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Page 1
Edurev123
7. Determination of Complete Solution
when one solution is known and Method of
Variation of Parameter
7.1 Using the method of variation of parameters, solve the second order
differential equation
?? ?? ?? ?? ?? ?? +?? ?? =?????? ?? ??
(2011: 15 Marks)
Solution:
The given equation is
?? 2
?? ?? ?? 2
+4?? =tan 2?? ? (?? 2
+4)?? =tan 2?? ,?? =
?? ????
?????????????????? ???????????????? ???? ?? 2
+?? =0
? ?? =±2?? ? Complementary function is
?? ?? =?? 1
cos 2?? +?? 2
sin 2?? , where ?? 1
and ?? 2
are arbitrary constants.
Let ?? ?? =?? cos 2?? +?? sin 2?? be particular integral of (i) where ?? and ?? are functions of ?? .
Then, ?? (?? )=cos 2?? ,?? (?? )=sin 2?? and ?? (?? )=tan 2?? ?? =?
4?? ?? (?? ,?? )
????
=
1
2
? cos 2?? ·tan 2?????? =
1
2
? sin 2?????? =-
1
4
cos 2?? ?? ?? =-
1
4
[log |sec 2?? +tan 2?? |-sin 2?? ]cos 2?? -
1
4
cos 2?? ·sin 2?? =-
1
4
(log |sec 2?? +tan 2?? |)cos 2??
? The required general solution is
?? =?? ?? +?? ?? =?? 1
cos 2?? +?? 2
sin 2?? -
1
4
(log |sec 2?? +tan 2?? |)cos 2??
7.2 Using the method of variation of parameter solve the differential equation :
Page 2
Edurev123
7. Determination of Complete Solution
when one solution is known and Method of
Variation of Parameter
7.1 Using the method of variation of parameters, solve the second order
differential equation
?? ?? ?? ?? ?? ?? +?? ?? =?????? ?? ??
(2011: 15 Marks)
Solution:
The given equation is
?? 2
?? ?? ?? 2
+4?? =tan 2?? ? (?? 2
+4)?? =tan 2?? ,?? =
?? ????
?????????????????? ???????????????? ???? ?? 2
+?? =0
? ?? =±2?? ? Complementary function is
?? ?? =?? 1
cos 2?? +?? 2
sin 2?? , where ?? 1
and ?? 2
are arbitrary constants.
Let ?? ?? =?? cos 2?? +?? sin 2?? be particular integral of (i) where ?? and ?? are functions of ?? .
Then, ?? (?? )=cos 2?? ,?? (?? )=sin 2?? and ?? (?? )=tan 2?? ?? =?
4?? ?? (?? ,?? )
????
=
1
2
? cos 2?? ·tan 2?????? =
1
2
? sin 2?????? =-
1
4
cos 2?? ?? ?? =-
1
4
[log |sec 2?? +tan 2?? |-sin 2?? ]cos 2?? -
1
4
cos 2?? ·sin 2?? =-
1
4
(log |sec 2?? +tan 2?? |)cos 2??
? The required general solution is
?? =?? ?? +?? ?? =?? 1
cos 2?? +?? 2
sin 2?? -
1
4
(log |sec 2?? +tan 2?? |)cos 2??
7.2 Using the method of variation of parameter solve the differential equation :
?? ?? ?? ?? ?? ?? +?? ?? ?? =?????? ????
(2013 : 10 Marks)
Solution:
Homogeneous part is
?? 2
?? ?? ?? 2
+?? 2
?? =0?(?? 2
+?? 2
)?? =0
Auxiliary equation: ?? 2
+?? 2
=0??? =±????
? ?? =cos ????
?? =sin ????
are two independent solutions of homogeneous equation.
Let ?? =???? +???? complete the solution.
Then,
?? =?
-????
?? ?? =?
?? ?? ??
where
?? = Wronskian (?? ,?? )
=|
cos ???? sin ????
-?? sin ???? ?? cos ????
|=??
? ?? =-?
sin ???? ·sec ????
?? ???? =-
1
?? ?tan ????????
=
1
?? 2
?
-?? sin ????
cos ?? ?? ???? =
1
?? 2
ln (cos ???? )+?? 1
?? =?
cos ???? sec ????
?? ???? =
?? ?? +?? 2
? ?? =?? 1
cos ???? +
1
?? 2
cos ???? ×ln (cos (???? ))+?? 2
sin ???? +
?? ?? sin ????
7.3 Solve by the method of variation of parameters :
????
????
-?? ?? =?????? ??
(2014 : 10 Marks)
Page 3
Edurev123
7. Determination of Complete Solution
when one solution is known and Method of
Variation of Parameter
7.1 Using the method of variation of parameters, solve the second order
differential equation
?? ?? ?? ?? ?? ?? +?? ?? =?????? ?? ??
(2011: 15 Marks)
Solution:
The given equation is
?? 2
?? ?? ?? 2
+4?? =tan 2?? ? (?? 2
+4)?? =tan 2?? ,?? =
?? ????
?????????????????? ???????????????? ???? ?? 2
+?? =0
? ?? =±2?? ? Complementary function is
?? ?? =?? 1
cos 2?? +?? 2
sin 2?? , where ?? 1
and ?? 2
are arbitrary constants.
Let ?? ?? =?? cos 2?? +?? sin 2?? be particular integral of (i) where ?? and ?? are functions of ?? .
Then, ?? (?? )=cos 2?? ,?? (?? )=sin 2?? and ?? (?? )=tan 2?? ?? =?
4?? ?? (?? ,?? )
????
=
1
2
? cos 2?? ·tan 2?????? =
1
2
? sin 2?????? =-
1
4
cos 2?? ?? ?? =-
1
4
[log |sec 2?? +tan 2?? |-sin 2?? ]cos 2?? -
1
4
cos 2?? ·sin 2?? =-
1
4
(log |sec 2?? +tan 2?? |)cos 2??
? The required general solution is
?? =?? ?? +?? ?? =?? 1
cos 2?? +?? 2
sin 2?? -
1
4
(log |sec 2?? +tan 2?? |)cos 2??
7.2 Using the method of variation of parameter solve the differential equation :
?? ?? ?? ?? ?? ?? +?? ?? ?? =?????? ????
(2013 : 10 Marks)
Solution:
Homogeneous part is
?? 2
?? ?? ?? 2
+?? 2
?? =0?(?? 2
+?? 2
)?? =0
Auxiliary equation: ?? 2
+?? 2
=0??? =±????
? ?? =cos ????
?? =sin ????
are two independent solutions of homogeneous equation.
Let ?? =???? +???? complete the solution.
Then,
?? =?
-????
?? ?? =?
?? ?? ??
where
?? = Wronskian (?? ,?? )
=|
cos ???? sin ????
-?? sin ???? ?? cos ????
|=??
? ?? =-?
sin ???? ·sec ????
?? ???? =-
1
?? ?tan ????????
=
1
?? 2
?
-?? sin ????
cos ?? ?? ???? =
1
?? 2
ln (cos ???? )+?? 1
?? =?
cos ???? sec ????
?? ???? =
?? ?? +?? 2
? ?? =?? 1
cos ???? +
1
?? 2
cos ???? ×ln (cos (???? ))+?? 2
sin ???? +
?? ?? sin ????
7.3 Solve by the method of variation of parameters :
????
????
-?? ?? =?????? ??
(2014 : 10 Marks)
Solution:
Given that,
????
????
-5?? =sin ?? (??)
Differentiating (i) w.r.t. ?? , we get
?? 2
?? ?? ?? 2
-5
????
????
=cos ??
(?? 2
-5?? )?? =cos ??
Now consider auxiliary equation of (ii)
?? (?? -5)=0
? ?? =0,5
? C.F. of (ii) is. ?? ?? =?? 1
+?? 2
?? 5??
Let ?? ?? =???? +???? be a particular integral of (ii) where ?? and ?? are functions of ?? and ?? =
1,?? =?? 5??
Now ?? =|
?? ?? ?? '
?? '
|-|
1 ?? 5?? 0 5?? 5?? |=5?? 5?? ?0
? ?? =-?
?? ?? ?? =-?
?? 5?? ·cos ?????? 5?? 5?? =-
1
5
? cos ?????? =-
1
5
sin:??
?? =?
????
?? =?
1·cos ?? 5?? 5?? ???? =
1
5
? ?? -5?? cos ?????? =
1
5
?? -5?? 25+1
[-5cos ?? +sin ?? ]
=
1?? -5?? 5
-5cos ?? +sin ?? ]
26
[-5
? The general solution of (ii) is
?? =?? ?? +?? ?? ?? =?? 1
+?? 2
?? 5?? -
1
5
sin ?? ·1+
1
5
?? -5?? 26
[-5cos ?? +sin ?? ]
?? =?? 1
+?? 2
?? 5?? -
1
26
[cos ?? +5sin ?? ]
which is the required solution of given equation.
7.4 Solve the following differential equations:
Page 4
Edurev123
7. Determination of Complete Solution
when one solution is known and Method of
Variation of Parameter
7.1 Using the method of variation of parameters, solve the second order
differential equation
?? ?? ?? ?? ?? ?? +?? ?? =?????? ?? ??
(2011: 15 Marks)
Solution:
The given equation is
?? 2
?? ?? ?? 2
+4?? =tan 2?? ? (?? 2
+4)?? =tan 2?? ,?? =
?? ????
?????????????????? ???????????????? ???? ?? 2
+?? =0
? ?? =±2?? ? Complementary function is
?? ?? =?? 1
cos 2?? +?? 2
sin 2?? , where ?? 1
and ?? 2
are arbitrary constants.
Let ?? ?? =?? cos 2?? +?? sin 2?? be particular integral of (i) where ?? and ?? are functions of ?? .
Then, ?? (?? )=cos 2?? ,?? (?? )=sin 2?? and ?? (?? )=tan 2?? ?? =?
4?? ?? (?? ,?? )
????
=
1
2
? cos 2?? ·tan 2?????? =
1
2
? sin 2?????? =-
1
4
cos 2?? ?? ?? =-
1
4
[log |sec 2?? +tan 2?? |-sin 2?? ]cos 2?? -
1
4
cos 2?? ·sin 2?? =-
1
4
(log |sec 2?? +tan 2?? |)cos 2??
? The required general solution is
?? =?? ?? +?? ?? =?? 1
cos 2?? +?? 2
sin 2?? -
1
4
(log |sec 2?? +tan 2?? |)cos 2??
7.2 Using the method of variation of parameter solve the differential equation :
?? ?? ?? ?? ?? ?? +?? ?? ?? =?????? ????
(2013 : 10 Marks)
Solution:
Homogeneous part is
?? 2
?? ?? ?? 2
+?? 2
?? =0?(?? 2
+?? 2
)?? =0
Auxiliary equation: ?? 2
+?? 2
=0??? =±????
? ?? =cos ????
?? =sin ????
are two independent solutions of homogeneous equation.
Let ?? =???? +???? complete the solution.
Then,
?? =?
-????
?? ?? =?
?? ?? ??
where
?? = Wronskian (?? ,?? )
=|
cos ???? sin ????
-?? sin ???? ?? cos ????
|=??
? ?? =-?
sin ???? ·sec ????
?? ???? =-
1
?? ?tan ????????
=
1
?? 2
?
-?? sin ????
cos ?? ?? ???? =
1
?? 2
ln (cos ???? )+?? 1
?? =?
cos ???? sec ????
?? ???? =
?? ?? +?? 2
? ?? =?? 1
cos ???? +
1
?? 2
cos ???? ×ln (cos (???? ))+?? 2
sin ???? +
?? ?? sin ????
7.3 Solve by the method of variation of parameters :
????
????
-?? ?? =?????? ??
(2014 : 10 Marks)
Solution:
Given that,
????
????
-5?? =sin ?? (??)
Differentiating (i) w.r.t. ?? , we get
?? 2
?? ?? ?? 2
-5
????
????
=cos ??
(?? 2
-5?? )?? =cos ??
Now consider auxiliary equation of (ii)
?? (?? -5)=0
? ?? =0,5
? C.F. of (ii) is. ?? ?? =?? 1
+?? 2
?? 5??
Let ?? ?? =???? +???? be a particular integral of (ii) where ?? and ?? are functions of ?? and ?? =
1,?? =?? 5??
Now ?? =|
?? ?? ?? '
?? '
|-|
1 ?? 5?? 0 5?? 5?? |=5?? 5?? ?0
? ?? =-?
?? ?? ?? =-?
?? 5?? ·cos ?????? 5?? 5?? =-
1
5
? cos ?????? =-
1
5
sin:??
?? =?
????
?? =?
1·cos ?? 5?? 5?? ???? =
1
5
? ?? -5?? cos ?????? =
1
5
?? -5?? 25+1
[-5cos ?? +sin ?? ]
=
1?? -5?? 5
-5cos ?? +sin ?? ]
26
[-5
? The general solution of (ii) is
?? =?? ?? +?? ?? ?? =?? 1
+?? 2
?? 5?? -
1
5
sin ?? ·1+
1
5
?? -5?? 26
[-5cos ?? +sin ?? ]
?? =?? 1
+?? 2
?? 5?? -
1
26
[cos ?? +5sin ?? ]
which is the required solution of given equation.
7.4 Solve the following differential equations:
?? ?? ?? ?? ?? ?? ?? -?? (?? +?? )
????
????
+(?? +?? )?? =(?? -?? )?? ?? ??
when ?? ?? is a solution to its corresponding homogeneous differential equation.
(2014 : 15 Marks)
Solution:
Given equation is
?? ?? ''
-2(?? +1)?? +(?? +2)?? =(?? -2)?? 2?? (??)
It is given that ?? ?? is a solution to its corresponding homogeneous differential equation,
i.e., ?? =?? =?? ?? is the part of C.F. of (i).
Let the general solution of (i) is ?? =???? . Then ?? is given by
?? 2
?? ?? ?? 2
+(?? +
2????
?? ????
)
????
????
=
?? ??
where
?? =
-2(1+?? )
?? ,?? =
?? +2
?? ,?? =
(?? -2)
?? ?? 2??
?
?? 2
?? ?? ?? 2
+(-
2
?? (?? +?? )+
2
?? ?? (?? ?? ))
????
????
=
(?? -2)
?? ?? 2?? ?? ??
?
?? 2
?? ?? ?? 2
+(-
2
?? -2+2)
????
????
=
(?? -2)
?? ?? ?? (???? )
Let
????
????
=?? ?
?? 2
?? ?? ?? 2
=
????
????
? from (ii), we have
????
????
+(-
2
?? )?? =
(?? -2)
?? ?? ??
which lies linear in ?? .
Page 5
Edurev123
7. Determination of Complete Solution
when one solution is known and Method of
Variation of Parameter
7.1 Using the method of variation of parameters, solve the second order
differential equation
?? ?? ?? ?? ?? ?? +?? ?? =?????? ?? ??
(2011: 15 Marks)
Solution:
The given equation is
?? 2
?? ?? ?? 2
+4?? =tan 2?? ? (?? 2
+4)?? =tan 2?? ,?? =
?? ????
?????????????????? ???????????????? ???? ?? 2
+?? =0
? ?? =±2?? ? Complementary function is
?? ?? =?? 1
cos 2?? +?? 2
sin 2?? , where ?? 1
and ?? 2
are arbitrary constants.
Let ?? ?? =?? cos 2?? +?? sin 2?? be particular integral of (i) where ?? and ?? are functions of ?? .
Then, ?? (?? )=cos 2?? ,?? (?? )=sin 2?? and ?? (?? )=tan 2?? ?? =?
4?? ?? (?? ,?? )
????
=
1
2
? cos 2?? ·tan 2?????? =
1
2
? sin 2?????? =-
1
4
cos 2?? ?? ?? =-
1
4
[log |sec 2?? +tan 2?? |-sin 2?? ]cos 2?? -
1
4
cos 2?? ·sin 2?? =-
1
4
(log |sec 2?? +tan 2?? |)cos 2??
? The required general solution is
?? =?? ?? +?? ?? =?? 1
cos 2?? +?? 2
sin 2?? -
1
4
(log |sec 2?? +tan 2?? |)cos 2??
7.2 Using the method of variation of parameter solve the differential equation :
?? ?? ?? ?? ?? ?? +?? ?? ?? =?????? ????
(2013 : 10 Marks)
Solution:
Homogeneous part is
?? 2
?? ?? ?? 2
+?? 2
?? =0?(?? 2
+?? 2
)?? =0
Auxiliary equation: ?? 2
+?? 2
=0??? =±????
? ?? =cos ????
?? =sin ????
are two independent solutions of homogeneous equation.
Let ?? =???? +???? complete the solution.
Then,
?? =?
-????
?? ?? =?
?? ?? ??
where
?? = Wronskian (?? ,?? )
=|
cos ???? sin ????
-?? sin ???? ?? cos ????
|=??
? ?? =-?
sin ???? ·sec ????
?? ???? =-
1
?? ?tan ????????
=
1
?? 2
?
-?? sin ????
cos ?? ?? ???? =
1
?? 2
ln (cos ???? )+?? 1
?? =?
cos ???? sec ????
?? ???? =
?? ?? +?? 2
? ?? =?? 1
cos ???? +
1
?? 2
cos ???? ×ln (cos (???? ))+?? 2
sin ???? +
?? ?? sin ????
7.3 Solve by the method of variation of parameters :
????
????
-?? ?? =?????? ??
(2014 : 10 Marks)
Solution:
Given that,
????
????
-5?? =sin ?? (??)
Differentiating (i) w.r.t. ?? , we get
?? 2
?? ?? ?? 2
-5
????
????
=cos ??
(?? 2
-5?? )?? =cos ??
Now consider auxiliary equation of (ii)
?? (?? -5)=0
? ?? =0,5
? C.F. of (ii) is. ?? ?? =?? 1
+?? 2
?? 5??
Let ?? ?? =???? +???? be a particular integral of (ii) where ?? and ?? are functions of ?? and ?? =
1,?? =?? 5??
Now ?? =|
?? ?? ?? '
?? '
|-|
1 ?? 5?? 0 5?? 5?? |=5?? 5?? ?0
? ?? =-?
?? ?? ?? =-?
?? 5?? ·cos ?????? 5?? 5?? =-
1
5
? cos ?????? =-
1
5
sin:??
?? =?
????
?? =?
1·cos ?? 5?? 5?? ???? =
1
5
? ?? -5?? cos ?????? =
1
5
?? -5?? 25+1
[-5cos ?? +sin ?? ]
=
1?? -5?? 5
-5cos ?? +sin ?? ]
26
[-5
? The general solution of (ii) is
?? =?? ?? +?? ?? ?? =?? 1
+?? 2
?? 5?? -
1
5
sin ?? ·1+
1
5
?? -5?? 26
[-5cos ?? +sin ?? ]
?? =?? 1
+?? 2
?? 5?? -
1
26
[cos ?? +5sin ?? ]
which is the required solution of given equation.
7.4 Solve the following differential equations:
?? ?? ?? ?? ?? ?? ?? -?? (?? +?? )
????
????
+(?? +?? )?? =(?? -?? )?? ?? ??
when ?? ?? is a solution to its corresponding homogeneous differential equation.
(2014 : 15 Marks)
Solution:
Given equation is
?? ?? ''
-2(?? +1)?? +(?? +2)?? =(?? -2)?? 2?? (??)
It is given that ?? ?? is a solution to its corresponding homogeneous differential equation,
i.e., ?? =?? =?? ?? is the part of C.F. of (i).
Let the general solution of (i) is ?? =???? . Then ?? is given by
?? 2
?? ?? ?? 2
+(?? +
2????
?? ????
)
????
????
=
?? ??
where
?? =
-2(1+?? )
?? ,?? =
?? +2
?? ,?? =
(?? -2)
?? ?? 2??
?
?? 2
?? ?? ?? 2
+(-
2
?? (?? +?? )+
2
?? ?? (?? ?? ))
????
????
=
(?? -2)
?? ?? 2?? ?? ??
?
?? 2
?? ?? ?? 2
+(-
2
?? -2+2)
????
????
=
(?? -2)
?? ?? ?? (???? )
Let
????
????
=?? ?
?? 2
?? ?? ?? 2
=
????
????
? from (ii), we have
????
????
+(-
2
?? )?? =
(?? -2)
?? ?? ??
which lies linear in ?? .
I.F. =?? -?
2
?? ????
=?? -(2log ?? )
=?? log ?? -2
=?? -2
=
1
?? 2
?? (|?? ) =? (
?? -2
?? )?? ?? ·???? +?? 1
?? (
1
?? 2
) =? (
?? -2
?? )?? ?? ·
1
?? 2
???? +?? 1
=? ?? -2
?? ?? ???? -2? ?? -3
?? ?? ???? +?? 1
=?? -2
?? ?? -? (-2)?? -3
?? ?? -2? ?? -3
?? ?? ???? +?? 1
?? ?? 2
=?? -2
?? ?? +2? ?? -3
?? ?? ???? -2? ?? -3
?? ?? ???? +?? 1
?? =?? ?? +?? 1
?? 2
???? =(?? ?? +?? 1
?? 2
)????
?? =?? ?? +
1
3
?? 1
?? 3
+?? 2
? ?? =???? =?? ?? (?? ?? +
1
3
?? 1
?? 3
+?? 2
) which is the required solution.
7.5 Using the method of variation of parameters, solve the ????
(?? ?? +?? ?? +?? )?? =?? -?? ?????? (?? )
(2016 : 15 Marks)
Solution:
For complimentary function (CF)
Auxiliary equation:
?? 2
+2?? +1 =0?(?? +1)
2
=0
?? =-1,-1
C.F. =(?? 1
+?? 2
?? )?? -?? =?? 1
?? -?? +?? 2
?? ?? -??
Taking ?? -?? and ?? ·?? -?? as ?? 1
and ?? 2
For Particular Integral (PI)
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FAQs on Determination of Complete Solution when one solution is known and Method of Variation of Parameter - Mathematics Optional Notes for UPSC
| 1. How can one determine the complete solution when one solution is known? |  |
Ans. One can determine the complete solution by using the method of Variation of Parameters, which involves finding a particular solution by assuming it to be of a specific form and then solving for the unknown parameters.
| 2. What is the significance of knowing one solution in finding the complete solution? | |
Ans. Knowing one solution helps in finding the complete solution as it provides a starting point for applying the method of Variation of Parameters to determine the general solution to the differential equation.
| 3. Can the method of Variation of Parameters be used for all types of differential equations? | |
Ans. The method of Variation of Parameters is typically used for linear differential equations, especially when finding the complete solution when one solution is known.
| 4. How does the method of Variation of Parameters differ from other methods of solving differential equations? | |
Ans. The method of Variation of Parameters differs from other methods as it involves assuming a particular solution form and solving for the unknown parameters, whereas other methods may involve different techniques such as separation of variables or substitution.
| 5. Are there any limitations to using the method of Variation of Parameters for finding the complete solution? | |
Ans. While the method of Variation of Parameters is effective for finding the complete solution when one solution is known, it may become complex for higher-order differential equations or equations with non-constant coefficients. In such cases, other methods may be more suitable.
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Determination of Complete Solution when one solution is known and Method of Variation of Parameter UPSC Questions
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The content of Determination of Complete Solution when one solution is known and Method of Variation of Parameter is prepared as per the latest UPSC syllabus.
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