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Page 1
Edurev123
8. Laplace and inverse Laplace
Transformation and Properties
8.1 Find the inverse Laplace transform of ?? (?? )=???? (
?? +?? ?? +?? ) .
(2009: 20 Marks)
Solution:
Approach: Use the differentiation property of inverse Laplace transform.
Given:
?? (?? )=ln (
?? +1
?? +?? )
By differentiation property we have
?? -1
[?? (?? )] =
1
?? ?? -1
[?? -1
(?? )]
? ?? -1
[ln (
?? +1
?? +?? )] =
1
?? ?? -1
[
?? ????
ln (
?? +1
?? +?? )]
=
1
?? ?? -1
[
?? ????
ln (?? +1)-ln (?? +?? )]
=
1
?? ?? -1
[
1
?? +1
-
1
?? +?? ]
=
1
?? {?? -1
(
1
?? +1
)-?? -1
(
1
?? +?? )} (Linearity)
=
1
?? {?? -?? -?? -5?? }.
8.2 Use Laplace transform method to solve the following initial value problem:
?? ?? ?? ?? ?? ?? -?? ????
????
+?? =?? ?? ,?? (?? )=?? and
????
????
|
?? =?? =-??
(2011: 15 Marks)
Solution:
The given initial value problem is
Page 2
Edurev123
8. Laplace and inverse Laplace
Transformation and Properties
8.1 Find the inverse Laplace transform of ?? (?? )=???? (
?? +?? ?? +?? ) .
(2009: 20 Marks)
Solution:
Approach: Use the differentiation property of inverse Laplace transform.
Given:
?? (?? )=ln (
?? +1
?? +?? )
By differentiation property we have
?? -1
[?? (?? )] =
1
?? ?? -1
[?? -1
(?? )]
? ?? -1
[ln (
?? +1
?? +?? )] =
1
?? ?? -1
[
?? ????
ln (
?? +1
?? +?? )]
=
1
?? ?? -1
[
?? ????
ln (?? +1)-ln (?? +?? )]
=
1
?? ?? -1
[
1
?? +1
-
1
?? +?? ]
=
1
?? {?? -1
(
1
?? +1
)-?? -1
(
1
?? +?? )} (Linearity)
=
1
?? {?? -?? -?? -5?? }.
8.2 Use Laplace transform method to solve the following initial value problem:
?? ?? ?? ?? ?? ?? -?? ????
????
+?? =?? ?? ,?? (?? )=?? and
????
????
|
?? =?? =-??
(2011: 15 Marks)
Solution:
The given initial value problem is
?? 2
?? ?? ?? 2
-2
????
????
+?? =?? ?? ,?? (0)=2 (??)
(
????
????
)
?? =0
=-1
Taking Laplace transform of both sides of (i), we get
?? (
?? 2
?? ?? ?? 2
-2
????
????
+?? )=?? [?? '
]
? ?? {?? ''
(?? )}-2?? {?? '
(?? )}+?? {?? (?? )}=?? {?? ?? }
? ?? 2
?? {?? (?? )}-???? (0)-?? '
(0)-2[???? {?? (?? )}-?? (0)]+?? {?? (?? )}=
1
??
? ?? (?? )][?? 2
-2?? +1]-?? (2)-(-1)+2(2)=
1
??
? ?? {?? (?? )}(?? -1)
2
=
1
?? +2?? -5
=
1+2?? 2
-5?? ??
? ?? {?? (?? )} =
1+2?? 2
-5?? ?? (?? -1)
2
=
1
?? -1
-
2
(?? -1)
2
+
1
?? … (ii) (By using Partial Fractions)
Taking inverse Laplace of (ii), we get
?? (?? )=?? -1
{
1
?? -1
-
2
(?? -1)
2
+
1
?? }
? ?? (?? )=?? ?? -2?? ?? ?? +1
8.3 Using Laplace transforms, solve the initial value problem
?? ''
+?? ?? +?? =?? -?? ,?? (?? )=-?? ,?? (?? )=??
(2012: 12 Marks)
Solution:
Given :
?? '
+2?? +?? =?? -?? (??)
?? (0) =-1,?? '
(0)=1
Page 3
Edurev123
8. Laplace and inverse Laplace
Transformation and Properties
8.1 Find the inverse Laplace transform of ?? (?? )=???? (
?? +?? ?? +?? ) .
(2009: 20 Marks)
Solution:
Approach: Use the differentiation property of inverse Laplace transform.
Given:
?? (?? )=ln (
?? +1
?? +?? )
By differentiation property we have
?? -1
[?? (?? )] =
1
?? ?? -1
[?? -1
(?? )]
? ?? -1
[ln (
?? +1
?? +?? )] =
1
?? ?? -1
[
?? ????
ln (
?? +1
?? +?? )]
=
1
?? ?? -1
[
?? ????
ln (?? +1)-ln (?? +?? )]
=
1
?? ?? -1
[
1
?? +1
-
1
?? +?? ]
=
1
?? {?? -1
(
1
?? +1
)-?? -1
(
1
?? +?? )} (Linearity)
=
1
?? {?? -?? -?? -5?? }.
8.2 Use Laplace transform method to solve the following initial value problem:
?? ?? ?? ?? ?? ?? -?? ????
????
+?? =?? ?? ,?? (?? )=?? and
????
????
|
?? =?? =-??
(2011: 15 Marks)
Solution:
The given initial value problem is
?? 2
?? ?? ?? 2
-2
????
????
+?? =?? ?? ,?? (0)=2 (??)
(
????
????
)
?? =0
=-1
Taking Laplace transform of both sides of (i), we get
?? (
?? 2
?? ?? ?? 2
-2
????
????
+?? )=?? [?? '
]
? ?? {?? ''
(?? )}-2?? {?? '
(?? )}+?? {?? (?? )}=?? {?? ?? }
? ?? 2
?? {?? (?? )}-???? (0)-?? '
(0)-2[???? {?? (?? )}-?? (0)]+?? {?? (?? )}=
1
??
? ?? (?? )][?? 2
-2?? +1]-?? (2)-(-1)+2(2)=
1
??
? ?? {?? (?? )}(?? -1)
2
=
1
?? +2?? -5
=
1+2?? 2
-5?? ??
? ?? {?? (?? )} =
1+2?? 2
-5?? ?? (?? -1)
2
=
1
?? -1
-
2
(?? -1)
2
+
1
?? … (ii) (By using Partial Fractions)
Taking inverse Laplace of (ii), we get
?? (?? )=?? -1
{
1
?? -1
-
2
(?? -1)
2
+
1
?? }
? ?? (?? )=?? ?? -2?? ?? ?? +1
8.3 Using Laplace transforms, solve the initial value problem
?? ''
+?? ?? +?? =?? -?? ,?? (?? )=-?? ,?? (?? )=??
(2012: 12 Marks)
Solution:
Given :
?? '
+2?? +?? =?? -?? (??)
?? (0) =-1,?? '
(0)=1
Taking Laplace transform of both sides of (i),
?? (?? '
(?? ))+2?? (?? (?? ))+?? (?? )=?? (?? -?? )
??? 2
?? (?? )-???? (0)-?? (0)+2(???? (?? )-?? (0))+?? (?? )=
1
?? +1
??? (?? )(?? 2
+2?? +1)-?? (-1)-1-2(-1)=
1
?? +1
? ?? (?? )(?? +1)
2
+(?? +1)=
1
?? +1
? ?? (?? )=
1
(?? +1)
3
-
1
(?? +1)
(???? )
Taking inverse Laplace transform on both sides of (ii), we have
?? =?? -1
[
1
(?? +1)
3
-
1
(?? +1)
]
? ?? =?? -?? ·
?? 2
2
-?? -??
8.4 By using Laplace transform method solve the differential equation :
(?? ?? +?? ?? )?? =?? ?????? (???? +?? ) , ?? ?? =
?? ?? ?? ?? ?? subject to the initial conditions ?? =?? and
????
????
=??
at ?? =?? in which ?? ,?? and ?? are constants.
(2013 : 15 Marks)
Solution:
(?? 2
+?? 2
)?? =?? sin (???? +?? )
Taking Laplace transform on both sides
?? (
?? 2
?? ?? ?? 2
)+?? 2
?? (?? )=???? [sin ???? cos ?? +cos ???? sin ?? ]
? ?? 2
?? (?? )-???? (0)-?? '
(0)+?? 2
?? (?? )=?? [
?? ?? 2
+?? 2
cos ?? +
?? ?? 2
+?? 2
sin ?? ]
? ?? (?? )=?? [
?? (?? 2
+?? 2
)
2
cos ?? +
?? (?? 2
+?? 2
)
2
sin ?? ]
? ?? =?? {?? -1
[
?? (?? 2
+?? 2
)
2
]cos ?? +?? -1
[
?? (?? 2
+?? 2
)
2
]sin ?? }
Page 4
Edurev123
8. Laplace and inverse Laplace
Transformation and Properties
8.1 Find the inverse Laplace transform of ?? (?? )=???? (
?? +?? ?? +?? ) .
(2009: 20 Marks)
Solution:
Approach: Use the differentiation property of inverse Laplace transform.
Given:
?? (?? )=ln (
?? +1
?? +?? )
By differentiation property we have
?? -1
[?? (?? )] =
1
?? ?? -1
[?? -1
(?? )]
? ?? -1
[ln (
?? +1
?? +?? )] =
1
?? ?? -1
[
?? ????
ln (
?? +1
?? +?? )]
=
1
?? ?? -1
[
?? ????
ln (?? +1)-ln (?? +?? )]
=
1
?? ?? -1
[
1
?? +1
-
1
?? +?? ]
=
1
?? {?? -1
(
1
?? +1
)-?? -1
(
1
?? +?? )} (Linearity)
=
1
?? {?? -?? -?? -5?? }.
8.2 Use Laplace transform method to solve the following initial value problem:
?? ?? ?? ?? ?? ?? -?? ????
????
+?? =?? ?? ,?? (?? )=?? and
????
????
|
?? =?? =-??
(2011: 15 Marks)
Solution:
The given initial value problem is
?? 2
?? ?? ?? 2
-2
????
????
+?? =?? ?? ,?? (0)=2 (??)
(
????
????
)
?? =0
=-1
Taking Laplace transform of both sides of (i), we get
?? (
?? 2
?? ?? ?? 2
-2
????
????
+?? )=?? [?? '
]
? ?? {?? ''
(?? )}-2?? {?? '
(?? )}+?? {?? (?? )}=?? {?? ?? }
? ?? 2
?? {?? (?? )}-???? (0)-?? '
(0)-2[???? {?? (?? )}-?? (0)]+?? {?? (?? )}=
1
??
? ?? (?? )][?? 2
-2?? +1]-?? (2)-(-1)+2(2)=
1
??
? ?? {?? (?? )}(?? -1)
2
=
1
?? +2?? -5
=
1+2?? 2
-5?? ??
? ?? {?? (?? )} =
1+2?? 2
-5?? ?? (?? -1)
2
=
1
?? -1
-
2
(?? -1)
2
+
1
?? … (ii) (By using Partial Fractions)
Taking inverse Laplace of (ii), we get
?? (?? )=?? -1
{
1
?? -1
-
2
(?? -1)
2
+
1
?? }
? ?? (?? )=?? ?? -2?? ?? ?? +1
8.3 Using Laplace transforms, solve the initial value problem
?? ''
+?? ?? +?? =?? -?? ,?? (?? )=-?? ,?? (?? )=??
(2012: 12 Marks)
Solution:
Given :
?? '
+2?? +?? =?? -?? (??)
?? (0) =-1,?? '
(0)=1
Taking Laplace transform of both sides of (i),
?? (?? '
(?? ))+2?? (?? (?? ))+?? (?? )=?? (?? -?? )
??? 2
?? (?? )-???? (0)-?? (0)+2(???? (?? )-?? (0))+?? (?? )=
1
?? +1
??? (?? )(?? 2
+2?? +1)-?? (-1)-1-2(-1)=
1
?? +1
? ?? (?? )(?? +1)
2
+(?? +1)=
1
?? +1
? ?? (?? )=
1
(?? +1)
3
-
1
(?? +1)
(???? )
Taking inverse Laplace transform on both sides of (ii), we have
?? =?? -1
[
1
(?? +1)
3
-
1
(?? +1)
]
? ?? =?? -?? ·
?? 2
2
-?? -??
8.4 By using Laplace transform method solve the differential equation :
(?? ?? +?? ?? )?? =?? ?????? (???? +?? ) , ?? ?? =
?? ?? ?? ?? ?? subject to the initial conditions ?? =?? and
????
????
=??
at ?? =?? in which ?? ,?? and ?? are constants.
(2013 : 15 Marks)
Solution:
(?? 2
+?? 2
)?? =?? sin (???? +?? )
Taking Laplace transform on both sides
?? (
?? 2
?? ?? ?? 2
)+?? 2
?? (?? )=???? [sin ???? cos ?? +cos ???? sin ?? ]
? ?? 2
?? (?? )-???? (0)-?? '
(0)+?? 2
?? (?? )=?? [
?? ?? 2
+?? 2
cos ?? +
?? ?? 2
+?? 2
sin ?? ]
? ?? (?? )=?? [
?? (?? 2
+?? 2
)
2
cos ?? +
?? (?? 2
+?? 2
)
2
sin ?? ]
? ?? =?? {?? -1
[
?? (?? 2
+?? 2
)
2
]cos ?? +?? -1
[
?? (?? 2
+?? 2
)
2
]sin ?? }
Now
?? ????
1
(?? 2
+?? 2
)
=-
2?? (?? 2
+?? 2
)
2
? ?? -1
(
2?? (?? 2
+?? 2
)
2
)=(-1)?? ?? -1
(
1
?? 2
+?? 2
)
=
-?? ?? ·sin ????
? ?? -1
(
?? (?? 2
+?? 2
)
2
)=
?? 2?? sin ????
Let
?? (?? ) =
1
?? 2
+?? 2
;?? (?? )=
1
?? 2
+?? 2
?? (?? ) =?? -1
(?? (?? ))=
1
?? sin ????
?? (?? ) =?? -1
(?? (?? ))=
1
?? sin ????
?? -1
(?? (?? )?? (?? )) =?? *?? =??
?? 0
??? (?? )?? (?? -?? )????
=
1
?? 2
??
?? 0
?sin ???? sin ?? (?? -?? )????
=
1
2?? 2
??
?? 0
?(cos ?? (?? -2?? )-cos ???? )????
=
1
2?? 2
[
sin ?? (?? -2?? )
-2?? -?? cos ???? ]
0
?? =
1
2?? 2
[
-sin ????
-2?? +
sin ????
2?? -?? cos ???? ]=
1
2?? 2
[
sin ????
?? -?? cos ???? ]
?? =?? {
1
2?? [
sin ????
?? -?? cos ???? ]cos ?? +(
?? 2?? sin ???? sin ?? )}
=?? {
1
2?? 2
sin ???? cos ?? -
?? 2?? cos (???? +?? )}
=
?? 2?? 2
{sin ???? cos ?? -???? cos (???? +?? )}
8.5 Solve the initial value problem
?? ?? ?? ?? ?? ?? +?? =?? ?? -?? ?? ?????? ?? ,?? (?? )=?? ,?? (?? )=??
by using Laplace-transform.
(2014 : 20 Marks)
Solution:
Given equation is
Page 5
Edurev123
8. Laplace and inverse Laplace
Transformation and Properties
8.1 Find the inverse Laplace transform of ?? (?? )=???? (
?? +?? ?? +?? ) .
(2009: 20 Marks)
Solution:
Approach: Use the differentiation property of inverse Laplace transform.
Given:
?? (?? )=ln (
?? +1
?? +?? )
By differentiation property we have
?? -1
[?? (?? )] =
1
?? ?? -1
[?? -1
(?? )]
? ?? -1
[ln (
?? +1
?? +?? )] =
1
?? ?? -1
[
?? ????
ln (
?? +1
?? +?? )]
=
1
?? ?? -1
[
?? ????
ln (?? +1)-ln (?? +?? )]
=
1
?? ?? -1
[
1
?? +1
-
1
?? +?? ]
=
1
?? {?? -1
(
1
?? +1
)-?? -1
(
1
?? +?? )} (Linearity)
=
1
?? {?? -?? -?? -5?? }.
8.2 Use Laplace transform method to solve the following initial value problem:
?? ?? ?? ?? ?? ?? -?? ????
????
+?? =?? ?? ,?? (?? )=?? and
????
????
|
?? =?? =-??
(2011: 15 Marks)
Solution:
The given initial value problem is
?? 2
?? ?? ?? 2
-2
????
????
+?? =?? ?? ,?? (0)=2 (??)
(
????
????
)
?? =0
=-1
Taking Laplace transform of both sides of (i), we get
?? (
?? 2
?? ?? ?? 2
-2
????
????
+?? )=?? [?? '
]
? ?? {?? ''
(?? )}-2?? {?? '
(?? )}+?? {?? (?? )}=?? {?? ?? }
? ?? 2
?? {?? (?? )}-???? (0)-?? '
(0)-2[???? {?? (?? )}-?? (0)]+?? {?? (?? )}=
1
??
? ?? (?? )][?? 2
-2?? +1]-?? (2)-(-1)+2(2)=
1
??
? ?? {?? (?? )}(?? -1)
2
=
1
?? +2?? -5
=
1+2?? 2
-5?? ??
? ?? {?? (?? )} =
1+2?? 2
-5?? ?? (?? -1)
2
=
1
?? -1
-
2
(?? -1)
2
+
1
?? … (ii) (By using Partial Fractions)
Taking inverse Laplace of (ii), we get
?? (?? )=?? -1
{
1
?? -1
-
2
(?? -1)
2
+
1
?? }
? ?? (?? )=?? ?? -2?? ?? ?? +1
8.3 Using Laplace transforms, solve the initial value problem
?? ''
+?? ?? +?? =?? -?? ,?? (?? )=-?? ,?? (?? )=??
(2012: 12 Marks)
Solution:
Given :
?? '
+2?? +?? =?? -?? (??)
?? (0) =-1,?? '
(0)=1
Taking Laplace transform of both sides of (i),
?? (?? '
(?? ))+2?? (?? (?? ))+?? (?? )=?? (?? -?? )
??? 2
?? (?? )-???? (0)-?? (0)+2(???? (?? )-?? (0))+?? (?? )=
1
?? +1
??? (?? )(?? 2
+2?? +1)-?? (-1)-1-2(-1)=
1
?? +1
? ?? (?? )(?? +1)
2
+(?? +1)=
1
?? +1
? ?? (?? )=
1
(?? +1)
3
-
1
(?? +1)
(???? )
Taking inverse Laplace transform on both sides of (ii), we have
?? =?? -1
[
1
(?? +1)
3
-
1
(?? +1)
]
? ?? =?? -?? ·
?? 2
2
-?? -??
8.4 By using Laplace transform method solve the differential equation :
(?? ?? +?? ?? )?? =?? ?????? (???? +?? ) , ?? ?? =
?? ?? ?? ?? ?? subject to the initial conditions ?? =?? and
????
????
=??
at ?? =?? in which ?? ,?? and ?? are constants.
(2013 : 15 Marks)
Solution:
(?? 2
+?? 2
)?? =?? sin (???? +?? )
Taking Laplace transform on both sides
?? (
?? 2
?? ?? ?? 2
)+?? 2
?? (?? )=???? [sin ???? cos ?? +cos ???? sin ?? ]
? ?? 2
?? (?? )-???? (0)-?? '
(0)+?? 2
?? (?? )=?? [
?? ?? 2
+?? 2
cos ?? +
?? ?? 2
+?? 2
sin ?? ]
? ?? (?? )=?? [
?? (?? 2
+?? 2
)
2
cos ?? +
?? (?? 2
+?? 2
)
2
sin ?? ]
? ?? =?? {?? -1
[
?? (?? 2
+?? 2
)
2
]cos ?? +?? -1
[
?? (?? 2
+?? 2
)
2
]sin ?? }
Now
?? ????
1
(?? 2
+?? 2
)
=-
2?? (?? 2
+?? 2
)
2
? ?? -1
(
2?? (?? 2
+?? 2
)
2
)=(-1)?? ?? -1
(
1
?? 2
+?? 2
)
=
-?? ?? ·sin ????
? ?? -1
(
?? (?? 2
+?? 2
)
2
)=
?? 2?? sin ????
Let
?? (?? ) =
1
?? 2
+?? 2
;?? (?? )=
1
?? 2
+?? 2
?? (?? ) =?? -1
(?? (?? ))=
1
?? sin ????
?? (?? ) =?? -1
(?? (?? ))=
1
?? sin ????
?? -1
(?? (?? )?? (?? )) =?? *?? =??
?? 0
??? (?? )?? (?? -?? )????
=
1
?? 2
??
?? 0
?sin ???? sin ?? (?? -?? )????
=
1
2?? 2
??
?? 0
?(cos ?? (?? -2?? )-cos ???? )????
=
1
2?? 2
[
sin ?? (?? -2?? )
-2?? -?? cos ???? ]
0
?? =
1
2?? 2
[
-sin ????
-2?? +
sin ????
2?? -?? cos ???? ]=
1
2?? 2
[
sin ????
?? -?? cos ???? ]
?? =?? {
1
2?? [
sin ????
?? -?? cos ???? ]cos ?? +(
?? 2?? sin ???? sin ?? )}
=?? {
1
2?? 2
sin ???? cos ?? -
?? 2?? cos (???? +?? )}
=
?? 2?? 2
{sin ???? cos ?? -???? cos (???? +?? )}
8.5 Solve the initial value problem
?? ?? ?? ?? ?? ?? +?? =?? ?? -?? ?? ?????? ?? ,?? (?? )=?? ,?? (?? )=??
by using Laplace-transform.
(2014 : 20 Marks)
Solution:
Given equation is
?? 2
?? ?? ?? 2
+?? =8?? -2?? sin ??
? ?? '
+?? =8?? -2?? sin?? (??)
? Taking Laplace transform of both sides of (??) , we get
?? (?? '
)+?? (?? )=8?? (?? -2?? sin ?? )
? ?? 2
?? {(?? )}-???? (0)-?? '
(0)+?? (?? (?? )=
8
(?? +2)
2
+1
? ?? 2
?? {(?? )}+?? {?? (?? )}=
8
?? 2
+4?? +5
? ?? (?? (?? ))(?? 2
+1)=
8
?? 2
+4?? +5
? ?? {?? (?? )}=
8
(?? 2
+1)(?? 2
+4?? +5)
? ?? (?? )=?? -1
8
(?? 2
+1)(?? 2
+4?? +5)
?? (?? )=?? -1
[
-?? +1
?? 2
+1
+
?? +3
?? 2
+4?? +5
]
=?? -1
(
-?? ?? 2
+1
)+?? -1
(
1
?? 2
+1
)+?? -1
(
(?? +2)+1
(?? +2)
2
+1
)
=-cos ?? +sin ?? +?? -2?? ?? -1
(
?? +1
?? 2
+1
)
=-cos ?? +sin ?? +?? -2?? {?? -1
(
?? ?? 2
+1
)··-1(
1
?? 2
+1
)}
=-cos ?? +sin ?? +?? -2?? cos ?? +?? -2?? sin ?? =(?? -2?? -1)cos ?? +(?? -2?? +1)sin ??
which is the required solution.
8.6
(i) Obtain Laplace Inverse transform of
?? (?? )={???? (?? +
?? ?? ?? )+
?? ?? ?? +????
?? -????
}
(2015 : 6 Marks)
(ii) Using Laplace transform, solve:
?? ''
+?? =?? ,?? (?? )=?? ,?? (?? )=-??
(2015 : 6 Marks)
Solution:
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Nov 21, 2024 Last updated |
Document Description: Laplace and inverse Laplace Transformation and Properties for UPSC 2024 is part of Mathematics Optional Notes for UPSC preparation.
The notes and questions for Laplace and inverse Laplace Transformation and Properties have been prepared according to the UPSC exam syllabus. Information about Laplace and inverse Laplace Transformation and Properties covers topics
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The content of Laplace and inverse Laplace Transformation and Properties is prepared as per the latest UPSC syllabus.
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