
cos ?? +(
?? 2
2
-?? +1)?? ??
So, general solution of given equation is
?? =?? cos ?? +?? sin ?? -
?? 2
cos ?? +(
?? 2
2
-?? +1)?? ??
4.2 Obtain the general solution of the second order ordinary differential equation
?? ''
-?? ?? '
+?? ?? =?? +?? ?? ?????? ??
where dashes denote derivatives w.r.t. ?? .
(2011 : 15 Marks)
Solution:
The given differential equation is
?? ''
+2?? +2?? =?? +?? ?? cos ??
The auxiliary equation is
?? 2
+2?? +2=0
? ?? =-1±??
? The complementary function is
?? ?? =?? -?? (?? 1
cos ?? +?? 2
sin ?? ) , where ?? 1
and ?? 2
are arbitrary constants.
For particular integral,
?? ?? =
?? +?? ?? cos ?? ?? 2
+2?? +2
,?? =
?? ????
=
1
?? 2
+2?? +2
?? +
1
?? 2
+2?? +2
·?? ?? cos ?? =
1
2[1+
?? 2
+2?? 2
]
]
?? +?? ?? 1
(?? +1)
2
+2(?? +1)+2
·cos ?? =
1
2
[1+
?? 2
+2?? 2
]
-1
·?? +?? ?? ·
1
?? 2
+4?? +5
·cos ?? =
1
2
[1-
?? 2
+2?? 2
]·?? +?? ?? ·
1
-1+4?? +5
·cos ?? =
1
2
[?? -1]+
?? ?? 4
·
1
?? +1
·cos ??
1
2
(?? -1)+
?? ?? 4
(?? -1)
?? 2
-1
cos ?? =
?? -1
2
+
?? ?? 4
·
(-1)
2
(-sin ?? -cos ?? )
=
?? -1
2
+
?? ?? 8
(sin ?? +cos ?? )
?? =?? ?? +?? ?? ?? =?? -?? (?? 1
cos ?? +?? 2
sin ?? )+
?? -1
2
+
?? ?? 8
(sin ?? +cos ?? )
is the required solution.
4.3 Find the general solution of the equation ?? '''
-?? ''
=???? ?? ?? +?? ?? .
(2012 : 20 Marks)
Solution:
The given equation is
?? ''
-?? ''
=12?? 2
+6?? (??)
Denote
?? ????
as Detc., the given equation reduces to
(?? 3
-?? 2
)?? =12?? 2
+?? ?? (???? )
The auxiliary equation of (ii) is :
?? 3
-?? 2
=0??? =0,0,1
4.4 Find a particular integral of
?? ?? ?? ?? ?? ?? +?? =?? ?? /?? ·??????
?? v?? ??
(2016: 10 marks)
Solution:
We write the given ODE as
(?? 2
+1)?? =?? ?? 2
·sin (
v3
2
?? )
P.I. =
1
?? 2
+1
(?? ?? 2
·sin
v3?? 2
)
=?? ?? 2
·
1
(?? +
1
2
)
2
+1
sin
v3
2
?? [
1
?? (?? )
?? ????
·?? =?? ????
??? ·?? (?? +?? )]
=?? ?? 2
·
1
?? 2
+?? +
5
4
·sin
v3?? 2
[?
1
?? (?? )
sin (???? )=
1
?? (-?? 2
)
sin (???? )]
=?? ?? 2
·
1
?? +
1
2
·sin
v3?? 2
=?? ?? 2
·
?? -
1
2
?? 2
-
1
4
·sin
v3?? 2
=?? ?? 2
·
?? -
1
2
-
3
4
-
1
4
·sin
v3?? 2
=?? ?? 2
·(
1
2
-?? )·sin
v3?? 2
=?? ?? 2
[
1
2
sin
v3
2
?? -
v3
2
cos
v3
2
?? ]
=?? ?? 2
·sin (
v3?? 2
-
?? 3
)
4.5 Solve: ?? ''
-?? =?? ?? ?? ?? ??
(2018 : 10 Marks)
Solution:
?????????? ?????????? ?????? ???? ?? ''
-?? =?? 2
?? 2??
or (?? 2
-1)?? =?? 2
?? 2??
C.F. : Auxiliary equation is (?? 2
-1)?? =0
???? ?? 2
=1??? =±1
?? ?? =?? 1
?? ?? +?? 2
?? -?? , where ?? 1
and ?? 2
are constants.
?? =
1
?? 2
-1
?? 2
?? 2?? =
1
?? 2
-1
?? 2?? ·?? 2
=?? 2?? 1
(?? +2)
2
-1
·?? 2
=?? 2?? ·
1
?? 2
+4?? +3
·?? 2
=
?? 2?? 3
·
1
1+
?? 2
+4?? 3
·?? 2
=
?? 2?? 3
[1+
?? 2
+4?? 3
]
-1
?? 2
=
?? 2?? 3
[1-
?? 2
3
-
4?? 3
+
(?? 2
+4?? )
2
9
+?.]?? 2
=
?? 2?? 3
[?? 2
-
2
3
-
8?? 3
+
32
9
]
=
?? 2?? 3
[?? 2
-
8?? 3
+
26
3
]
?? =?? ?? +?? ?? =?? 1
?? ?? +?? 2
?? -?? +
?? 2?? 3
[?? 2
-
8?? 3
+
26
9
]
4.6 Solve ?? ''
-?? ?? ''
+????
'
-?? ?? =???? ?? ?? ?? +???? ?? -?? .
(2018: 10 marks)
Solution:
Given equation is ?? ''
-6?? '
+12
'
-8?? -12?? 2?? +27?? -??
or
(?? 3
-6?? 2
+12?? -8)?? =12?? 2?? +27?? -??
C.F. : Auxiliary equation is
?? 3
-6?? 2
+12?? -8=0
? (?? -2)
3
=0
? ?? =2,2,2
? ?? ?? =4?? 2?? +?? 2
?? ?? 2?? +?? 3
?? 2
?? ?? , where ?? 1
,?? 2
,?? 3
are constants.
P.I.: ?? ?? =
1
(?? -2)
3
(12?? 2?? +27?? -?? )
=12×
1
(?? -2)
3
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-
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(-3)
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Document Description: Second and Higher Order Linear Equation with Constant coefficients for UPSC 2024 is part of Mathematics Optional Notes for UPSC preparation.
The notes and questions for Second and Higher Order Linear Equation with Constant coefficients have been prepared according to the UPSC exam syllabus. Information about Second and Higher Order Linear Equation with Constant coefficients covers topics
like and Second and Higher Order Linear Equation with Constant coefficients Example, for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Second and Higher Order Linear Equation with Constant coefficients.
Introduction of Second and Higher Order Linear Equation with Constant coefficients in English is available as part of our Mathematics Optional Notes for UPSC
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Additional Information about Second and Higher Order Linear Equation with Constant coefficients for UPSC Preparation
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The content of Second and Higher Order Linear Equation with Constant coefficients is prepared as per the latest UPSC syllabus.
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