Post on 24-Feb-2023
Engineering Mathematics - I Semester – 1 By Dr N V Nagendram
UNIT – IV Laplace Transformations and ItsApplications
. First shifting / Translation Lemma
. Second shifting / Translation Lemma
. Change of scale property
. Multiplication by tn
. Problems Vs Solutions
Engineering Mathematics - I Semester – 1 By Dr N V Nagendram
UNIT – IV Laplace Transformations Class 1
Lemma: First shifting / Translation Lemma.
If L ( F(t) ) where s > then L ( eat F(t) ) , s > a or If is a Laplace transformation of
F(t) then is a Laplace transformation of eat F(t).
Proof: Y
O L ( eat F(t) ) L
( F(t) ) X
Figure
Given is a Laplace transformation. Then =
= L { F(t) }
So, = = = eat
= eat
= eat since, = = L
{ F(t) }
is a Laplace transformation of eat L{ F(t) }.
This completes the proof of the theorem.
Lemma: Second Translation / shifting Lemma.
If L { F(t) } and G(t) = then, L{ G(t) } e-a s .
Proof:
a
|------------------------ t < a ------------|0------|--- t > a -----|
Given = = L { F(t) } and given G(t) =
So that, = =
Put t – a = x implies dt = dx , since da/dx = 0 implies a = c.
= = = e-a s = e – a s
This completes the proof of lemma.
Note: for every x (a, ) such that x (0, ) and for every t (0, ), a = 0
=
For any t, e-as
L { G(t) } = e-as L { F(t) }
e-as = L { G(t) } since, =
L { G(t) } = e-as
L { G(t) } = e-as L { F(t) }
This completes the proof of lemma.
Lemma: Change of scale property
If L { F(t) } then L { F(at) } = .
Proof: Given = = L { F(t) }
Let us consider L { F(at) } =
Put at = x implies t = x/a so dt = (1/a) dx
L { F(at) } = =
=
= [since,
]
= [since, =
]
L { F(at) } = or L { F(at) } =
This completes the proof of the lemma.
Lemma: Multiplication by tn
If L { F(t) } then L { tn F(t) } (-1)n for
n = 1,2,3,………………We have =
On differentiation w.r.t. s, = { }
By Leibnitz principle for differentiation under integral sign,
= { } = { }
= { }
for n = m, = ( 1)m { }
This completes the proof of lemma on multiplication by tn.
Lemma: Division by t
If L { F(t) } then L { F(t) } provided
integral exists.
Proof: since =
On integrating both sides w.r.t. s from s we get,
=
= = here, since t is
independent of s
= = = L { F(t) } by
definition of Laplace transformation.
L { F(t) }
This completes the proof of lemma.
Engineering Mathematics - I Semester – 1 By Dr N V Nagendram
UNIT – IV Laplace Transformations Class 2
Problem #1: Find L { t sin at }
Problem #2: Find L { t2 sin at }
Problem #3: Find L { t3 e
-3t }
Problem #4: Find L { t e-t sin 3t
}
Problem #5: Find L
Problem #6: Find L
Problem #7: Find and evaluate at s = 2
*Problem #8: Find
Problem #9: Find L
Problem #1: Find L { t sin at }
Solution: Since L { sin at } =
L { t sin at } = { } =
L { t sin at } = is required solution.
Problem #2: Find L { t2 sin at }
Solution: Since L { sin at } =
L { t2 sin at } = ( 1)2 { }
= =
= =
= = =
= =
L { t2 sin at } = = is required solution.
Problem #3: Find L { t3 e
-3t }
Solution: since, L { e-3t } =
L { t3 e
-at } = ( 1)3
= -
= -
= -
= -
=
L { t3 e
-at } = is required solution.
Problem #4: Find L { t e-t
Sin 3t }
Solution: since, L { Sin 3t } =
And L { t Sin 3t } = - =
So, L { t e-t
Sin 3t } =
L { t e-t
Sin 3t } = is required solution.
Problem #5: Find L
Solution: since L { 1- et } = =
Now L = = =
=
=
=
L = is required solution.
Problem #6: Find L
Solution : since L { Cos at – Cos bt } =
L = =
=
= 0 =
L = is required solution.
Problem #7: Find and evaluate at s = 2
Solution : Let
= = - 1. = at s = 2
=
= and s = 2 value is Is required
solution.
Problem #8: Find
Solution: L { Sin mt } =
Now =
= as s 0
= if m > 0
= if m < 0
= is required solution.
Problem #9: Find L
Solution: since L { } =
L { } =L { et Cot
-1s} = Cot
-1 (s – 1) [
by shifting lemma ]
L = .Cot-1 (s 1). Hence the solution.
Engineering Mathematics - I Semester – 1 By Dr N V Nagendram
UNIT – IV Laplace Transformations Class 8
Section II
Problems Vs SolutionsS
NO
F(t) L { F(t) } Solution
01 Sin t Cos t L{Sin t Cos t}
= ½ L{ 2 sin t cos t}
= ½ L { Sin 2t }
, for s>|2|
02 Cosh2 2t L{Cosh2 2t }
= ½. L{1+Cosh 4t} ,s > 0
03 Sinh at – Sin at L{Sinh at – Sin at}, for s>|a|
04 Sin (at + b) L{ Sin (at+b) }
=L{Sin at cos b +
Cos at Sin b}
,
for s>|a|
05 Cos( t + ) L { Cos( t + ) }
for s>||
06 Cos3 3t = 1/4 cos3t + 3/4
cos t
L{ Cos3 3t }
= L{ 1/4 cos3t + 3/4 cos
t}
07 Cos3 t L{ Cos3 t }
08 Sin3 2t
(Sin 3t = 3 sin t – 4 sin3
t)
L { Sin3 2t }
09 Sin 2t Sin 3t L { Sin 2t Sin 3t }
=½ L{ Cos t - Cos 5t }
10 Cos 5t Cos 2t L { Cos 5t Cos 2t }
=½ L{ Cos 7t + Cos 3t }
11F(t) =
L { F(t) }
12F(t) =
L { F(t) }
13 F(t) = L { F(t) }
14 e-t[3 cos 5t – 4 sin 5t] L{e-t[3 cos 5t – 4 sin
5t] }
15 e2t(3 sinh 2t – 5 cosh 2t) L{ e2t(3 sinh 2t – 5 cosh
2t)}
16L { }
17 e-t cos2 t L (e-t cos2 t } =
L { e-t }
18 e-at Sin bt L { e-at Sin bt }
19 Cosh at – Cos at L { Cosh at – Cos at }
20 ( 1 + t e-t )3 L { ( 1 + t e-t )3 }
21 t e-4t Sin 3t L [t e-4t Sin 3t }
22 (t-1)3[u(t – 1) L { (t-1)3 [u(t – 1) }6
23 eat [ u( t – a ) } L [eat [ u( t – a ) }
24 e-2t {1- u(t – 1) } L { e-2t } – L { e-2t (1-
u(t – 1) )}
25 If L { F(t) }
then
L { F(t/a) } a. Change of scale
property.
26 If L [ F(t) } L[3e2t sin
t – 4 e2t cos 4t }
L[ F(3t) }
27If L { } Tan-1 L { } Tan-1
28 F(t) L { F(t) }
29 F(t) L { F(t) }
30 F(t) t2 at b L { F(t) }
31 F(t) t3 5 Cos t L { F(t) }
32. Evaluate [Ans. L{
}
33. Evaluate [Ans. L{
}
34. Evaluate [Ans. L{ }
35. Evaluate [Ans. L{ }
36. Evaluate [Ans. L{ }
36. Evaluate [Ans. L{ }
37. Evaluate [Ans. L{ }
UNIT – IV Inverse Laplace Transformations Class 9
Section III
Having found Laplace Transformation of a new functions let us now
determine the inverse Laplace Transformations of given functions
of S.
We have seen L { F(t) } is an algebraic function which is rational.
Hence to find inverse laplace transforms, we have to express the given
function of S into partial fractions which will, then to recognize as
one of the following standard forms:
Sl
.
NO
Inverse Laplace Function Solution
1 L-1 { } 1
2 L-1 { } e
at
3 L-1 { } n1,2,3
..
4 L-1 { } e
atn1,2
,3..
5 L-1 { }
6 L-1 { }
7 L-1 { }
8 L-1 { }
9 L-1 { }
Cosh at
10 L-1 { } e
at Cos bt
11 L-1 { }
12 L-1 { }
Engineering Mathematics - I Semester – 1 By Dr N V Nagendram
UNIT – IV Inverse Laplace Transformations Class 10
Section III
Problems Vs Solutions
Problem #1: Evaluate L-1
Problem #02: Evaluate L-1
Problem #03: Evaluate L-1
Problem #04: Evaluate L-1
Problem #05: Evaluate L-1
Problem #06: Evaluate L-1
Problem #07: Evaluate L-1
Problem #08: Evaluate L-1
Problem #09: Evaluate L-1
Problem #10: Evaluate L-1
Problem #11: Evaluate L-1
Problem #1: Evaluate L-1
Solution: L-1
L-1
L-1
L-1
L-1
L-1
L-1
L-1
1 – 3 t 4
L-1
1 – 3 t 4 is required solution.
Problem #02: Evaluate L-1
Solution: L-1
L-1
L-1
L-1
L-1
L-1
is required solution.
Problem #03: Evaluate L-1
Solution: L-1
By using synthetic division method, we can get factors as
S3 s
2 s c
S - 1 1 - 6 11 - 6
0 1 -5 6
1 -5 6 0
S 2 0 2 - 6
1 - 3 0
(S – 1) (S – 2) (S – 3) factor so by partial fractions
on solving We get A ½, B -
1, C 5/2
L-1 L
-1- L
-1 L
-1
½ et - e
-2t 5/2 e
3t
L-1 ½ e
t - e
-2t 5/2 e
3t is required
solution.
Problem #04: Evaluate L-1
Solution: To find L-1 by using partial fractions
4s5 A( s – 1 ) ( s 2 ) B ( s 2 )Put s 1 9 3B B 3 and co efficient of s
2, 0 A – 1/3
A 1/3
L-1 L
-1
L-1
L-1 L-1
1/3 et 3 t e
t – 1/3 e
-2t
L-1 1/3 e
t 3 t e
t – 1/3 e
-2t is required
solution.
Problem #05: Evaluate L-1
Solution: L-1
5s 3 A (s 1)( )
On solving we get, s 1 8 8A A 1
s 0 3 5A C C 2
L-1 L
-1
L-1 L
-1 L-1
et - e
- t Cos 2t 3/2 e
-t Sin 2t
L-1 e
t - e
- t Cos 2t 3/2 e
-t Sin 2t
Is required solution.
Problem #06: Evaluate L-1
Solution: To find L-1
For that by known formula, s4 4a
4 (s
22a
2)2-(2as)
2(s
22a
22as)
(s22a
2-2as)
By partial fractions
s (As b) (Cs d) Co efficient of s3 A – C 0Co efficient of s2 B D 0
Co efficient of s A C 0 and on solving, B ; D –
B so D
L-1 L
-1 L
-1
L-1 e-a t sin at ea t sin at is required
solution.
Problem #07: Evaluate L-1
Solution: L-1 L
-16 L
-1 3 L
-1
6
L-1 6 Is required solution.
Problem #08: Evaluate L-1
Solution: L-1
L-1 L
-1
L-1
Is required solution.
Problem #09: Evaluate L-1
Solution: L-1
L-1
L-1
L-1
[Since, L-1
e-3t Cos 2t - 3/2 e
-3t Sin 2t
L-1
e-3t Cos 2t - 3/2 e
-3t Sin 2t is required
solution.
Problem #10: Evaluate L-1
Solution: L-1
L-1
L-1
L-1
3 e-t Cosh 3t 3. .Sinh 3t.e
-t
3 e-t
e-t
e-t
Engineering Mathematics - I Semester – 1 By Dr N V Nagendram
UNIT – IV Inverse Laplace Transformations Class 11
Section III
Problems Vs Solutions
Problem #12: Evaluate L-1
Solution: To find L-1
By partial fractions, L-1
L-1
L-1
L-1
We get A = ; B = ; C =
L-1
L-1
L-1
L-1
L-1
L-1
L-1
e-3t
e2t
L-1
e-3t
e2t is required solution.
Problem #13: Evaluate L-1
Solution: L-1
L-1
By partial fractions =
We get, A = ; B = 2 ; C =
L-1
L-1
+ 2 L-1
L-1
Problem #14: Evaluate L-1
Solution: L-1 L
-1
L-1 L
-1
L-1 L
-1
= L-1 L
-1
= t et - t e-t
L-1 = t et - t e-t is required solution.
= L-1 -L
-1 L
-1 -L
-1
= L-1 - L
-1 L
-1 - L
-1
= L-1 - L
-1 L
-1 - L
-1
= t et - t e-2t
L-1 = = t et - t e-2t is required
solution.
Problem #16: Evaluate L-1
Solution: To find L-1
By partial fractions =
Put s = 3 A =
Put s = 0 C =
Put s = 1 B =
L-1 L
-1 L
-1
L-1 L
-1 L
-1
L-1 L
-1 L
-1
= L-1 L
-1 L
-1
= e3t Cos 2t Sin 2t
L-1 = e
3t Cos 2t Sin 2t is required
solution.
Problem #17: Evaluate L-1
Solution: L-1 = L
-1
= L-1
= L-1 L
-1
= L-1 L
-1
= L-1 L
-1
= Sin t - t e-t
L-1 = Sin t - t e-t is required solution.
Solution: L-1 = L
-1 = L
-1
L-1 = L
-1 by partial
fractions
Here 1= A[(s-a)2+3as] + (Bs + c) (s-a) ; Put s = a A = ;
Put s = 0 C = ; Put s = 1 B =
L-1 = L
-1
= L-1
= L-1 + L
-1 - L
-1
= + L-1 L
-1
= + L-1 L
-1 L
-1
+ L-1
= e-at t e-at e-at t e-at Cos at
L-1 = e
-at t e-at e-at
t e-at Cos at
Is required solution.
Problem #20: Evaluate L-1
Solution: L-1
= L-1
= L-1
+ L-1
+ L-1
= L-1
+ L-1
+ L-1
= L-1
+ L-1
+ L-1
= L-1
+ L-1
+ L-1
L-1
= L-1
+ L-1
+ L-1
L-1
= Sin 2t + Sin t + Sin t - Sin 2t
= Sin 2t + Sin t + Sin t - Sin 2t
= Sin t - Sin 2t
= [5 Sin t - Sin 2t]
L-1
= [5 Sin t - Sin 2t] is required
solution.
Problem #21: Evaluate L-1
Solution: L-1
= L-1
= L-1
= L-1
L-1
= L-1
L-1
= 2 e-2t Cos 3t 7 e
-2t sin 3t
L-1
= 2 e-2t Cos 3t 7 e
-2t sin 3t is required
solution.
Problem #22: Evaluate L-1
Solution: L-1
= L-1
= e-2t Cos t
L-1
= e-2t Cos t is required solution.
Problem #23: Evaluate L-1
Solution: L-1
= L-1
= L-1
= L-1
+ L-1
- L-1
= L-1
+ L-1
- 3 L-1
= Sin t + e-t Sin t - 3 L
-1
L-1
= Sin t + e-t Sin t - 3 L
-1
Is required solution.
= e-1/2t
Cos e-1/2t
Sin
L-1
= e-1/2t
Cos e-1/2t
Sin
is required solution.
Problem #25: Evaluate L-1
Solution: L-1
= L-1
= L-1
L-1
= L-1
L-1
Problem #26: Evaluate L-1
Solution: L-1 = L
-1
= L-1
= L-1 + L
-1 + L
-1
= L-1 + L
-1 + L
-1
= e2t + 4 t e
2t + 4 t
2 e
2t
L-1 = e
2t + 4 t e
2t + 4 t
2 e
2t is required solution.
Problem #27: Evaluate L-1
Solution: L-1 = L
-1
= L-1 = L
-1 +L
-1
= e2t Cos 3t + 5/3 e
2t Sin 3t
L-1 = e
2t Cos 3t + 5/3 e
2t Sin 3t is required
solution.
Problem #28: Evaluate L-1
Solution: Since, L-1 =
L-1 = = =
L-1 = is required solution.
Problem #29: Evaluate L-1
Solution: L-1 = L
-1 = e
-at L
-1
Here we have, L-1
= = ;
L-1
= = and
L-1
= =
L-1 = =
L-1 = is required solution.
Problem #30: Evaluate L-1
Solution: Let F (t) = L-1
Hence, L = = = since,
after applying limits. So, = =
L-1 = is required solution.
Problem #31: Evaluate L-1
Solution: L-1
= L-1
=
=
=
=
=
L-1 = is required solution.
Problem #32: Evaluate L-1
Solution: Let F (t) = L-1
So, t. F(t) = - L-1
= - L-1
= - L-1
= - L-1 + L
-1
= - L-1 + L
-1
= - e-t
+ et
t. F(t) = - e-t
+ et
F(t) = L-1 =2 is required solution.
Problem #33: Evaluate L-1
Solution: t. F(t) L-1
L-1
L-1 L
-1 L
-1
L-1 L
-1 L
-1
t. F(t) 2 Cos t 1 t e-t
F(t) e-t
L-1 L
-1 L
-1 L
-1
et Sin t e
- t Sin t 2 Sin t 2 sinh t
sin t
t F(t) 2 Sinh t sin t
F(t) is required solution.
Problem #35: Evaluate L-1
Solution: Let t F(t) L-1
L-1
L-1
Engineering Mathematics - I Semester – 1 By Dr N V Nagendram
UNIT – IV Inverse Laplace Transformations Class 12
Section III Convolution theorem:
If L-1{ } F(t) and L
-1{ } G(t) then L
-1{ }
F * G is called the convolution or falting of
F and G.
u
u t
tu t
O u 0
t
Let, (t)
L{ (t) }
here put t-u v
then dv dt
F * G.
L-1{ } F * G
This completes the proof of the theorem.Exercises: Try yourself….. By Dr N V Nagendram
Problem #01 Evaluate L-1 [Ans. t Sin at
]
Problem #02 Evaluate L-1 [Ans. t Sin at
]
Problem #03 Evaluate L-1
[Ans.
]
Problem #04 Evaluate L-1
[Ans.
]
Problem #05 Evaluate L-1 [Ans. Cos t
]
Problem #06 Evaluate L-1 [Ans. ( t- Sin at)
]
Problem #07 Evaluate L-1 [Ans.
]
Problem #08 Evaluate L-1 [Ans. e
-at(1 – at)
]
Problem #09 Evaluate L-1 [Ans.
]
Problem #10 Evaluate L-1 [Ans. e-at
]
Problem #11 Evaluate L-1 [Ans. (1 - e-at)
]
Problem #12 Evaluate L-1 [Ans. (e-b t - e-at)
]
Problem #13 Evaluate L-1 [Ans. e
- t e-2t e-
3t ]
Problem #14 Evaluate L-1 [Ans. (Cos at – Cos
bt) ]
Problem #15 Evaluate L-1 [Ans. (1 – Cosh
at) ]
Problem #16 Evaluate L-1 [Ans. (et – Cos t)
]
Problem #17 Evaluate L-1 [Ans.
]
Problem #18 Evaluate L-1 [Ans.
]
Problem #19 Evaluate L-1 [Ans.
]
Problem #1 Evaluate L-1
Solution: L-1 = L
-1
[Since, L-1 and L
-1 ]
By convolution theorem,
L-1 = =
=
L-1 = = t Sin at
L-1 = t Sin at is required solution.
Problem #2 Evaluate L-1
Solution: To find L-1
Engineering Mathematics - I Semester – 1 By Dr N V Nagendram
UNIT – IV Application of Laplace Transform Class 13
Section III
Application of Laplace Transform to differential equations with
constant coefficients:
Laplace transform is especially suitable to obtain the solutionof linear non-homogeneous ordinary differential equations withconstant coefficients, when all the boundary conditions arespecified for the unknown function and its derivatives at asingle point.
Consider the initial value problem …………………………..
(1)
y(t = 0) = k0, y1(t=0) = k1 ………………………………………………………………… (2)
Where a,b,k0,k1 are all constants and r(t) is a function of t.
Method of Solution to differential equation (D.E) by Laplace Transform(L.T.):
1. Apply Laplace transform on both sides of the given differential equation (1), resulting in a subsidiary equation as
[s2Y – s y(0) - y(0)] +a[sY-y(0)]+bY = R(s) …………………………………………. (3) where Y = L{y(t)} and R(s) = L{r(t)}.
Replace y(0), y(0) using given initial conditions (2).
2. Solve (3) algebraically for Y(s), usually to a sum of partialfractions.
3. Apply inverse Laplace transform to Y(s) obtained in 2. Thisyields the solution of O.D.E. (1) satisfying the initialconditions (2) as y(t) = L-1[Y(s)].
Problem #1 Solve the differential equation by using Laplacetransformation y-2y-8y = 0, y(0) = 3, y(0) = 6.
Solution: On applying Laplace transform (s2Y - 3s - 6) – 2( sY –3 ) - 8Y = 0
On solving, Y(s) = =
By using partial fractions, Y(s) = +
Applying 1 L.T. y(t) = L-1(Y(s)) = 2 L-1 +L-1 = 2 e4t +
e-2t.
y(t) = 2 e4t + e-2t is required solution.
Problem #2 Solve the differential equation by using Laplacetransformation y+2y+5y = e-t Sin t, y(0) = 0, y(0) = 1.Solution: Using L.T. On applying Laplace transform we get thegiven differential equation as, (s2Y - 0 - 1) + 2( sY – 0 ) =
L{ e-t Sin t } =
On solving Y =
By partial fractions, = +
= (As + B) ( )+ (Cs + D ) ( )= s3(A + C) + s2 (2A + 2C + B + D ) + s (2A + 5C + 2B +
2D )+ 2B + 5DEquating coefficients of s on either side,A + C = 0 ; 2A + 2C + B + D = 1; 2A + 5C + 2B + 2D = 2; 2B + 5D= 3
A = 0, B = , C = 0, D =
Y(s)= + = +
Applying I.T. y(t) = L-1{Y}= L-1 + L-1
using first shifting theorem, y(t)= e-t L-1 + e-t L-1
y(t)= is required solution.
Problem #3 Solve the differential equation by using Laplacetransformation y+n2y = a Sin (nt+2), y(0) = 0, y(0) = 0.
Solution: y+n2y = a Sin (nt+2)
= a [Sin nt Cos 2 + Cos nt Sin 2]
Applying L.T. L{ y} + n2 L { y } = a cos 2. L{Sin nt} + a Sin 2L{Cos nt}
Using L.T. On applying Laplace transform we get the givendifferential equation as,
[s2Y – sy(0) - y(0)] + n2Y = .a.Cos 2 + .a. Sin 2
Solving Y, Y(s) = .a. Cos 2 + .a. Sin 2
Applying I.T. we get
y(t) = n. a. Cos 2. L-1 + a. Sin 2. L-1 …… (1)
From I.L.T. tables, we know that 2nd term in R.H.S.
L-1 = …………………………………………………………………… (2) to find
first term in R.H.S.
L-1 = L-1 = =
….(3)
Thus substituting (2) and (3) in (1), we get
y(t) = a.n. Cos 2. + a Sin 2
= [ + Cos 2. Sin nt + nt. Sin 2 .
Sin nt]
= [ ]
= [ ]
y(t) = [ ] is required
solution.Problem #4 Find the general solution of the differential equationby using Laplace transformation y - 3y +3y - y =t2 et,y(0) = 1, y(0) = 0,y(0) = - 2.
Solution: Since the initial conditions are arbitrary assume y(0)= a, y(0) =b, y(0) =c.
Then Using L.T. On applying Laplace transform we get the givendifferential equation as, (s2Y – as2 – bs - c) - 3( s2Y – as -b )
+ 3(sY – a) - Y =
Y =
By partial fractions Y = where c1,
c2, c3 are constants depending on a, b, c.
Applying I.T. and using first shifting theorem
y(t) = c1 et + c2 t et + c3 et + is required
solution.
Problem #5 Solve the differential equation by using Laplacetransformation y - 3y +3y - y =t2 et, y(0) = 1, y(0) =0,y(0) = - 2.
Solution: Applying L.T. to D.E. L{ y - 3y +3y - y} =L{ t2 et }
L{ y} - 3L{ y} +3 L { y} - L{ y} =L{ t2 et
}
[s3Y – s2y(0) - sy(0) - y(0) ] – 3 [s2Y – sy(0) - y(0)] + 3[sY
– y(0)] – Y =
Using the initial condition y(0) = 1, y(0) = 0,y(0) = - 2, andsolving for Y
(s3 – 3a2 + 3s – 1)Y – s2 + 3s – 1=
Y = = =
Y =
On applying I L.T. we get
y(t) = L-1 { Y } = L-1 L-1 L-1 +2 L-1
y(t) = L-1 { Y } = et t et + is required solution.
Engineering Mathematics - I Semester – 1 By Dr N V Nagendram
UNIT – IV Application of Laplace Transform Class 14
Section III Exercise Try yourself……..
Use L.T. to solve each of the following I.V.P. consisting of aD.E. with I.C.
01. , general solution [Ans. Assumey(0)=0=A=constant; y=Aet]02. [Ans. y=(3et +e3t)/2]03. [Ans.G.S. y=C + De-t, C=A +B,D= B ]
04. [Ans.G.S. y=et+ Cos t + Sin t]05. [Ans.G.S. y=(3t + 2)e3t
]
06. [Ans.G.S. y=
]
07. [Ans.G.S. y=e-2t-2e-3t+e-5t
]
08. [Ans.G.S.
y=4e2t+3te2t+3e3t-2e5t ]09. [Ans.G.S. y=
]
10. [Ans. y=
]
11. [Ans.G.S. y=
]
12. [Ans.y=
]
13. [Ans.G.S. y=e-2t + 2e-t -
2te-t – Cos t + 2 Sin t]14. [Ans.
]
15. [Ans.G.S. y= 2(Sin 2t –
Sin t ]
Engineering Mathematics - I Semester – 1 By Dr N V Nagendram
UNIT – IV Application of Laplace Transform Class 14
Section IV
Application of Laplace Transform to system of simultaneous
differential equations:
Laplace transform can also be used to solve a system or family ofm simultaneous equations in m dependent variables which arefunctions of the independent variable t. consider a family of twosimultaneous differential equations in the two dependentvariables x, y which are functions of t.
……………………………………….. (1)
……………………………………….. (2)
Initial conditions x(0) = c1; y(0) = c2; x(0) = c3; y(0) = c4
………………………… (3)
Here a1, a2, a3, a4, a5, a6,b1, b2, b3, b4, b5, b6 and c1, c2, c3, c4
are all constants and R1(t), R2(t) are functions of t.
Method of solution to system of differential equation (D.E.):
I. Apply Laplace transform on both sides of each of the twodifferential equation (1) and (2) above. This reduces (1) and (2)to two algebraic equations in X(s) and Y(s) where X(s) =L{ x(t) } and Y(s) = L{ y(t) }. a1[s
2X-sx(0) - x(0)] + a2[ s2Y – sy(0) - y(0)] + a3[sX – x(0) ] + a4[sY- y(0) ]+ a5 X + a6 Y = Q1(s)…..(4)
b1[s2X-sx(0) - x(0)] + b2[ s2Y – sy(0) - y(0)] + b3[sX – x(0) ] + b4[sY- y(0) ]
+ b5 X + b6 Y = Q2(s)… (5)
use the initial conditions (3) and substitute for x(0) , y(0) ,x(0) , y(0). II. Solve (4) and (5) for X(s) and Y(s).
III. The required solution is obtained by taking the inverseLaplace Transform of X(s) and Y(s) as
x(t) = L-1{ X(s) } and y(t) = L-1 { Y(s) }.
Problem #1 Solve ;
x(0) = 0; y(0) = 6.5; x(0) = 0;
Solution: Taking Laplace Transformation of the given differentialequation we have
[s2X-sx(0) - x(0)] -3 [sX – x(0) ] - [sY- y(0) ]+ 2Y = 14 +3
[sX – x(0) ] – 3X + [sY- y(0) ] =
Use given initial conditions (I.C.) x(0) = 0; y(0) = 6.5=13/2;x(0) = 0;
S(s - 3)X +(2 – s)Y = and (s – 3 )X + sY =
On solving we get Y(s) = and X(s) =
Taking inverse Laplace Transform y(t) = L-1{Y(s)} = L-1
= L-1
= L-1
y(t) = 7t + 5 – et + e-2t
Similarly, x(t) = L-1{X(s)} = L-1 by partial
fractions
= L-1
= 2 et e3t e-2t
x(t) =2 et e3t e-2t Is required solution.
Problem #2 Solve ;
x(0) = 2; y(0) = 1;
Solution: Taking Laplace Transformation of the given differentialequation we have
2[sX(s) - x(0)] + sY(s)- y(0) –X(s)- Y(s) =
sX(s) – x(0) + sY(s) - y(0)+ 2 X(s) + Y(s) =
Use given initial conditions (I.C.) x(0) = 2; y(0) = 1.
(2s - 1)X(s) +(s – 1)Y(s) = and (s + 2 )X(s) + (s+1)Y(s)
=
On solving we get Y(s) = =
and X(s) =
Taking inverse Laplace Transform y(t) = L-1{Y(s)} = L-1
= Cos t – 13 Sin t + Sinht
y(t) = Cos t – 13 Sin t + Sinh t
Similarly, x(t) = L-1{X(s)} = L-1 by partial fractions
= 2 Cos t + 8 Sin t
x(t) =2 Cos t + 8 Sin t Is required solution.
Engineering Mathematics - I Semester – 1 By Dr N V Nagendram
UNIT – IV Application of Laplace Transform Class 15
Exercise Solve the following system ofequations Try yourself ……..
01. ; , x(0) = 8, y(0) = 3
[Ans. x(t) = 5 e-t + 3 e4t; y(t) = 5 e-t – 2 e4t]
02. ; ,
x(0) = 35,y(0) = 27; x(0) = -48,y(0) = -55.
[Ans. x(t) = 30 Cos t – 15 Sin 3t + 3 e-t + 2 Cos 2t; y(t) = 30 Cos t – 60 Sin t - 3 e-t + Sin 2t]
03. ; , x(0) = y(0) = 1; x(0) =
y(0) = 0
[Ans. x(t) = ;
y(t) = ]
04. ; , x(0) = 1, y(0) = 0
[Ans. x(t) = ; y(t) =
]
05. ; , x(0) = 3, y(0) = 0
[Ans.x(t)= ; y(t) = ]
06. ; , x(0) = y(0)= y(0) = 0
[Ans. x(t)=1+ e-t – e-at-e-bt; y(t)=1+e-t– be-at a e-bt ]
Note: a = ; b =
07. ; , x(0) = 0, x(0) = 0
[Ans. x(t) = 22e-t (1+t); y(t) = 2t-2e-t (1+t)]
08. ; , x(0) =1, y(0) =0
[Ans. x(t) = e-2t - t et; y(t) = et +tet ]
09. ; ,x(0) =-1, y(0) = 0
[Ans. x(t) = -2 et + e4t; y(t)= et + e4t]
10. ; , x(0) = 2, y(0) = 0
[Ans. x(t)=e-t + et = 2Cosh t; y(t)=Sin t – 2 Sinh t]