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Linear P.D.E. of Second Order with Constant Coefficient | Mathematics Optional Notes for UPSC PDF Download

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 Page 2


 At ?? =0, ?? 0
=
-?? 2
4
?
-?? 2
4
=
-?? 2
4
+?? 5
??? 5
=0
?? =
-?? 2
4
?? 2?? (?????? ) 
From (x), (xi) and (xii), characteristics are 
?? =?? (2-?? ?? )
?? =v2?? (1-?? ?? )
?? =
-?? 2
4
?? 2?? 
To find the equation of integral surface, we eliminate ?? and ?? ?? from (x) and (xi) and put 
their values in (xii). 
and 
?? ?? ?=
v2(?? -v2?? )
v2?? -?? ?? ?=
v2?? -?? v2
 
Using these values of ?? and ?? ?? , we get 
?? =
-?? 2
4
?? 2?? =-
(v2?? -?? )
2
2
×
(v2?? -2?? )
2
(v2?? -?? )
2
???? =-
2
?? (?? -v2?? )
2
???? =-(?? -v2?? )
2
, which is the required surface. 
 
Edurev123 
4. Linear P.D.E. of Second Order with 
Constant Coefficient 
4.1 Solve : 
(?? ?? -?? ?? '
-?? ?? ?? )?? =(?? ?? ?? +???? -?? ?? )?????? ????? -?????? ????? 
where ?? and ?? '
 represent 
?? ?? ?? and 
?? ?? ?? . 
(2009 : 15 Marks) 
Page 3


 At ?? =0, ?? 0
=
-?? 2
4
?
-?? 2
4
=
-?? 2
4
+?? 5
??? 5
=0
?? =
-?? 2
4
?? 2?? (?????? ) 
From (x), (xi) and (xii), characteristics are 
?? =?? (2-?? ?? )
?? =v2?? (1-?? ?? )
?? =
-?? 2
4
?? 2?? 
To find the equation of integral surface, we eliminate ?? and ?? ?? from (x) and (xi) and put 
their values in (xii). 
and 
?? ?? ?=
v2(?? -v2?? )
v2?? -?? ?? ?=
v2?? -?? v2
 
Using these values of ?? and ?? ?? , we get 
?? =
-?? 2
4
?? 2?? =-
(v2?? -?? )
2
2
×
(v2?? -2?? )
2
(v2?? -?? )
2
???? =-
2
?? (?? -v2?? )
2
???? =-(?? -v2?? )
2
, which is the required surface. 
 
Edurev123 
4. Linear P.D.E. of Second Order with 
Constant Coefficient 
4.1 Solve : 
(?? ?? -?? ?? '
-?? ?? ?? )?? =(?? ?? ?? +???? -?? ?? )?????? ????? -?????? ????? 
where ?? and ?? '
 represent 
?? ?? ?? and 
?? ?? ?? . 
(2009 : 15 Marks) 
Solution: 
Clearly the Auxiliary equation is 
?? 2
-?? -2 =0
? ?? 2
-2?? +?? -2 =0
? ?? (?? -2)+1(?? -2) =0
? ?? =-1,2
?? ?? =?? 1
(?? +2?? )+?? 2
(?? -?? )
 
where ?? ?
1
 and ?? 2
 are Arbitrary function. 
Particular Integral 
=
1
(?? +?? '
)(?? -2?? '
)
[(2?? 2
+???? -2?? 2
)sin????? -cos????? ] 
Putting ?? -?? =?? we get ?? =?? +?? 
?=
1
(?? -2?? '
)
??[2?? 2
+?? (?? +?? )-2(?? +?? )
2
sin??? (?? +?? )-cos??? (?? +?? )]????
?=
1
(?? -2?? '
)
??[2?? 2
+(?? +?? )[?? -2?? -2?? |sin?(?? 2
+???? )-cos?(?? '
+???? )]????
?=
1
(?? -2?? '
)
??[2?? 2
+(?? +?? )(-?? -2?? )sin?(?? 2
+???? )-cos?(?? 2
+???? )]????
=
1
(?? -2?? '
)
??[2?? +?? )(?? -?? )sin?(?? 2
+???? )-cos?(?? 2
+???? )]????
?=
1
(?? -2?? '
)
??-(2?? +?? )(?? -?? )
cos?(?? 2
+???? )
(2?? +?? )
+??cos?(?? 2
+???? )???? -??cos?(?? 2
+???? )????
=
1
(?? -2?? '
)
(?? -2?? )cos????? ](?? '
-2?? -2?? )cos??? (?? '
-2?? )???? -?? =?? +2?? ????(?? '
-4?? )cos?(?? '
?? -2?? 2
)????
??
[sin?(?? '
?? -2?? 2
)]
(?? '
-4?? )
(?? '
-4?? )
?=sin?[(?? +2?? )?? -2?? 2
]
?=sin?????
 
Clearly, the required solution is ?? =?? 1
+?? 2
 
4.2 Solve the PDE: 
(?? ?? -?? '
)(?? -?? ?? '
)?? =?? ?? ?? +?? +???? 
(2010 : 12 Marks) 
Solution: 
Page 4


 At ?? =0, ?? 0
=
-?? 2
4
?
-?? 2
4
=
-?? 2
4
+?? 5
??? 5
=0
?? =
-?? 2
4
?? 2?? (?????? ) 
From (x), (xi) and (xii), characteristics are 
?? =?? (2-?? ?? )
?? =v2?? (1-?? ?? )
?? =
-?? 2
4
?? 2?? 
To find the equation of integral surface, we eliminate ?? and ?? ?? from (x) and (xi) and put 
their values in (xii). 
and 
?? ?? ?=
v2(?? -v2?? )
v2?? -?? ?? ?=
v2?? -?? v2
 
Using these values of ?? and ?? ?? , we get 
?? =
-?? 2
4
?? 2?? =-
(v2?? -?? )
2
2
×
(v2?? -2?? )
2
(v2?? -?? )
2
???? =-
2
?? (?? -v2?? )
2
???? =-(?? -v2?? )
2
, which is the required surface. 
 
Edurev123 
4. Linear P.D.E. of Second Order with 
Constant Coefficient 
4.1 Solve : 
(?? ?? -?? ?? '
-?? ?? ?? )?? =(?? ?? ?? +???? -?? ?? )?????? ????? -?????? ????? 
where ?? and ?? '
 represent 
?? ?? ?? and 
?? ?? ?? . 
(2009 : 15 Marks) 
Solution: 
Clearly the Auxiliary equation is 
?? 2
-?? -2 =0
? ?? 2
-2?? +?? -2 =0
? ?? (?? -2)+1(?? -2) =0
? ?? =-1,2
?? ?? =?? 1
(?? +2?? )+?? 2
(?? -?? )
 
where ?? ?
1
 and ?? 2
 are Arbitrary function. 
Particular Integral 
=
1
(?? +?? '
)(?? -2?? '
)
[(2?? 2
+???? -2?? 2
)sin????? -cos????? ] 
Putting ?? -?? =?? we get ?? =?? +?? 
?=
1
(?? -2?? '
)
??[2?? 2
+?? (?? +?? )-2(?? +?? )
2
sin??? (?? +?? )-cos??? (?? +?? )]????
?=
1
(?? -2?? '
)
??[2?? 2
+(?? +?? )[?? -2?? -2?? |sin?(?? 2
+???? )-cos?(?? '
+???? )]????
?=
1
(?? -2?? '
)
??[2?? 2
+(?? +?? )(-?? -2?? )sin?(?? 2
+???? )-cos?(?? 2
+???? )]????
=
1
(?? -2?? '
)
??[2?? +?? )(?? -?? )sin?(?? 2
+???? )-cos?(?? 2
+???? )]????
?=
1
(?? -2?? '
)
??-(2?? +?? )(?? -?? )
cos?(?? 2
+???? )
(2?? +?? )
+??cos?(?? 2
+???? )???? -??cos?(?? 2
+???? )????
=
1
(?? -2?? '
)
(?? -2?? )cos????? ](?? '
-2?? -2?? )cos??? (?? '
-2?? )???? -?? =?? +2?? ????(?? '
-4?? )cos?(?? '
?? -2?? 2
)????
??
[sin?(?? '
?? -2?? 2
)]
(?? '
-4?? )
(?? '
-4?? )
?=sin?[(?? +2?? )?? -2?? 2
]
?=sin?????
 
Clearly, the required solution is ?? =?? 1
+?? 2
 
4.2 Solve the PDE: 
(?? ?? -?? '
)(?? -?? ?? '
)?? =?? ?? ?? +?? +???? 
(2010 : 12 Marks) 
Solution: 
Given, the equation is 
(?? 2
-?? '
)(?? -2?? '
)?? =?? 2?? +?? +???? 
Complementary Function : 
The auxiliary equation is : 
(?? 2
-1)(?? -2) =0
? ?? =±1
? Solution is ?? ?? =?? 1
 
Particular Integral : 
?? ?? =
1
(?? 2
-?? '
)(?? -2?? '
)
?? 2?? +?? +
1
(?? 2
-?? '
)(?? -2?? '
)
????
? ?? ?? =
1
(2
2
-1)(?? -2?? '
)
?? 2?? +?? +
1
(?? 2
-?? '
)·?? ×
1
(1-
2?? '
?? )
????
? ?? ?? =
1
3
1
(?? -2?? '
)
?? 2?? +?? +
1
(?? 2
-?? '
)
·
1
?? (1+
2?? '
?? )????
?? ?? ?? =
1
3
?? 1·1!
?? 2?? +?? +
1
(?? 2
-?? '
)
·
1
?? (???? +?? 2
)
?? ?? ?? =
?? 3
?? 2?? +?? +
1
(?? 2
-?? '
)
×(
?? 2
?? 2
+
?? 3
3
)
 
? ?? ?? =
?? 3
?? 2?? +?? +
1
?? 2
×(1-
?? '
?? 2
)
-1
(?? 2
?? +
?? 3
3
)
? ?? ?? =
?? 3
?? 2?? +?? +
1
?? 2
×(1+
?? '
?? 2
)(?? 2
?? +
?? 3
3
)
? ?? ?? =
?? 3
?? 2?? +?? +
1
?? 2
×(?? 2
?? +
?? 3
3
+
?? 4
12
)
? ?? ?? =
?? 3
?? 2?? +?? +
1
?? (
?? 3
?? 3
+
?? 4
12
+
?? 5
60
)
? ?? ?? =
?? 3
?? 2?? +?? +
?? 4
?? 12
+
?? 5
60
+
?? 6
360
 Now, ?? =?? ?? +?? ?? ? ?? =?? 1
(?? +?? )+?? 2
(?? -?? )+?? 3
(?? +2?? )+
?? 3
?? 2?? +?? +
?? 4
?? 12
+
?? 5
60
+
?? 6
360
 
4.3 Solve the PDE 
(?? ?? -?? ?? +?? +?? ?? '
-?? )?? =?? (?? -?? )
-?? ?? ?? 
(2011 : 12 Marks) 
Page 5


 At ?? =0, ?? 0
=
-?? 2
4
?
-?? 2
4
=
-?? 2
4
+?? 5
??? 5
=0
?? =
-?? 2
4
?? 2?? (?????? ) 
From (x), (xi) and (xii), characteristics are 
?? =?? (2-?? ?? )
?? =v2?? (1-?? ?? )
?? =
-?? 2
4
?? 2?? 
To find the equation of integral surface, we eliminate ?? and ?? ?? from (x) and (xi) and put 
their values in (xii). 
and 
?? ?? ?=
v2(?? -v2?? )
v2?? -?? ?? ?=
v2?? -?? v2
 
Using these values of ?? and ?? ?? , we get 
?? =
-?? 2
4
?? 2?? =-
(v2?? -?? )
2
2
×
(v2?? -2?? )
2
(v2?? -?? )
2
???? =-
2
?? (?? -v2?? )
2
???? =-(?? -v2?? )
2
, which is the required surface. 
 
Edurev123 
4. Linear P.D.E. of Second Order with 
Constant Coefficient 
4.1 Solve : 
(?? ?? -?? ?? '
-?? ?? ?? )?? =(?? ?? ?? +???? -?? ?? )?????? ????? -?????? ????? 
where ?? and ?? '
 represent 
?? ?? ?? and 
?? ?? ?? . 
(2009 : 15 Marks) 
Solution: 
Clearly the Auxiliary equation is 
?? 2
-?? -2 =0
? ?? 2
-2?? +?? -2 =0
? ?? (?? -2)+1(?? -2) =0
? ?? =-1,2
?? ?? =?? 1
(?? +2?? )+?? 2
(?? -?? )
 
where ?? ?
1
 and ?? 2
 are Arbitrary function. 
Particular Integral 
=
1
(?? +?? '
)(?? -2?? '
)
[(2?? 2
+???? -2?? 2
)sin????? -cos????? ] 
Putting ?? -?? =?? we get ?? =?? +?? 
?=
1
(?? -2?? '
)
??[2?? 2
+?? (?? +?? )-2(?? +?? )
2
sin??? (?? +?? )-cos??? (?? +?? )]????
?=
1
(?? -2?? '
)
??[2?? 2
+(?? +?? )[?? -2?? -2?? |sin?(?? 2
+???? )-cos?(?? '
+???? )]????
?=
1
(?? -2?? '
)
??[2?? 2
+(?? +?? )(-?? -2?? )sin?(?? 2
+???? )-cos?(?? 2
+???? )]????
=
1
(?? -2?? '
)
??[2?? +?? )(?? -?? )sin?(?? 2
+???? )-cos?(?? 2
+???? )]????
?=
1
(?? -2?? '
)
??-(2?? +?? )(?? -?? )
cos?(?? 2
+???? )
(2?? +?? )
+??cos?(?? 2
+???? )???? -??cos?(?? 2
+???? )????
=
1
(?? -2?? '
)
(?? -2?? )cos????? ](?? '
-2?? -2?? )cos??? (?? '
-2?? )???? -?? =?? +2?? ????(?? '
-4?? )cos?(?? '
?? -2?? 2
)????
??
[sin?(?? '
?? -2?? 2
)]
(?? '
-4?? )
(?? '
-4?? )
?=sin?[(?? +2?? )?? -2?? 2
]
?=sin?????
 
Clearly, the required solution is ?? =?? 1
+?? 2
 
4.2 Solve the PDE: 
(?? ?? -?? '
)(?? -?? ?? '
)?? =?? ?? ?? +?? +???? 
(2010 : 12 Marks) 
Solution: 
Given, the equation is 
(?? 2
-?? '
)(?? -2?? '
)?? =?? 2?? +?? +???? 
Complementary Function : 
The auxiliary equation is : 
(?? 2
-1)(?? -2) =0
? ?? =±1
? Solution is ?? ?? =?? 1
 
Particular Integral : 
?? ?? =
1
(?? 2
-?? '
)(?? -2?? '
)
?? 2?? +?? +
1
(?? 2
-?? '
)(?? -2?? '
)
????
? ?? ?? =
1
(2
2
-1)(?? -2?? '
)
?? 2?? +?? +
1
(?? 2
-?? '
)·?? ×
1
(1-
2?? '
?? )
????
? ?? ?? =
1
3
1
(?? -2?? '
)
?? 2?? +?? +
1
(?? 2
-?? '
)
·
1
?? (1+
2?? '
?? )????
?? ?? ?? =
1
3
?? 1·1!
?? 2?? +?? +
1
(?? 2
-?? '
)
·
1
?? (???? +?? 2
)
?? ?? ?? =
?? 3
?? 2?? +?? +
1
(?? 2
-?? '
)
×(
?? 2
?? 2
+
?? 3
3
)
 
? ?? ?? =
?? 3
?? 2?? +?? +
1
?? 2
×(1-
?? '
?? 2
)
-1
(?? 2
?? +
?? 3
3
)
? ?? ?? =
?? 3
?? 2?? +?? +
1
?? 2
×(1+
?? '
?? 2
)(?? 2
?? +
?? 3
3
)
? ?? ?? =
?? 3
?? 2?? +?? +
1
?? 2
×(?? 2
?? +
?? 3
3
+
?? 4
12
)
? ?? ?? =
?? 3
?? 2?? +?? +
1
?? (
?? 3
?? 3
+
?? 4
12
+
?? 5
60
)
? ?? ?? =
?? 3
?? 2?? +?? +
?? 4
?? 12
+
?? 5
60
+
?? 6
360
 Now, ?? =?? ?? +?? ?? ? ?? =?? 1
(?? +?? )+?? 2
(?? -?? )+?? 3
(?? +2?? )+
?? 3
?? 2?? +?? +
?? 4
?? 12
+
?? 5
60
+
?? 6
360
 
4.3 Solve the PDE 
(?? ?? -?? ?? +?? +?? ?? '
-?? )?? =?? (?? -?? )
-?? ?? ?? 
(2011 : 12 Marks) 
Solution: 
The given partial differential equation is 
(?? 2
-?? 2
+?? +3?? -2)?? ?=?? ?? -?? -?? 2
?? 
? the compiementary function of (??) is 
?? ?? =?? -2?? ?? 1
(?? +?? )+?? ?? ?? 2
(?? -?? ) (???? ) 
The particular integral of (i) is 
?? ?? =
1
(?? -?? '
+2)(?? +?? '
-1)
(?? ?? -?? -?? 2
?? ) 
=
1
(?? -?? '
+2)(?? +?? '
-1)
?? ?? -?? -
1
(?? -?? '
+2)(?? +?? '
-1)
·?? 2
?? 
=
1
(1+1+2)(1-1-1)
?? ?? -?? -
1
?? 2
-?? 2
+?? +3?? '
-2
·?? 2
?? 
=
-1
4
?? ?? -?? +
1
2[1-
?? 2
-?? 2
+?? +3?? 2
]
·?? 2
?? 
=
-1
4
?? ?? -?? +
1
2
[1+(
?? 2
-?? 2
+?? +3?? '
2
)+(
?? 2
-?? 2
+?? +3?? '
2
)
2
+?..]?? 2
?? 
=
-1
4
?? ?? -?? +
1
2
[1+
?? 2
+
3?? '
2
+(
1
2
+
1
4
)?? 2
+(
9
4
-
1
2
)?? '2
+
3
2
?? ?? '
+
21
8
?? 2
?? '
-
23
8
?? ?? '2
…..]?? 2
?? 
=
-1
4
?? ?? -?? +
1
2
[?? 2
?? +???? +
3
2
?? 2
+
3
2
?? +3?? +
21
4
] 
??? =?? ?? +?? ?? =?? -2?? ?? 1
(?? +?? )+?? ?? ?? 2
(?? +?? )-
1
4
?? ?? -?? +
1
2
[?? 2
?? +???? +
3
2
?? 2
+
3
2
?? +3?? +
21
4
] 
4.4 Solve the partial differential equation: 
(?? -?? ?? '
)(?? -?? '
)
?? ?? =?? ?? +?? 
(2011 : 12 Marks) 
Solution:
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Dec 31, 2024 Last updated
Document Description: Linear P.D.E. of Second Order with Constant Coefficient for UPSC 2024 is part of Mathematics Optional Notes for UPSC preparation. The notes and questions for Linear P.D.E. of Second Order with Constant Coefficient have been prepared according to the UPSC exam syllabus. Information about Linear P.D.E. of Second Order with Constant Coefficient covers topics like and Linear P.D.E. of Second Order with Constant Coefficient Example, for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Linear P.D.E. of Second Order with Constant Coefficient.
Introduction of Linear P.D.E. of Second Order with Constant Coefficient in English is available as part of our Mathematics Optional Notes for UPSC for UPSC & Linear P.D.E. of Second Order with Constant Coefficient 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: Linear P.D.E. of Second Order with Constant Coefficient | Mathematics Optional Notes for UPSC
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