UPSC Exam > UPSC Notes > Mathematics Optional Notes for UPSC > Second and Higher Order Linear Equation with Constant coefficients
Second and Higher Order Linear Equation with Constant coefficients | Mathematics Optional Notes for UPSC PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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
?? 2?? +27×
1
(?? -2)
-
?? -??
=12×
?? 3
3!
?? 2?? +27×
1
(-3)
3
?? -??
Read More
|
387 videos|203 docs
|
Related Exams
About this Document
|
Nov 23, 2024 Last updated |
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
for UPSC & Second and Higher Order Linear Equation with Constant coefficients in Hindi for Mathematics Optional Notes for UPSC course.
Download more important topics related with notes, lectures and mock test series for UPSC
Exam by signing up for free. UPSC: Second and Higher Order Linear Equation with Constant coefficients | Mathematics Optional Notes for UPSC
Description
Full syllabus notes, lecture & questions for Second and Higher Order Linear Equation with Constant coefficients | Mathematics Optional Notes for UPSC - UPSC | Plus excerises question with solution to help you revise complete syllabus for Mathematics Optional Notes for UPSC | Best notes, free PDF download
Information about Second and Higher Order Linear Equation with Constant coefficients
In this doc you can find the meaning of Second and Higher Order Linear Equation with Constant coefficients defined & explained in the simplest way possible. Besides explaining types of
Second and Higher Order Linear Equation with Constant coefficients theory, EduRev gives you an ample number of questions to practice Second and Higher Order Linear Equation with Constant coefficients tests, examples and also practice UPSC
tests
|
Explore Courses for UPSC exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
study material
,MCQs
,mock tests for examination
,Sample Paper
,Important questions
,Exam
,Free
,ppt
,Second and Higher Order Linear Equation with Constant coefficients | Mathematics Optional Notes for UPSC
,Summary
,Second and Higher Order Linear Equation with Constant coefficients | Mathematics Optional Notes for UPSC
,past year papers
,Objective type Questions
,shortcuts and tricks
,practice quizzes
,Second and Higher Order Linear Equation with Constant coefficients | Mathematics Optional Notes for UPSC
,Semester Notes
,video lectures
,Previous Year Questions with Solutions
,Extra Questions
,Viva Questions
;
Additional Information about Second and Higher Order Linear Equation with Constant coefficients for UPSC Preparation
Second and Higher Order Linear Equation with Constant coefficients Free PDF Download
The Second and Higher Order Linear Equation with Constant coefficients is an invaluable resource that delves deep into the core of the UPSC exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Second and Higher Order Linear Equation with Constant coefficients now and kickstart your journey towards success in the UPSC exam.
Importance of Second and Higher Order Linear Equation with Constant coefficients
The importance of Second and Higher Order Linear Equation with Constant coefficients cannot be overstated, especially for UPSC aspirants.
This document holds the key to success in the UPSC exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Second and Higher Order Linear Equation with Constant coefficients Notes
Second and Higher Order Linear Equation with Constant coefficients Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Second and Higher Order Linear Equation with Constant coefficients.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Second and Higher Order Linear Equation with Constant coefficients Notes on EduRev are your ultimate resource for success.
Second and Higher Order Linear Equation with Constant coefficients UPSC Questions
The "Second and Higher Order Linear Equation with Constant coefficients UPSC Questions" guide is a valuable resource for all aspiring students preparing for the
UPSC exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Second and Higher Order Linear Equation with Constant coefficients on the App
Students of UPSC can study Second and Higher Order Linear Equation with Constant coefficients alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Second and Higher Order Linear Equation with Constant coefficients,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Second and Higher Order Linear Equation with Constant coefficients is prepared as per the latest UPSC syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup