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Page 1
Edurev123
6. Second Order Linear Equations with
Variable Coefficient
6.1 Solve the ordinary differential equation
?? (?? -?? )?? '
-(?? ?? -?? )?? +?? ?? =?? ?? (?? ?? -?? )
(2012: 20 marks)
Solution:
The given equation is :
?? (?? -1)?? '
-(2?? -1)?? +2?? =?? 2
(2?? -3)
? ?? ''
-
(2?? -1)
?? (?? -1)
?? '
+
2?? ?? (?? -1)
=
?? 2
(2?? -3)
?? (?? -1)
(??)
which is of the form of
?? ''
+???? +???? =?? ?? =
-(2?? -1)
?? (?? -1)
,?? =
2
?? '
(?? -1)
2+2???? +?? ?? 2
=2+2?? ·
[-(2?? -1)]
?? (?? -1)
+?? 2
·
2
?? (?? -1)
=0
?4=?? 2
is a part of the complementary function.
Also ?? = Second part of ?? .?? .
=?? ?
?? -? ?????? ?? 2
????
=?? 2
?
1
?? 4
?? ?
2?? -1
?? (?? -1)
????
???? =?? 2
?
1
?? 4
·?? ? (
1
?? +
1
?? -1
)????
?? ?? =?? 2
?
1
?? 4
·(?? 2
-?? )???? =?? 2
? (
1
?? 2
-
1
?? 3
)????
=?? 2
(
?? -1
-1
-
?? -2
-2
)
=-?? +2
So, the complementary function is
?? ?? =?? 1
+?? 2
?? +?? 3
?? ?? where ?? 1
,?? 2
and ?? 3
are arbitrary constants.
Particular integral is
Page 2
Edurev123
6. Second Order Linear Equations with
Variable Coefficient
6.1 Solve the ordinary differential equation
?? (?? -?? )?? '
-(?? ?? -?? )?? +?? ?? =?? ?? (?? ?? -?? )
(2012: 20 marks)
Solution:
The given equation is :
?? (?? -1)?? '
-(2?? -1)?? +2?? =?? 2
(2?? -3)
? ?? ''
-
(2?? -1)
?? (?? -1)
?? '
+
2?? ?? (?? -1)
=
?? 2
(2?? -3)
?? (?? -1)
(??)
which is of the form of
?? ''
+???? +???? =?? ?? =
-(2?? -1)
?? (?? -1)
,?? =
2
?? '
(?? -1)
2+2???? +?? ?? 2
=2+2?? ·
[-(2?? -1)]
?? (?? -1)
+?? 2
·
2
?? (?? -1)
=0
?4=?? 2
is a part of the complementary function.
Also ?? = Second part of ?? .?? .
=?? ?
?? -? ?????? ?? 2
????
=?? 2
?
1
?? 4
?? ?
2?? -1
?? (?? -1)
????
???? =?? 2
?
1
?? 4
·?? ? (
1
?? +
1
?? -1
)????
?? ?? =?? 2
?
1
?? 4
·(?? 2
-?? )???? =?? 2
? (
1
?? 2
-
1
?? 3
)????
=?? 2
(
?? -1
-1
-
?? -2
-2
)
=-?? +2
So, the complementary function is
?? ?? =?? 1
+?? 2
?? +?? 3
?? ?? where ?? 1
,?? 2
and ?? 3
are arbitrary constants.
Particular integral is
?? ?? =
1
(?? 3
-?? 2
)
(12?? 2
+6?? )
=
-1
?? 2
(1-?? )
(12?? 2
+6?? )
=
-1
?? 2
(1-?? )
-1
(12?? 2
+6?? )
=
-1
?? 2
(1+?? +?? 2
+?.)(12?? 2
+6?? )
=
-1
?? 2
(12?? 2
+30?? +30)
=
-1
?? (4?? 3
+15?? 2
+30?? )
=-(?? 4
+5?? 3
+15?? 2
)
? The required general solution is :
?? =?? ?? +?? ?? =?? 1
+?? 2
?? +?? 3
?? ?? -(?? 4
+5?? 3
+15?? 2
)
Particular integral,
P.I. =?? ?
?? -??????? ?? 2
(??? 4
?? -??????? ???? )????
=?? 2
?
(?? 2
-?? )
?? 4
(?
?? (2?? -3)
(?? -1)
·?? (?? -1)???? )???? [??? -? ?????? =?? (1)
1]
=?? 2
?
?? 2
-?? ?? 4
[? ?? 2
(2?? -3)???? ]????
=?? 2
?
?? -1
?? 3
[? (2?? 3
-3?? 2
)???? ]????
=
?? 2
2
? (?? -1)(?? -2)????
=
?? 2
2
? (?? 2
-3?? -2)???? =
?? 2
2
(
?? 3
3
-
3?? 2
2
+2?? )
? Complete solution of (i) is
?? =?? 1
?? 2
+?? 2
(2-?? )+
?? 2
2
(
?? 3
3
-
3?? 2
2
+2?? )
6.2 Find the general solution of the equation:
?? ?? ?? ?? ?? ?? ?? ?? +?? ????
????
+?? =???? ?? ?????? (???? ?? )
(2013: 15 Marks)
Page 3
Edurev123
6. Second Order Linear Equations with
Variable Coefficient
6.1 Solve the ordinary differential equation
?? (?? -?? )?? '
-(?? ?? -?? )?? +?? ?? =?? ?? (?? ?? -?? )
(2012: 20 marks)
Solution:
The given equation is :
?? (?? -1)?? '
-(2?? -1)?? +2?? =?? 2
(2?? -3)
? ?? ''
-
(2?? -1)
?? (?? -1)
?? '
+
2?? ?? (?? -1)
=
?? 2
(2?? -3)
?? (?? -1)
(??)
which is of the form of
?? ''
+???? +???? =?? ?? =
-(2?? -1)
?? (?? -1)
,?? =
2
?? '
(?? -1)
2+2???? +?? ?? 2
=2+2?? ·
[-(2?? -1)]
?? (?? -1)
+?? 2
·
2
?? (?? -1)
=0
?4=?? 2
is a part of the complementary function.
Also ?? = Second part of ?? .?? .
=?? ?
?? -? ?????? ?? 2
????
=?? 2
?
1
?? 4
?? ?
2?? -1
?? (?? -1)
????
???? =?? 2
?
1
?? 4
·?? ? (
1
?? +
1
?? -1
)????
?? ?? =?? 2
?
1
?? 4
·(?? 2
-?? )???? =?? 2
? (
1
?? 2
-
1
?? 3
)????
=?? 2
(
?? -1
-1
-
?? -2
-2
)
=-?? +2
So, the complementary function is
?? ?? =?? 1
+?? 2
?? +?? 3
?? ?? where ?? 1
,?? 2
and ?? 3
are arbitrary constants.
Particular integral is
?? ?? =
1
(?? 3
-?? 2
)
(12?? 2
+6?? )
=
-1
?? 2
(1-?? )
(12?? 2
+6?? )
=
-1
?? 2
(1-?? )
-1
(12?? 2
+6?? )
=
-1
?? 2
(1+?? +?? 2
+?.)(12?? 2
+6?? )
=
-1
?? 2
(12?? 2
+30?? +30)
=
-1
?? (4?? 3
+15?? 2
+30?? )
=-(?? 4
+5?? 3
+15?? 2
)
? The required general solution is :
?? =?? ?? +?? ?? =?? 1
+?? 2
?? +?? 3
?? ?? -(?? 4
+5?? 3
+15?? 2
)
Particular integral,
P.I. =?? ?
?? -??????? ?? 2
(??? 4
?? -??????? ???? )????
=?? 2
?
(?? 2
-?? )
?? 4
(?
?? (2?? -3)
(?? -1)
·?? (?? -1)???? )???? [??? -? ?????? =?? (1)
1]
=?? 2
?
?? 2
-?? ?? 4
[? ?? 2
(2?? -3)???? ]????
=?? 2
?
?? -1
?? 3
[? (2?? 3
-3?? 2
)???? ]????
=
?? 2
2
? (?? -1)(?? -2)????
=
?? 2
2
? (?? 2
-3?? -2)???? =
?? 2
2
(
?? 3
3
-
3?? 2
2
+2?? )
? Complete solution of (i) is
?? =?? 1
?? 2
+?? 2
(2-?? )+
?? 2
2
(
?? 3
3
-
3?? 2
2
+2?? )
6.2 Find the general solution of the equation:
?? ?? ?? ?? ?? ?? ?? ?? +?? ????
????
+?? =???? ?? ?????? (???? ?? )
(2013: 15 Marks)
Solution:
We have
(?? 2
?? 2
+???? +1)?? =ln ?? sin (ln ?? )
This is in the Cauchy-Euler form.
Let
?? =ln ??
?
????
????
=
1
?? ??? ?? ????
=
?? ????
????? =?? 1
where
?? ????
=?? 1
Similarly,
?? 2
?? 2
=?? 1
(?? 1
-1)
? (?? 1
(?? 1
-1)+?? 1
+1)?? =?? sin ??
? (?? 1
2
+1)?? =?? sin ??
Auxiliary equation is ?? 2
+1=0
? Comfiementary function (C.F.)
?? =±?? =?? 1
cos ?? +?? 2
sin ?? P.I. =
1
(?? 2
+1)
?? sin ?? = Real part
1
?? 2
+1
?? ?? ????
Now,
1
?? 2
+1
?? ?? ????
=?? ????
1
(?? +??)
2
+1
?? =?? ????
1
?? 2
+2????
?? =
?? ????
2?? 1
?? (1+
?? 2?? )
-1
?? =
?? ????
2?? ·
1
?? (1-
?? 2?? +
?? 2
4?? 2
)??
Page 4
Edurev123
6. Second Order Linear Equations with
Variable Coefficient
6.1 Solve the ordinary differential equation
?? (?? -?? )?? '
-(?? ?? -?? )?? +?? ?? =?? ?? (?? ?? -?? )
(2012: 20 marks)
Solution:
The given equation is :
?? (?? -1)?? '
-(2?? -1)?? +2?? =?? 2
(2?? -3)
? ?? ''
-
(2?? -1)
?? (?? -1)
?? '
+
2?? ?? (?? -1)
=
?? 2
(2?? -3)
?? (?? -1)
(??)
which is of the form of
?? ''
+???? +???? =?? ?? =
-(2?? -1)
?? (?? -1)
,?? =
2
?? '
(?? -1)
2+2???? +?? ?? 2
=2+2?? ·
[-(2?? -1)]
?? (?? -1)
+?? 2
·
2
?? (?? -1)
=0
?4=?? 2
is a part of the complementary function.
Also ?? = Second part of ?? .?? .
=?? ?
?? -? ?????? ?? 2
????
=?? 2
?
1
?? 4
?? ?
2?? -1
?? (?? -1)
????
???? =?? 2
?
1
?? 4
·?? ? (
1
?? +
1
?? -1
)????
?? ?? =?? 2
?
1
?? 4
·(?? 2
-?? )???? =?? 2
? (
1
?? 2
-
1
?? 3
)????
=?? 2
(
?? -1
-1
-
?? -2
-2
)
=-?? +2
So, the complementary function is
?? ?? =?? 1
+?? 2
?? +?? 3
?? ?? where ?? 1
,?? 2
and ?? 3
are arbitrary constants.
Particular integral is
?? ?? =
1
(?? 3
-?? 2
)
(12?? 2
+6?? )
=
-1
?? 2
(1-?? )
(12?? 2
+6?? )
=
-1
?? 2
(1-?? )
-1
(12?? 2
+6?? )
=
-1
?? 2
(1+?? +?? 2
+?.)(12?? 2
+6?? )
=
-1
?? 2
(12?? 2
+30?? +30)
=
-1
?? (4?? 3
+15?? 2
+30?? )
=-(?? 4
+5?? 3
+15?? 2
)
? The required general solution is :
?? =?? ?? +?? ?? =?? 1
+?? 2
?? +?? 3
?? ?? -(?? 4
+5?? 3
+15?? 2
)
Particular integral,
P.I. =?? ?
?? -??????? ?? 2
(??? 4
?? -??????? ???? )????
=?? 2
?
(?? 2
-?? )
?? 4
(?
?? (2?? -3)
(?? -1)
·?? (?? -1)???? )???? [??? -? ?????? =?? (1)
1]
=?? 2
?
?? 2
-?? ?? 4
[? ?? 2
(2?? -3)???? ]????
=?? 2
?
?? -1
?? 3
[? (2?? 3
-3?? 2
)???? ]????
=
?? 2
2
? (?? -1)(?? -2)????
=
?? 2
2
? (?? 2
-3?? -2)???? =
?? 2
2
(
?? 3
3
-
3?? 2
2
+2?? )
? Complete solution of (i) is
?? =?? 1
?? 2
+?? 2
(2-?? )+
?? 2
2
(
?? 3
3
-
3?? 2
2
+2?? )
6.2 Find the general solution of the equation:
?? ?? ?? ?? ?? ?? ?? ?? +?? ????
????
+?? =???? ?? ?????? (???? ?? )
(2013: 15 Marks)
Solution:
We have
(?? 2
?? 2
+???? +1)?? =ln ?? sin (ln ?? )
This is in the Cauchy-Euler form.
Let
?? =ln ??
?
????
????
=
1
?? ??? ?? ????
=
?? ????
????? =?? 1
where
?? ????
=?? 1
Similarly,
?? 2
?? 2
=?? 1
(?? 1
-1)
? (?? 1
(?? 1
-1)+?? 1
+1)?? =?? sin ??
? (?? 1
2
+1)?? =?? sin ??
Auxiliary equation is ?? 2
+1=0
? Comfiementary function (C.F.)
?? =±?? =?? 1
cos ?? +?? 2
sin ?? P.I. =
1
(?? 2
+1)
?? sin ?? = Real part
1
?? 2
+1
?? ?? ????
Now,
1
?? 2
+1
?? ?? ????
=?? ????
1
(?? +??)
2
+1
?? =?? ????
1
?? 2
+2????
?? =
?? ????
2?? 1
?? (1+
?? 2?? )
-1
?? =
?? ????
2?? ·
1
?? (1-
?? 2?? +
?? 2
4?? 2
)??
=
?? ????
2?? 1
?? (?? -
1
2?? )=
?? ????
2?? (
?? 2
2
-
?? 2?? )
=
?? ????
4
(-???? 2
+?? )
? R.P. of
1
?? 2
+1
?? ?? ????
= R.P. of {
?? ????
4
(-???? 2
+?? )}
=
1
4
(?? cos ?? +?? 2
sin ?? )
? ???? =
1
4
(?? cos ?? +?? 2
sin ?? )
?? =?? 1
cos ?? +?? 2
sin ?? +
?? 4
cos ?? +
?? 2
4
sin ?? =(?? 1
+
ln ?? 4
)cos (ln ?? )+{?? 2
+
(ln ?? )
2
4
}sin (ln ?? )
6.3 Solve the differential equation:
?? ?? ?? ?? ?? ?? ?? ?? +?? ?? ?? ?? ?? ?? ?? ?? ?? +?? ????
????
+?? ?? =???? ?????? (??????
?? ?? )
(2014 : 20 Marks)
Solution:
Given that
(?? 3
?? 3
+3?? 2
?? 2
+???? +8)?? =65cos log ?? (??)
?????? ?? =?? ??
? log ?? =??
?????? ?????? ?? 1
=
?? ????
Then
2?? =?? 1
and ?? 2
?? 2
=?? 1
(?? 1
-1)
?? 3
=?? 1
(?? 1
-1)(?? 1
-2)
? (i), we have
(?? 1
(?? 1
-1)(?? 1
-2)+3(?? 1
)(?? 1
-1)+?? 1
+8)?? =65cos ??
? (?? 1
3
+8)?? =65cos?? (???? )
Auxiliary equation of (ii) is
Page 5
Edurev123
6. Second Order Linear Equations with
Variable Coefficient
6.1 Solve the ordinary differential equation
?? (?? -?? )?? '
-(?? ?? -?? )?? +?? ?? =?? ?? (?? ?? -?? )
(2012: 20 marks)
Solution:
The given equation is :
?? (?? -1)?? '
-(2?? -1)?? +2?? =?? 2
(2?? -3)
? ?? ''
-
(2?? -1)
?? (?? -1)
?? '
+
2?? ?? (?? -1)
=
?? 2
(2?? -3)
?? (?? -1)
(??)
which is of the form of
?? ''
+???? +???? =?? ?? =
-(2?? -1)
?? (?? -1)
,?? =
2
?? '
(?? -1)
2+2???? +?? ?? 2
=2+2?? ·
[-(2?? -1)]
?? (?? -1)
+?? 2
·
2
?? (?? -1)
=0
?4=?? 2
is a part of the complementary function.
Also ?? = Second part of ?? .?? .
=?? ?
?? -? ?????? ?? 2
????
=?? 2
?
1
?? 4
?? ?
2?? -1
?? (?? -1)
????
???? =?? 2
?
1
?? 4
·?? ? (
1
?? +
1
?? -1
)????
?? ?? =?? 2
?
1
?? 4
·(?? 2
-?? )???? =?? 2
? (
1
?? 2
-
1
?? 3
)????
=?? 2
(
?? -1
-1
-
?? -2
-2
)
=-?? +2
So, the complementary function is
?? ?? =?? 1
+?? 2
?? +?? 3
?? ?? where ?? 1
,?? 2
and ?? 3
are arbitrary constants.
Particular integral is
?? ?? =
1
(?? 3
-?? 2
)
(12?? 2
+6?? )
=
-1
?? 2
(1-?? )
(12?? 2
+6?? )
=
-1
?? 2
(1-?? )
-1
(12?? 2
+6?? )
=
-1
?? 2
(1+?? +?? 2
+?.)(12?? 2
+6?? )
=
-1
?? 2
(12?? 2
+30?? +30)
=
-1
?? (4?? 3
+15?? 2
+30?? )
=-(?? 4
+5?? 3
+15?? 2
)
? The required general solution is :
?? =?? ?? +?? ?? =?? 1
+?? 2
?? +?? 3
?? ?? -(?? 4
+5?? 3
+15?? 2
)
Particular integral,
P.I. =?? ?
?? -??????? ?? 2
(??? 4
?? -??????? ???? )????
=?? 2
?
(?? 2
-?? )
?? 4
(?
?? (2?? -3)
(?? -1)
·?? (?? -1)???? )???? [??? -? ?????? =?? (1)
1]
=?? 2
?
?? 2
-?? ?? 4
[? ?? 2
(2?? -3)???? ]????
=?? 2
?
?? -1
?? 3
[? (2?? 3
-3?? 2
)???? ]????
=
?? 2
2
? (?? -1)(?? -2)????
=
?? 2
2
? (?? 2
-3?? -2)???? =
?? 2
2
(
?? 3
3
-
3?? 2
2
+2?? )
? Complete solution of (i) is
?? =?? 1
?? 2
+?? 2
(2-?? )+
?? 2
2
(
?? 3
3
-
3?? 2
2
+2?? )
6.2 Find the general solution of the equation:
?? ?? ?? ?? ?? ?? ?? ?? +?? ????
????
+?? =???? ?? ?????? (???? ?? )
(2013: 15 Marks)
Solution:
We have
(?? 2
?? 2
+???? +1)?? =ln ?? sin (ln ?? )
This is in the Cauchy-Euler form.
Let
?? =ln ??
?
????
????
=
1
?? ??? ?? ????
=
?? ????
????? =?? 1
where
?? ????
=?? 1
Similarly,
?? 2
?? 2
=?? 1
(?? 1
-1)
? (?? 1
(?? 1
-1)+?? 1
+1)?? =?? sin ??
? (?? 1
2
+1)?? =?? sin ??
Auxiliary equation is ?? 2
+1=0
? Comfiementary function (C.F.)
?? =±?? =?? 1
cos ?? +?? 2
sin ?? P.I. =
1
(?? 2
+1)
?? sin ?? = Real part
1
?? 2
+1
?? ?? ????
Now,
1
?? 2
+1
?? ?? ????
=?? ????
1
(?? +??)
2
+1
?? =?? ????
1
?? 2
+2????
?? =
?? ????
2?? 1
?? (1+
?? 2?? )
-1
?? =
?? ????
2?? ·
1
?? (1-
?? 2?? +
?? 2
4?? 2
)??
=
?? ????
2?? 1
?? (?? -
1
2?? )=
?? ????
2?? (
?? 2
2
-
?? 2?? )
=
?? ????
4
(-???? 2
+?? )
? R.P. of
1
?? 2
+1
?? ?? ????
= R.P. of {
?? ????
4
(-???? 2
+?? )}
=
1
4
(?? cos ?? +?? 2
sin ?? )
? ???? =
1
4
(?? cos ?? +?? 2
sin ?? )
?? =?? 1
cos ?? +?? 2
sin ?? +
?? 4
cos ?? +
?? 2
4
sin ?? =(?? 1
+
ln ?? 4
)cos (ln ?? )+{?? 2
+
(ln ?? )
2
4
}sin (ln ?? )
6.3 Solve the differential equation:
?? ?? ?? ?? ?? ?? ?? ?? +?? ?? ?? ?? ?? ?? ?? ?? ?? +?? ????
????
+?? ?? =???? ?????? (??????
?? ?? )
(2014 : 20 Marks)
Solution:
Given that
(?? 3
?? 3
+3?? 2
?? 2
+???? +8)?? =65cos log ?? (??)
?????? ?? =?? ??
? log ?? =??
?????? ?????? ?? 1
=
?? ????
Then
2?? =?? 1
and ?? 2
?? 2
=?? 1
(?? 1
-1)
?? 3
=?? 1
(?? 1
-1)(?? 1
-2)
? (i), we have
(?? 1
(?? 1
-1)(?? 1
-2)+3(?? 1
)(?? 1
-1)+?? 1
+8)?? =65cos ??
? (?? 1
3
+8)?? =65cos?? (???? )
Auxiliary equation of (ii) is
? ?? 1
3
+8 =0
? (?? 1
+2)(?? 1
-2?? 1
+4) =0
?? 1
=-2,1±v3?? ? C.F. =?? ?? =?? 1
?? -2?? +?? 2
?? ?? (?? 2
cos v3?? +?? 3
sin v3?? )
P.I. =
1
(?? 1
3
+8)
(65cos ?? )
=
65
-?? 1
+8
cos ?? =
(?? 1
+8)65
-?? 1
2
+64
cos ??
=
65
65
(?? 1
+8)cos ?? =(?? 1
+8)cos ?? =-sin ?? +8cos ?? ?? =?? ?? +?? ?? ?? =?? 1
?? -2?? +?? ?? (?? 2
cos v3?? +?? 3
sin v3?? )-sin ?? +8cos ?? =?? 1
?? -2
+?? (?? 2
cos v3log ?? +?? 3
sin v3log ?? )-sin (log ?? )+8cos (log ?? )
which is the required solution.
6.4 Solve : ?? ?? ?? ?? ?? ?? ?? ?? +?? ?? ?? ?? ?? ?? ?? ?? ?? +?? ?? ?? ?? ?? ?? ?? ?? ?? -?? ?? ????
????
-?? ?? =?? ?? +???????? (?????? ?? ) .
(2015 : 13 Marks)
Solution:
Putting ?? =?? ?? , equation becomes
[?? (?? -1)(?? -2)(?? -3)+6?? (?? -1)(?? -2)+4?? (?? -1)-2?? -4]=?? 2?? +2cos ?? [?? (?? -1)(?? -2)(?? -3)+6?? (?? -1)(?? -2)+2(?? -2)(?? +1)]=?? 2?? +2cos ?? ? (?? 2
+1)(?? 2
-4)?? =?? ?? +2cos ??
For homogeneous part
?? ?? =?? 1
?? 2?? +?? 2
?? -2?? +?? 3
sin ?? +?? 4
cos ??
Particular integral
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