Crescent Academy…….………………….…..For Research in Education
LAPLACE TRANSFORM
Type I: Basic
1.
2.
Find the Laplace Transforms of 4𝑡 2 + 𝑠𝑖𝑛3𝑡 + 𝑒 2𝑡
Ans.
𝑠3
+
3
𝑠 2 +9
+
1
𝑠−2
Find the Laplace Transforms of 1 + 𝑠𝑖𝑛𝑡
[M18/Extc/5M]
Solution:
𝐿 1 + 𝑠𝑖𝑛𝑡
𝑡
= 𝐿 cos + sin
2
=
𝑠
1
𝑠 2 +4
+
1
=
=
3.
8
1
2
1
𝑠 2 +4
𝑡
𝜃
𝜃
2
2
since 1 + 𝑠𝑖𝑛𝜃 = cos + sin
2
𝑠+2
1
𝑠 2 +4
4𝑠+2
4𝑠 2 +1
Find the LT of 𝑠𝑖𝑛2𝑡 − 𝑐𝑜𝑠2𝑡 2
[M19/Extc/5M]
Solution:
𝐿 𝑠𝑖𝑛2𝑡 − 𝑐𝑜𝑠2𝑡 2
= 𝐿{sin2 2𝑡 − 2𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠2𝑡 + cos 2 2𝑡}
= 𝐿 1 − 𝑠𝑖𝑛4𝑡
since, 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 = 𝑠𝑖𝑛2𝜃
=
1
𝑠
−
4
𝑠 2 +16
2
4.
Find the Laplace Transforms of sin2 𝑡
Ans.
5.
Find the Laplace Transforms of cos 2 3𝑡
Ans.
6.
7.
Evaluate using L.T.: 0 𝑒 −2𝑡 sin2 2𝑡 𝑑𝑡
Ans.
5
2
−2𝑡
2
Find the Laplace Transforms of 𝑡 − 𝑒
+ cosh 3𝑡
∞
Ans.
3
8.
Find the Laplace Transforms of sin 𝑡
Ans.
9.
Find the Laplace Transforms of cos 3 2𝑡
Ans.
S.E/Paper Solutions
1
𝑠 𝑠 2 +4
𝑠 2 +18
𝑠 𝑠 2 +36
1
2
𝑠3
−
1
𝑠+2
6
+
1 1
2 𝑠
𝑠 2 +1 𝑠 2 +9
𝑠 𝑠 2 +28
𝑠 2 +36 𝑠 2 +4
+
𝑠
𝑠 2 −36
Crescent Academy…….………………….…..For Research in Education
∞
10. Evaluate using L.T.: 0 𝑒 −2𝑡 sin3 𝑡 𝑑𝑡
[M17/ElexExtcElectBiomInst/5M]
Solution:
By definition,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = sin3 𝑡
∞ −𝑠𝑡
𝑒 sin3 𝑡 𝑑𝑡 = 𝐿 sin3 𝑡
0
= 𝐿
3𝑠𝑖𝑛𝑡 −𝑠𝑖𝑛 3𝑡
4
1
= 𝐿 3𝑠𝑖𝑛𝑡 − 𝑠𝑖𝑛3𝑡
4
1
1
3
= 3. 2 − 2
4
𝑠 +1
𝑠 +9
Put 𝑠 = 2
∞ −2𝑡
𝑒
0
sin3 𝑡 𝑑𝑡 =
1
3
4 4+1
−
3
=
4+9
6
65
11. Find the Laplace Transforms of 𝑠𝑖𝑛3𝑡𝑐𝑜𝑠4𝑡
Ans.
12. Find the Laplace Transforms of 𝑠𝑖𝑛𝑡𝑠𝑖𝑛5𝑡
Ans.
3𝑠 2 −21
𝑠 2 +49 𝑠 2 +1
10𝑠
𝑠 2 +16 𝑠 2 +36
13. Find the Laplace Transforms of 𝑐𝑜𝑠𝑡 𝑐𝑜𝑠2𝑡 𝑐𝑜𝑠3𝑡
[N15/ChemBiot/5M][M16/ChemBiot/5M]
Solution:
𝐿{𝑐𝑜𝑠2𝑡 𝑐𝑜𝑠𝑡 𝑐𝑜𝑠3𝑡}
1
= 𝐿 2𝑐𝑜𝑠2𝑡 𝑐𝑜𝑠𝑡 𝑐𝑜𝑠3𝑡
2
1
= 𝐿 𝑐𝑜𝑠3𝑡 + 𝑐𝑜𝑠𝑡 𝑐𝑜𝑠3𝑡
2
1
= 𝐿{cos 2 3𝑡 + 𝑐𝑜𝑠3𝑡 𝑐𝑜𝑠𝑡}
2
1 1
= . 𝐿 2 cos 2 3𝑡 + 2𝑐𝑜𝑠3𝑡 𝑐𝑜𝑠𝑡
=
=
2 2
1
4
1
4
𝐿 1 + 𝑐𝑜𝑠6𝑡 + 𝑐𝑜𝑠4𝑡 + 𝑐𝑜𝑠2𝑡
1
𝑠
+
𝑠
𝑠 2 +36
+
𝑠
𝑠 2 +16
+
𝑠
𝑠 2 +4
∞
1
14. If 0 𝑒 −2𝑡 sin 𝑡 + 𝛼 cos 𝑡 − 𝛼 𝑑𝑡 = , then find α
4
[M16/CompIT/5M]
Solution:
By definition,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = sin 𝑡 + 𝛼 cos 𝑡 − 𝛼
∞ −𝑠𝑡
𝑒 sin 𝑡 + 𝛼 cos 𝑡 − 𝛼 𝑑𝑡 = 𝐿 sin 𝑡 + 𝛼 cos 𝑡 − 𝛼
0
S.E/Paper Solutions
2
Crescent Academy…….………………….…..For Research in Education
=
=
∴
∞
0
1
𝐿 2 sin 𝑡 + 𝛼 cos 𝑡 − 𝛼
2
1
𝐿 sin 2𝑡 + sin 2𝛼
2
𝑒 −𝑠𝑡 sin 𝑡 + 𝛼 cos 𝑡 − 𝛼 𝑑𝑡 =
1
2
2
𝑠 2 +4
+
Put 𝑠 = 2,
∞ −2𝑡
𝑒
0
sin 𝑡 + 𝛼 cos 𝑡 − 𝛼 𝑑𝑡 =
1
4
1
4
1
4
1
8
=
1 2
2 8
1 1
𝑠𝑖𝑛 2𝛼
+
2
𝑠𝑖𝑛 2𝛼
2 4
2
1
𝑠𝑖𝑛 2𝛼
= +
8
1
− =
=
+
4
𝑠𝑖𝑛 2𝛼
8
4
𝑠𝑖𝑛 2𝛼
4
𝑠𝑖𝑛2𝛼 =
2𝛼 =
𝛼=
𝜋
1
2
6
𝜋
12
15. Find the Laplace Transform of 𝑠𝑖𝑛2𝑡𝑐𝑜𝑠3𝑡
[M17/AutoMechCivil/5M]
Solution:
𝐿{𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠3𝑡}
1
= 𝐿 2𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠3𝑡
2
1
= 𝐿 𝑠𝑖𝑛5𝑡 + sin(−𝑡)
2
1
= 𝐿{𝑠𝑖𝑛5𝑡 − 𝑠𝑖𝑛𝑡}
=
2
1
5
2 𝑠 2 +25
−
1
𝑠 2 +1
16. Find the Laplace Transform of 𝑠𝑖𝑛2𝑡𝑠𝑖𝑛𝑡𝑠𝑖𝑛3𝑡
[N13/Biot/5M]
Solution:
𝐿{𝑠𝑖𝑛2𝑡 𝑠𝑖𝑛𝑡 𝑠𝑖𝑛3𝑡}
1
= 𝐿 2𝑠𝑖𝑛2𝑡 𝑠𝑖𝑛𝑡 𝑠𝑖𝑛3𝑡
2
1
= 𝐿 𝑐𝑜𝑠𝑡 − 𝑐𝑜𝑠3𝑡 𝑠𝑖𝑛3𝑡
2
1
= 𝐿{𝑐𝑜𝑠𝑡𝑠𝑖𝑛3𝑡 − 𝑐𝑜𝑠3𝑡 𝑠𝑖𝑛3𝑡}
2
1 1
= . 𝐿 2𝑐𝑜𝑠𝑡 𝑠𝑖𝑛3𝑡 − 2𝑐𝑜𝑠3𝑡 𝑠𝑖𝑛3𝑡
=
2 2
1
4
𝐿 𝑠𝑖𝑛4𝑡 − sin −2𝑡 − 𝑠𝑖𝑛6𝑡
S.E/Paper Solutions
3
𝑠𝑖𝑛 2𝛼
𝑠
Crescent Academy…….………………….…..For Research in Education
=
=
1
4
1
𝐿 𝑠𝑖𝑛4𝑡 + 𝑠𝑖𝑛2𝑡 − 𝑠𝑖𝑛6𝑡
4
4 𝑠 2 +16
+
2
𝑠 2 +4
−
6
𝑠 2 +36
∞
17. Evaluate 0 𝑒 𝑡 𝑠𝑖𝑛2𝑡𝑐𝑜𝑠3𝑡𝑑𝑡
[N17/AutoMechCivil/6M]
Solution:
By definition,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = 𝑠𝑖𝑛2𝑡𝑐𝑜𝑠3𝑡
∞ −𝑠𝑡
𝑒 𝑠𝑖𝑛2𝑡𝑐𝑜𝑠3𝑡𝑑𝑡 = 𝐿 𝑠𝑖𝑛2𝑡𝑐𝑜𝑠3𝑡
0
=
1
2
1
𝐿 2𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠3𝑡
= 𝐿 𝑠𝑖𝑛5𝑡 + sin(−𝑡)
2
1
= 𝐿{𝑠𝑖𝑛5𝑡 − 𝑠𝑖𝑛𝑡}
=
2
1
5
2
𝑠 2 +25
−
1
𝑠 2 +1
Put 𝑠 = −1
∞ 𝑡
𝑒 𝑠𝑖𝑛2𝑡𝑐𝑜𝑠3𝑡𝑑𝑡
0
=
1
5
2 1+25
−
1
1+1
=−
∞
2
13
3
18. If 0 𝑒 −2𝑡 sin 𝑡 + 𝛼 cos 𝑡 − 𝛼 𝑑𝑡 = , then find α
8
[N14/CompIT/6M]
Solution:
By definition,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = sin 𝑡 + 𝛼 cos 𝑡 − 𝛼
∞ −𝑠𝑡
𝑒 sin 𝑡 + 𝛼 cos 𝑡 − 𝛼 𝑑𝑡 = 𝐿 sin 𝑡 + 𝛼 cos 𝑡 − 𝛼
0
=
=
∴
∞
0
1
2
1
2
𝐿 2 sin 𝑡 + 𝛼 cos 𝑡 − 𝛼
𝐿 sin 2𝑡 + sin 2𝛼
1
𝑒 −𝑠𝑡 sin 𝑡 + 𝛼 cos 𝑡 − 𝛼 𝑑𝑡 =
2
Put 𝑠 = 2,
∞ −2𝑡
𝑒
0
sin 𝑡 + 𝛼 cos 𝑡 − 𝛼 𝑑𝑡 =
3
8
3
8
3
8
S.E/Paper Solutions
4
=
1 2
2 8
1 1
2
𝑠 2 +4
+
+
+
𝑠𝑖𝑛 2𝛼
2
𝑠𝑖𝑛 2𝛼
2 4
2
1
𝑠𝑖𝑛 2𝛼
= +
8
1
4
𝑠𝑖𝑛 2𝛼
8
4
− =
𝑠𝑖𝑛 2𝛼
𝑠
Crescent Academy…….………………….…..For Research in Education
1
𝑠𝑖𝑛 2𝛼
=
4
4
𝑠𝑖𝑛2𝛼 = 1
𝜋
2𝛼 =
𝛼=
19. Obtain 𝐿 𝑓 3𝑡
if 𝐿 𝑓 𝑡
=
𝜋
2
4
20−4𝑠
𝑠 2 −4𝑠+20
Ans.
60−4𝑠
𝑠 2 −12𝑠+180
20. State and Prove change of scale property of L.T. If 𝐿 𝑠𝑖𝑛 𝑡 =
then find 𝐿 𝑠𝑖𝑛2 𝑡
21. Evaluate using L.T.:
S.E/Paper Solutions
Ans.
∞
𝑒𝑟𝑓
0
2 𝑡 𝑒 −5𝑡 𝑑𝑡
5
Ans.
𝜋
𝑠 𝑠
2
15
.𝑒
1
−𝑠
𝜋
2𝑠 𝑠
1
.𝑒
−4𝑠
,
Crescent Academy…….………………….…..For Research in Education
Type II: First Shifting Property
1.
State the first shifting theorem for Laplace Transforms. Use the theorem
to obtain 𝐿 𝑒 −𝑡 sin2 𝑡
2.
3.
Find L.T. of 𝑒
4𝑡
Ans.
3
sin 𝑡
Ans.
1
1
−
2 𝑠+1
3
1
4
𝑠+1
𝑠 2 +2𝑠+5
𝑠 2 −8𝑠+17
−
−2𝑡
Find L.T. of 𝑒
1 − 𝑠𝑖𝑛𝑡
[N17/IT/4M]
Solution:
𝐿 𝑒 −2𝑡 1 − 𝑠𝑖𝑛𝑡
𝑡
= 𝐿 𝑒 −2𝑡 cos − sin
=𝐿 𝑒
𝑡
2
cos − 𝑒
2
𝑠+2
=
=
−2𝑡
−
1
4
𝑠+2 2 +
1
2
𝑡
2
−2𝑡
1
2
sin
𝑡
𝑡
2
2
since, 1 − 𝑠𝑖𝑛𝑡 = cos − sin
𝑡
2
1
4
𝑠+2 2 +
𝑠+2−
1
𝑠 2 +4𝑠+4+4
3
=
4.
5.
6.
𝑠+2
17
𝑠 2 +4𝑠+ 4
Find L.T. of 𝑒 𝑡 𝑐𝑜𝑠𝑡
2
Ans.
𝑠𝑖𝑛𝑡 2
Find L.T. of
Ans.
𝑒𝑡
𝑐𝑜𝑠 2𝑡 𝑠𝑖𝑛𝑡
Find L.T. of
𝑒𝑡
[M15/ChemBiot/5M][M18/AutoMechCivil/5M]
Solution:
𝐿
𝑐𝑜𝑠 2𝑡 𝑠𝑖𝑛𝑡
𝑒𝑡
−𝑡
= 𝐿 𝑒 𝑐𝑜𝑠2𝑡 𝑠𝑖𝑛𝑡
1
= 𝐿 𝑒 −𝑡 2𝑐𝑜𝑠2𝑡 𝑠𝑖𝑛𝑡
=
2
1
2
1
𝐿 𝑒 −𝑡 𝑠𝑖𝑛3𝑡 − 𝑠𝑖𝑛𝑡
= 𝐿 𝑒 −𝑡 𝑠𝑖𝑛3𝑡 − 𝑒 −𝑡 𝑠𝑖𝑛𝑡
=
=
7.
2
1
2
1
3
𝑠+1
3
2 +9
2 𝑠 2 +2𝑠+10
−
−
1
𝑠+1 2 +1
1
𝑠 2 +2𝑠+2
Find L.T. of 𝑠𝑖𝑛𝑡𝑐𝑜𝑠2𝑡𝑐𝑜𝑠𝑕𝑡
[N14/CompIT/5M]
Solution:
S.E/Paper Solutions
6
1
1
2 𝑠−2
1 1
2 𝑠+2
+
−
1
𝑠 2 −8𝑠+25
𝑠−2
𝑠 2 −4𝑠+8
𝑠+2
𝑠 2 +4𝑠+8
Crescent Academy…….………………….…..For Research in Education
𝐿 𝑠𝑖𝑛𝑡 𝑐𝑜𝑠2𝑡 𝑐𝑜𝑠𝑕𝑡
𝑒 𝑡 +𝑒 −𝑡
= 𝐿 𝑠𝑖𝑛𝑡 𝑐𝑜𝑠2𝑡
2
𝑡
1
= 𝐿 𝑠𝑖𝑛𝑡 𝑐𝑜𝑠2𝑡 𝑒 + 𝑒 −𝑡
=
=
=
=
=
2
1
4
1
4
1
4
1
4
1
𝐿 2𝑠𝑖𝑛𝑡 𝑐𝑜𝑠2𝑡 𝑒 𝑡 + 𝑒 −𝑡
𝑒 𝑡 + 𝑒 −𝑡
𝐿 𝑠𝑖𝑛3𝑡 + sin −𝑡
𝐿 𝑒 𝑡 𝑠𝑖𝑛3𝑡 + 𝑒 −𝑡 𝑠𝑖𝑛3𝑡 − 𝑒 𝑡 𝑠𝑖𝑛𝑡 − 𝑒 −𝑡 𝑠𝑖𝑛𝑡
3
𝑠−1
3
2 +9
4 𝑠 2 −2𝑠+10
3
+
+
2 +9
𝑠+1
−
3
𝑠 2 +2𝑠+10
−
1
𝑠−1
1
𝑠 2 −2𝑠+2
8.
Find L.T. of 𝑒 𝑡 𝑠𝑖𝑛2𝑡 𝑠𝑖𝑛3𝑡
9.
Find L.T. of 𝑠𝑖𝑛2𝑡𝑐𝑜𝑠𝑡𝑐𝑜𝑠𝑕2𝑡
[N18/IT/5M]
Solution:
𝐿 𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠𝑡 𝑐𝑜𝑠𝑕2𝑡
1
= 𝐿 𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠𝑡 𝑒
=
=
=
=
=
4
1
4
1
4
1
4
1
4
2
2𝑡
=
2
1
1
𝑠−1
2
𝑠 2 −2𝑠+2
𝑒 2𝑡 + 𝑒 −2𝑡
𝐿 𝑠𝑖𝑛3𝑡 + sin 𝑡
𝐿 𝑒 2𝑡 𝑠𝑖𝑛3𝑡 + 𝑒 −2𝑡 𝑠𝑖𝑛3𝑡 + 𝑒 2𝑡 𝑠𝑖𝑛𝑡 + 𝑒 −2𝑡 𝑠𝑖𝑛𝑡
3
𝑠−2
3
2 +9
𝑠 2 −4𝑠+13
+
+
3
2 +9
𝑠+2
+
3
𝑠 2 +4𝑠+13
𝑒 𝑡 −𝑒 −𝑡
1
=
𝑠 2 +2𝑠+2
+ 𝑒 −2𝑡
+
1
𝑠−2
2 +1
1
𝑠 2 −4𝑠+5
2
𝑠𝑖𝑛𝑡
= 𝐿 𝑒 −3𝑡 − 𝑒 −5𝑡 𝑠𝑖𝑛𝑡
2
1
𝑠+1 2 +1
1
𝐿 2𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠𝑡 𝑒 2𝑡 + 𝑒 −2𝑡
= 𝐿 𝑒 −4𝑡
=
−
1
Ans.
10. Find L.T. of 𝑒 −4𝑡 𝑠𝑖𝑛𝑕𝑡 𝑠𝑖𝑛𝑡
[N14/ChemBiot/5M]
Solution:
𝐿 𝑒 −4𝑡 𝑠𝑖𝑛𝑕𝑡 𝑠𝑖𝑛𝑡
2
1
−
𝑒 2𝑡 +𝑒 −2𝑡
= 𝐿 𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠𝑡
2
1
2 +1
𝐿 𝑒 −3𝑡 𝑠𝑖𝑛𝑡 − 𝑒 −5𝑡 𝑠𝑖𝑛𝑡
1
𝑠+3
1
2 +1
2 𝑠 2 +6𝑠+10
S.E/Paper Solutions
−
−
1
𝑠+5 2 +1
1
𝑠 2 +10𝑠+26
7
+
+
1
𝑠+2 2 +1
1
𝑠 2 +4𝑠+5
−
𝑠−1
𝑠 2 −2𝑠+26
Crescent Academy…….………………….…..For Research in Education
2𝑎 2 𝑠
11. Find L.T. of 𝑠𝑖𝑛𝑎𝑡 𝑠𝑖𝑛𝑕𝑎𝑡
Ans. 4 4
𝑠 +4𝑎
12. State and prove first shifting property of Laplace transform and also find
𝐿 𝑡𝑒 3𝑡 𝑠𝑖𝑛𝑕𝑡
Ans.
13. If Laplace Transform of erf
[N18/Comp/5M]
Solution:
We have, 𝐿 erf 𝑡
𝑡 =
4
2
𝑠−1
∞
𝑠−1+4
1
𝑠
+1
4
𝑠−1
3𝑡
𝑡
=
2
𝑠 𝑠+4
2
=
𝑠+3
3
14. Show that 0 𝑒 −𝑡 . sin . sinh
𝑑𝑡 =
2
2
2
−𝑡
15. Find the Laplace Transform of 𝑡𝑒 𝑐𝑜𝑠𝑕2𝑡
[M15/AutoMechCivil/5M]
Solution:
𝐿 𝑡𝑒 −𝑡 𝑐𝑜𝑠𝑕2𝑡
𝑒 2𝑡 +𝑒 −2𝑡
= 𝐿 𝑡 𝑒 −𝑡
=
=
=
1
2
1
2
1
2
2
−3𝑡
𝑡
𝐿 𝑡 𝑒 +𝑒
𝐿 𝑒 𝑡 𝑡 + 𝑒 −3𝑡 𝑡
1
𝑠−1
2
1
+
𝑠+3
2
16. Prove that 𝐿 𝑐𝑜𝑠2𝑡. 𝑐𝑜𝑠𝑕2𝑡 =
17. Prove that 𝐿{𝑐𝑜𝑠𝑎𝑡. 𝑐𝑜𝑠𝑕𝑎𝑡} =
𝑡
18. Find L.T. of sinh sin
2
[N17/Extc/5M]
Solution:
𝑡
3𝑡
2
2
𝐿 sinh sin
𝑡
=𝐿
1
𝑒 2 −𝑒
−
. sin
2
𝑡
2
= 𝐿 𝑒 sin
2
𝑡
2
S.E/Paper Solutions
3𝑡
2
−𝑒
𝑠3
𝑠 4 +64
𝑠3
𝑠 4 +4𝑎 4
3𝑡
2
3𝑡
2
𝑡
−2
sin
3𝑡
2
8
2
1
𝑠−4
𝑡
2
−
1
𝑠−2
then find 𝐿 𝑒 . erf 2 𝑡
𝑠 𝑠+1
4
=
𝑠 𝑠+1
1
=
By change of scale property,
1
𝐿 erf 2 𝑡 = 𝐿 𝑒𝑟𝑓 4𝑡 = . 𝑠
Now,
𝐿 𝑒 𝑡 erf 2 𝑡
1
1
2
Crescent Academy…….………………….…..For Research in Education
=
=
=
=
=
=
=
3
2
1 2 3
𝑠−2 +4
1
2
3
4
1
1 3
𝑠 2 −𝑠+4 +4
3
1
4
3
𝑠 2 −𝑠+1
𝑠 2 +𝑠+1
4
3
4
−
−
3
2
1 2 3
𝑠+2 +4
1
1 3
𝑠 2 +𝑠+4 +4
−
1
𝑠 2 +𝑠+1
−(𝑠 2 −𝑠+1)
𝑠 2 −𝑠+1 (𝑠 2 +𝑠+1)
2𝑠
𝑠 2 +1 2 −𝑠 2
3𝑠
2 𝑠 4 +2𝑠 2 +1−𝑠 2
3𝑠
2 𝑠 4 +𝑠 2 +1
19. Find L.T. of 𝑒 −5𝑡 . erf 𝑡
[M19/Elect/5M]
Solution:
We have,
1
𝐿 erf 𝑡 =
𝑠 𝑠+1
∴𝐿 𝑒
−5𝑡
erf 𝑡 =
1
=
𝑠+5+1
𝑠+5
1
(𝑠+5) 𝑠+6
∞
20. Evaluate 0 𝑒 −2𝑡 . 𝑡 5 𝑐𝑜𝑠𝑕𝑡 𝑑𝑡
[N16/ElexExtcElectBiomInst/5M]
Solution:
By definition,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = 𝑡 5 𝑐𝑜𝑠𝑕𝑡
∞ −𝑠𝑡 5
𝑒 𝑡 𝑐𝑜𝑠𝑕𝑡𝑑𝑡 = 𝐿 𝑡 5 𝑐𝑜𝑠𝑕𝑡
0
𝑒 𝑡 +𝑒 −𝑡
= 𝐿 𝑡5
=
=
1
2
1
2
𝐿 𝑒 𝑡 + 𝑒 −𝑡 𝑡 5
2
𝑡 5
5!
𝑠−1
6
+
5!
𝑠+1
6
Put 𝑠 = 2,
∞ −2𝑡 5
𝑒 t 𝑐𝑜𝑠𝑕𝑡𝑑𝑡
0
S.E/Paper Solutions
=
1 5!
2 1
+
5!
36
=
9
14600
243
Crescent Academy…….………………….…..For Research in Education
Type III: Multiplication by 𝒕𝒏
1.
Find L.T. of 𝑡𝑠𝑖𝑛𝑎𝑡
Ans.
2.
Find L.T. of 𝑡𝑐𝑜𝑠𝑎𝑡
Ans.
3.
3
= −
= −
= −
= −
=
=
5.
𝑠 2 +𝑎 2
2
Find L.T. of 𝑡 sin 𝑡
[N15/AutoMechCivil/5M]
Solution:
𝑑
𝐿 𝑡𝑠𝑖𝑛3 𝑡 = −1
𝐿 sin3 𝑡
𝑑𝑠
𝑑
= −1
4.
2𝑎𝑠
𝑠 2 +𝑎 2 2
𝑠 2 −𝑎 2
𝑑𝑠
𝑑
1
4
𝐿 3𝑠𝑖𝑛𝑡 − 𝑠𝑖𝑛3𝑡
4 𝑑𝑠
1 𝑑
1
3.
4 𝑑𝑠
3 𝑑
1
4
3
1
𝑠 2 +1
1
4
3𝑠
𝑠 2 +1
2
2
−
𝑠 2 +9
1
−
𝑠 2 +9
−1[2𝑠]
𝑠 2 +1
4
3
−
𝑠 2 +1
𝑑𝑠 𝑠 2 +1
𝑠 2 +1 0
6𝑠
2
3𝑡
3𝑠𝑖𝑛𝑡 −𝑠𝑖𝑛 3𝑡
𝐿
−
+
2
𝑠 2 +9 0 −1[2𝑠]
𝑠 2 +9
2
1
𝑠 2 +9
1
2
2
𝑠 2 +9
Find L.T. of 𝑡𝑒 𝑠𝑖𝑛𝑡
Ans.
−3𝑡
2𝑠−6
𝑠 2 −6𝑠+10
Find L.T. of 𝑡𝑒 𝑐𝑜𝑠2𝑡𝑐𝑜𝑠3𝑡
[M14/ElexExtcElectBiomInst/4M]
Solution:
We have,
1
𝐿 𝑐𝑜𝑠2𝑡 𝑐𝑜𝑠3𝑡 = 𝐿 2𝑐𝑜𝑠2𝑡 𝑐𝑜𝑠3𝑡
=
=
2
1
2
1
2
𝐿 𝑐𝑜𝑠5𝑡 + cos −𝑡
𝑠
𝑠 2 +25
𝑑
𝐿 𝑡 𝑐𝑜𝑠2𝑡 𝑐𝑜𝑠3𝑡 = −1
= −1
=−
=−
=
1
2
1
2
1
2
𝑑𝑠
𝑑
𝑠
𝑠 2 +1
𝐿 𝑐𝑜𝑠2𝑡 𝑐𝑜𝑠3𝑡
1
𝑠
𝑠 2 +25
+
𝑑𝑠 2
𝑠 2 +25 1 −𝑠 2𝑠
–𝑠 2 +25
𝑠 2 +25
2
𝑠 −25
𝑠 2 +25
1
=
2
+
2
+
𝑠
𝑠 2 +1
𝑠 2 +1
+
𝑠 2 +25 2
∴ 𝐿 𝑒 −3𝑡 𝑡 𝑐𝑜𝑠2𝑡 𝑐𝑜𝑠3𝑡 =
S.E/Paper Solutions
+
−𝑠 2 +1
𝑠 2 +1 2
2
𝑠 −1
𝑠 2 +1 2
𝑠+3 2 −25
2
1
𝑠+3 2 +25
𝑠 2 +6𝑠−16
2
𝑠 2 +6𝑠+34
10
1 −𝑠 2𝑠
𝑠 2 +1 2
+
2
2
+
𝑠+3 2 −1
𝑠+3 2 +1
𝑠 2 +6𝑠+8
2
𝑠 2 +6𝑠+10
2
2
Crescent Academy…….………………….…..For Research in Education
6.
∞
Evaluate using L.T.: 0 𝑒 −3𝑡 . 𝑡𝑐𝑜𝑠𝑡𝑑𝑡
[M15/ElexExtcElectBiomInst/5M]
Solution:
By definition,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = 𝑡 𝑐𝑜𝑠𝑡
∞ −𝑠𝑡
𝑒 𝑡 𝑐𝑜𝑠𝑡𝑑𝑡 = 𝐿 𝑡 𝑐𝑜𝑠𝑡
0
∞ −𝑠𝑡
𝑒 𝑡
0
𝑐𝑜𝑠𝑡𝑑𝑡 = 𝐿 𝑡 𝑐𝑜𝑠𝑡
𝑑
= −1
𝑑𝑠
𝑑
= −1
𝑠
𝑑𝑠 𝑠 2 +1
𝑠 2 +1 1
= −1
−𝑠 2𝑠
𝑠 2 +1
𝑠 2 +1−2𝑠 2
= −1
𝑠 2 +1
−𝑠 2 +1
= −1
=
𝐿 𝑐𝑜𝑠𝑡
𝑠 2 +1
𝑠 2 −1
𝑠 2 +1
2
2
2
2
Put 𝑠 = 3,
∴
7.
∞
0
𝑒 −3𝑡 𝑡 𝑐𝑜𝑠𝑡𝑑𝑡 =
Evaluate using L.T.:
32 −1
32 +1 2
=
∞
𝑠𝑖𝑛𝑡 2
𝑡
0
𝑒𝑡
8
100
=
2
25
𝑑𝑡
[N13/Chem/6M]
Solution:
We have,
∞
𝑡. 𝑒 −2𝑡 sin2 𝑡 𝑑𝑡 =?
0
By definition,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = 𝑡 sin2 𝑡
∞ −𝑠𝑡
𝑒 𝑡 sin2 𝑡 𝑑𝑡 = 𝐿 𝑡 sin2 𝑡
0
∞ −𝑠𝑡
𝑒 𝑡 sin2
0
1−𝑐𝑜𝑠 2𝑡
𝑡 𝑑𝑡 = 𝐿 𝑡
2
1
= 𝐿 𝑡 − 𝑡𝑐𝑜𝑠2𝑡
=
=
S.E/Paper Solutions
2
1 1
𝑑
2 𝑠
1 1
𝑑𝑠
𝑑
− −1
2
2
𝑠2
− −1
11
𝑑𝑠
𝐿 𝑐𝑜𝑠2𝑡
𝑠
𝑠 2 +4
Crescent Academy…….………………….…..For Research in Education
=
=
=
=
1 1
𝑠 2 +4 1 −𝑠 2𝑠
2 𝑠
𝑠 2 +4
− −1
2
1 1
𝑠 2 +4−2𝑠 2
2 𝑠
𝑠 2 +4
− −1
2
1 1
−𝑠 2 +4
2 𝑠
𝑠 2 +4
− −1
2
1 1
𝑠 2 −4
2 𝑠
𝑠 2 +4
−
2
2
2
2
2
Put 𝑠 = 2,
∴
8.
9.
∞
0
𝑒
−2𝑡
𝑡 sin2 𝑡 𝑑𝑡 =
1 1
−0 =
2 4
1
8
Find L.T. of 𝑡𝑒 3𝑡 𝑐𝑜𝑠𝑡
Ans.
Find Laplace Transform of 𝑡𝑒
[N17/AutoMechCivil/5M]
Solution:
𝑑
𝐿 𝑡 𝑠𝑖𝑛𝑡 = −1
𝐿 𝑠𝑖𝑛𝑡
𝑑𝑠
𝑑
= −1
=
−1[2𝑠]
𝑠 2 +1
2𝑠
𝑠 2 +1
𝑠𝑖𝑛𝑡
1
𝑑𝑠 𝑠 2 +1
𝑠 2 +1 0
= −1
−3𝑡
2
2
2 𝑠+3
∴ 𝐿 𝑒 −3𝑡 𝑡 𝑠𝑖𝑛𝑡 =
𝑠+3 2 +1
2
=
2 𝑠+3
𝑠 2 +6𝑠+10
2
10. Find L.T. of 𝑡𝑒 3𝑡 𝑠𝑖𝑛4𝑡
[N16/CompIT/5M][M18/IT/4M]
Solution:
𝐿 𝑡𝑒 3𝑡 𝑠𝑖𝑛4𝑡 =?
We have,
4
𝐿 𝑠𝑖𝑛4𝑡 = 2
𝑠 +16
𝐿 𝑡 𝑠𝑖𝑛4𝑡 = −1
= −1
= −4
=
𝐿 𝑠𝑖𝑛4𝑡
4
𝑠 2 +16
𝑑𝑠
𝑠 2 +16 0 −1[2𝑠]
𝑠 2 +16
8𝑠
𝑠 2 +16
∴ 𝐿 𝑒 3𝑡 𝑡 𝑠𝑖𝑛4𝑡 =
S.E/Paper Solutions
𝑑
𝑑𝑠
𝑑
2
2
8 𝑠−3
𝑠−3
2 +16 2
=
12
8 𝑠−3
𝑠 2 −6𝑠+25 2
𝑠 2 −6𝑠+8
𝑠 2 −6𝑠+10
2
Crescent Academy…….………………….…..For Research in Education
11. Find the Laplace Transform of 𝑒 −2𝑡 𝑡 𝑐𝑜𝑠𝑡
[M18/Comp/5M]
Solution:
𝑑
𝐿 𝑐𝑜𝑠𝑡
𝐿 𝑡 𝑐𝑜𝑠𝑡 = −1
= −1
𝑑𝑠
𝑑
= −1
−𝑠 2𝑠
𝑠 2 +1 2
2
𝑠 +1−2𝑠 2
= −1
𝑠 2 +1
−𝑠 2 +1
= −1
=
𝑠
𝑑𝑠 𝑠 2 +1
𝑠 2 +1 1
𝑠 2 +1
𝑠 2 −1
𝑠 2 +1
2
2
2
𝑠+2 2 −1
∴ 𝐿 𝑒 −2𝑡 𝑡 𝑐𝑜𝑠𝑡 =
𝑠+2 2 +1
2
𝑠 2 +4𝑠+4−1
=
𝑠 2 +4𝑠+4+1
2
=
𝑠 2 +4𝑠+3
𝑠 2 +4𝑠+5
12. Find L.T. of 𝑒 −𝑡 𝑡 𝑐𝑜𝑠2𝑡 𝑠𝑖𝑛4𝑡
[M18/Elect/5M]
Solution:
𝐿 𝑡𝑒 −𝑡 𝑐𝑜𝑠2𝑡 𝑠𝑖𝑛4𝑡 =?
We have,
1
𝐿 𝑐𝑜𝑠2𝑡 𝑠𝑖𝑛4𝑡 = 𝐿 2𝑐𝑜𝑠2𝑡 𝑠𝑖𝑛4𝑡
=
=
2
1
2
1
2
𝐿 𝑠𝑖𝑛6𝑡 + sin −2𝑡
6
𝑑
𝐿 𝑡 𝑐𝑜𝑠2𝑡 𝑠𝑖𝑛4𝑡 = −1
𝑑𝑠
𝑑
= −1
=−
=−
=
−
𝑠 2 +36
𝐿 𝑐𝑜𝑠2𝑡 𝑠𝑖𝑛4𝑡
1
6
𝑠 2 +36
12𝑠
−
𝑠 2 +36
2
6𝑠
𝑠 2 +36 2
∴ 𝐿 𝑡𝑒 −𝑡 𝑐𝑜𝑠2𝑡 𝑠𝑖𝑛4𝑡 =
=
−
2
2
−
2
𝑠 2 +4
4𝑠
+
𝑠 2 +4
2𝑠
2
𝑠 2 +4 2
6 𝑠+1
2 +36 2
𝑠+1
6𝑠+6
𝑠 2 +2𝑠+37
13. Find L.T. of 𝑒 𝑡 𝑡 𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠𝑡
[M19/Comp/5M]
Solution:
S.E/Paper Solutions
−
𝑑𝑠 2 𝑠 2 +36
𝑠 2 +4
2
𝑠 +36 0 −6[2𝑠]
𝑠 2 +4 0 −2[2𝑠]
1
2
1
2
𝑠 2 +4
13
2
−
−
2 𝑠+1
𝑠+1 2 +4
2𝑠+2
𝑠 2 +2𝑠+5
2
2
2
2
Crescent Academy…….………………….…..For Research in Education
𝐿 𝑒 𝑡 𝑡 𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠𝑡 =?
We have,
1
𝐿 𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠𝑡 = 𝐿 2𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠𝑡
2
=
=
1
𝐿 𝑠𝑖𝑛3𝑡 + sint
2
1
3
𝑠 2 +9
2
𝐿 𝑡 𝑐𝑜𝑠2𝑡 𝑠𝑖𝑛4𝑡 = −1
=−
=
2
3𝑠
𝑑
𝐿 𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠𝑡
1
3
𝑠 2 +9
6𝑠
−
𝑠 2 +9 2
𝑠 2 +9
+
=
−
𝑠
𝑠 2 +1
−
3 𝑠−1
𝑠 2 −2𝑠+10 2
𝑠−1 2 +1 2
𝑠−1
+
𝑠 2 −2𝑠+2
∞
𝑠𝑖𝑛𝑡𝑑𝑡 = 𝐿 𝑡 𝑠𝑖𝑛𝑡
𝑑
= −1
𝑑𝑠
𝑑
= −1
=
Put 𝑠 = 3,
∞
∴ 0 𝑒 −3𝑡 𝑡 𝑠𝑖𝑛𝑡𝑑𝑡 =
15. Evaluate using L.T.:
[N18/Extc/5M]
Solution:
By definition,
S.E/Paper Solutions
∞
0
𝑠 2 +1
2
2
6
32 +1 2
𝑒
1
𝑠 2 +1
2𝑠
−2𝑡
𝐿 𝑠𝑖𝑛𝑡
𝑑𝑠 𝑠 2 +1
𝑠 2 +1 [0]−1[2𝑠]
= −1
=
2
2 𝑠−1
14. Evaluate using L.T.: 0 𝑒 −3𝑡 𝑡 𝑠𝑖𝑛𝑡 𝑑𝑡
[M17/AutoMechCivil/6M]
Solution:
By definition,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = 𝑡 𝑠𝑖𝑛𝑡
∞ −𝑠𝑡
𝑒 𝑡 𝑠𝑖𝑛𝑡𝑑𝑡 = 𝐿 𝑡 𝑠𝑖𝑛𝑡
0
∞ −𝑠𝑡
𝑒 𝑡
0
𝑠 2 +1
2𝑠
𝑠 2 +1 2
2 +9 2
𝑠−1
+
2
2
3 𝑠−1
∴ 𝐿 𝑒 𝑡 𝑡 𝑠𝑖𝑛2𝑡 𝑐𝑜𝑠𝑡 =
1
+
𝑑𝑠 2 𝑠 2 +9
𝑠 2 +1
2
𝑠 +9 [0]−3[2𝑠]
𝑠 2 +1 [0]−1[2𝑠]
1
2
1
𝑠 2 +1
𝑑𝑠
𝑑
= −1
=−
1
+
6
100
=
𝑡 𝑠𝑖𝑛𝑡 𝑑𝑡
14
3
50
2
2
Crescent Academy…….………………….…..For Research in Education
∞ −𝑠𝑡
𝑒 𝑓
0
𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
Put 𝑓 𝑡 = 𝑡 𝑠𝑖𝑛𝑡
∞ −𝑠𝑡
𝑒 𝑡 𝑠𝑖𝑛𝑡𝑑𝑡 = 𝐿 𝑡 𝑠𝑖𝑛𝑡
0
∞ −𝑠𝑡
𝑒 𝑡
0
𝑠𝑖𝑛𝑡𝑑𝑡 = 𝐿 𝑡 𝑠𝑖𝑛𝑡
𝑑
= −1
= −1
𝑠 2 +1
2𝑠
=
2
𝑠 2 +1 2
4
∞
0
1
𝑑𝑠 𝑠 2 +1
𝑠 2 +1 [0]−1[2𝑠]
= −1
Put 𝑠 = 2,
∞
∴ 0 𝑒 −2𝑡 𝑡 𝑠𝑖𝑛𝑡𝑑𝑡 =
𝐿 𝑠𝑖𝑛𝑡
𝑑𝑠
𝑑
22 +1 2
4
=
𝑡
25
2
16. Evaluate using L.T.:
𝑒 𝑡 cos 𝑡 𝑑𝑡
[M19/AutoMechCivil/6M]
Solution:
We have,
∞ 𝑡
𝑒 𝑡 cos 2 𝑡 𝑑𝑡 =?
0
By definition,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = 𝑡 cos 2 𝑡
∞ −𝑠𝑡
𝑒 𝑡 cos 2 𝑡 𝑑𝑡 = 𝐿 𝑡 cos 2 𝑡
0
∞ −𝑠𝑡
𝑒 𝑡 cos 2
0
1+𝑐𝑜𝑠 2𝑡
𝑡 𝑑𝑡 = 𝐿 𝑡
2
1
= 𝐿 𝑡 + 𝑡𝑐𝑜𝑠2𝑡
=
=
=
=
=
=
2
1 1
𝑠2
2
1 1
2
𝑠2
+ −1
+ −1
𝑑
𝑑𝑠
𝑑
𝑑𝑠
𝐿 𝑐𝑜𝑠2𝑡
𝑠
𝑠 2 +4
1 1
𝑠 2 +4 1 −𝑠 2𝑠
2 𝑠
𝑠 2 +4
+ −1
2
1 1
𝑠 2 +4−2𝑠 2
2 𝑠
𝑠 2 +4
+ −1
2
1 1
−𝑠 2 +4
2 𝑠
𝑠 2 +4
+ −1
2
1 1
𝑠 2 −4
2 𝑠
𝑠 2 +4
+
2
2
2
2
2
Put 𝑠 = −1,
∴
∞
0
𝑒 𝑡 𝑡 cos 2 𝑡 𝑑𝑡 =
S.E/Paper Solutions
1
2
1
−1
+
2
−1 2 −4
−1
15
2 +4 2
=
11
25
Crescent Academy…….………………….…..For Research in Education
17. Find L.T. of 𝑡 1 + 𝑠𝑖𝑛𝑡
[M15/ElexExtcElectBiomInst/4M][N15/ElexExtcElectBiomInst/4M]
[M16/ElexExctElectBiomInst/4M][N17/AutoMechCivil/6M]
[N17/Extc/4M][M18/Extc/4M][M19/Extc/4M]
Solution:
𝑑
𝐿 𝑡 1 + 𝑠𝑖𝑛𝑡 = −1
𝐿 1 + 𝑠𝑖𝑛𝑡
= −1
= −1
= −1
= −1
𝑑𝑠
𝑑
𝑑𝑠
𝑡
𝑡
2
2
𝐿 cos + sin
𝑑
𝑠
1
2
1
𝑠 2 +4
+
1
𝑑𝑠 𝑠 2 +
4
1
𝑑 𝑠+2
1
𝑑𝑠 𝑠 2 +
4
1
𝑠 2 +4 1
1
− 𝑠+2 2𝑠
1 2
𝑠 2 +4
1
𝑠 2 +4 −2𝑠 2 −𝑠
= −1
1 2
𝑠 2 +4
1
–𝑠 2 −𝑠+4
= −1
1 2
1
=
=
𝑠 2 +4
𝑠 2 +𝑠−4
2
4𝑠2 +1
4
4 4𝑠 2 +4𝑠−1
18. Evaluate using L.T.:
4𝑠 2 +1
∞
0
2
𝑡 1 + 𝑠𝑖𝑛𝑡𝑑𝑡
Ans. −4
19. Find L.T. of 𝑡 1 + 𝑠𝑖𝑛2𝑡
Ans.
20. Find L.T. of 𝑡 1 − 𝑠𝑖𝑛𝑡
Ans.
∞
0
21. Evaluate using L.T.:
𝑒 −𝑡 . 𝑡 1 + 𝑠𝑖𝑛𝑡𝑑𝑡
[N16/AutoMechCivil/5M]
Solution:
By definition,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = 𝑡 1 + 𝑠𝑖𝑛𝑡
∞ −𝑠𝑡
𝑒 𝑡 1 + 𝑠𝑖𝑛𝑡𝑑𝑡 = 𝐿 𝑡 1 + 𝑠𝑖𝑛𝑡
0
= −1
= −1
S.E/Paper Solutions
𝑑
𝑑𝑠
𝑑
𝑑𝑠
16
𝐿
1 + 𝑠𝑖𝑛𝑡
𝑡
𝑡
2
2
𝐿 cos + sin
𝑠 2 +2𝑠−1
𝑠 2 +1 2
4 4𝑠 2 −4𝑠−1
4𝑠 2 +1
2
Crescent Academy…….………………….…..For Research in Education
𝑑
= −1
𝑠
1
𝑑𝑠 𝑠 2 +
4
+
1
𝑠+2
𝑑
= −1
1
2
1
𝑠 2 +4
1
𝑑𝑠 𝑠 2 +
4
1
1
𝑠 2 +4 1 − 𝑠+ 2𝑠
2
= −1
1 2
𝑠 2 +4
1
𝑠 2 +4 −2𝑠 2 −𝑠
= −1
1 2
𝑠 2 +4
1
–𝑠 2 −𝑠+4
= −1
1 2
𝑠 2 +4
1
𝑠 2 +𝑠−4
=
2
4𝑠2 +1
4
4 4𝑠 2 +4𝑠−1
=
4𝑠 2 +1
2
Put 𝑠 = 1,
∞
28
∴ 0 𝑒 −𝑡 . 𝑡 1 + 𝑠𝑖𝑛𝑡𝑑𝑡 =
25
22. Find L.T. of 𝑡 𝑒 3𝑡 erf 5 𝑡
[M19/IT/4M]
Solution:
We have,
1
𝐿 erf 𝑡 =
𝑠 𝑠+1
By change of scale property,
1
1
𝐿 erf 5 𝑡 = . 𝑠 𝑠 =
25
25
25
Now,
𝐿 𝑡 erf 5 𝑡 = −1
= −1
= −1
= (−5)
= −5
𝑑𝑠
𝑑
𝐿 erf 5 𝑡
5
𝑑𝑠 𝑠 𝑠+25
𝑑
5
𝑑𝑠
𝑠 3 +25𝑠 2
𝑑
𝑠 3 + 25𝑠 2
𝑑𝑠
−
1
5
2
3𝑠 2 +50𝑠
2
𝑠 3 +25𝑠 2
= .
S.E/Paper Solutions
𝑠 𝑠+25
+1
𝑑
5
1
2
1
2
−
𝑠 3 + 25𝑠 2
3
2
17
3
−2
× 3𝑠 2 + 50𝑠
Crescent Academy…….………………….…..For Research in Education
5
𝑠 3𝑠+50
= .
2 𝑠 3 𝑠+25
5
3𝑠+50
3
2
2 𝑠 2 𝑠+25
3
2
= .
Thus,
5
3 𝑠−3 +50
𝐿 𝑒 3𝑡 𝑡 erf 5 𝑡 = .
2
𝑠−3
1
23. Given 𝐿 erf 𝑡 =
𝑠 𝑠+1
2
5
3
2
𝑠−3+25
2
∞
𝑡
0
, evaluate
3𝑠+41
= .
𝑠−3
2
𝑠+22
3
2
𝑒 −𝑡 erf 𝑡 𝑑𝑡
[M14/CompIT/6M]
Solution:
By definition,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = 𝑡 erf 𝑡
∞ −𝑠𝑡
𝑒 𝑡 erf 𝑡 𝑑𝑡 = 𝐿 𝑡 erf 𝑡
0
𝑑
= −1
= −1
= −1
𝑑
1
𝑑𝑠
𝑠 3 +𝑠 2
𝑑
𝑠3 + 𝑠2
𝑑𝑠
= −1
𝑡 𝑑𝑡 =
25. Prove that
=
1
−
3𝑠 2 +2𝑠
2 𝑠 3 +𝑠 2
Put 𝑠 = 1,
∞
∴ 0 𝑒 −𝑡 𝑡 erf 𝑡 𝑑𝑡 =
24. If 𝐿 𝐽0 𝑡
1
𝑑𝑠 𝑠 𝑠+1
= −1
∞ −𝑠𝑡
𝑒 𝑡 erf
0
𝐿 erf 𝑡
𝑑𝑠
𝑑
5
2 2
3
2
1
2
3
−2
× 3𝑠 2 + 2𝑠
3
2
=
5
4 2
, find 𝐿 𝑡𝐽0 3𝑡
Ans.
4𝑡 𝑑𝑡 =
3
125
[M18/IT/6M]
Solution:
By definition,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = 𝑡 𝐽0 4𝑡
∞ −𝑠𝑡
𝑒 𝑡 𝐽0 4𝑡 𝑑𝑡 = 𝐿 𝑡 𝐽0 4𝑡
0
We have,
S.E/Paper Solutions
1
−2
𝑠3 + 𝑠2
𝑠 2 +1
∞
𝑡𝑒 −3𝑡 𝐽0
0
1
2
18
where 𝐿 𝐽0 𝑡
𝑠
𝑠 2 +9
=
3
2
1
1+𝑠 2
Crescent Academy…….………………….…..For Research in Education
𝐿 𝐽0 𝑡
1
=
(given)
𝑠 2 +1
By change of scale property,
1
1
1
= .
𝐿 𝐽0 4𝑡 = .
2
4
∴
∞
0
𝑒
−𝑠𝑡
𝑠
4
+1
1
4
𝑠2 +16
16
.
𝑠 2 +16
=
1
𝑠 2 +16
𝑡 𝐽0 4𝑡 𝑑𝑡 = 𝐿 𝑡 𝐽0 4𝑡
𝑑
𝑠 2 +16
2
−
𝑠
1
−2
𝑠 2 + 16
2
3
−2
× 2𝑠
3
2
𝑠 2 +16
3
9+16
1
𝑠 + 16
𝑑𝑠
= −1
Put 𝑠 = 3,
∞
∴ 0 𝑒 −3𝑡 𝑡 𝐽0 4𝑡 𝑑𝑡 =
1
𝑑𝑠
𝑑
= −1
𝑒 −𝑠𝑡 𝑡 𝐽0 4𝑡 𝑑𝑡 =
𝐿 𝐽0 4𝑡
𝑑𝑠
𝑑
= −1
∞
0
1
16
16
= −1
∴
=
3
2
3
=
25
3
2
=
3
125
3𝑠+8
26. Find L.T. of 𝑡 erf 2 𝑡
Ans.
27. Find L.T. of 𝑡𝑒𝑟𝑓 3 𝑡
Ans. .
28. Show that
∞
0
𝑡𝑒 −6𝑡 𝐽0 8𝑡 𝑑𝑡 =
3
500
where 𝐿 𝐽0 𝑡
[N14/ElexExtcElectBiomInst/6M]
Solution:
By definition,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = 𝑡 𝐽0 8𝑡
∞ −𝑠𝑡
𝑒 𝑡 𝐽0 8𝑡 𝑑𝑡 = 𝐿 𝑡 𝐽0 8𝑡
0
We have,
1
(given)
𝐿 𝐽0 𝑡 = 2
𝑠 +1
By change of scale property,
1
1
1
1
1
𝐿 𝐽0 8𝑡 = .
= . 2 = .
2
8
∴
∞
0
𝑠
8
+1
8
8
𝑠 +64
64
8
𝑠 2 +64
𝑒 −𝑠𝑡 𝑡 𝐽0 8𝑡 𝑑𝑡 = 𝐿 𝑡 𝐽0 8𝑡
= −1
= −1
= −1
S.E/Paper Solutions
𝑑
𝑑𝑠
𝑑
𝑑𝑠
𝑑
𝑑𝑠
𝐿 𝐽0 8𝑡
1
𝑠 2 +64
2
𝑠 + 64
19
1
2
−
=
3
𝑠 2 𝑠+4 2
9
𝑠+6
2 𝑠 2 𝑠+9
1
=
𝑠 2 +1
1
𝑠 2 +64
3
2
Crescent Academy…….………………….…..For Research in Education
= −1
∴
∞
0
𝑒 −𝑠𝑡 𝑡 𝐽0 8𝑡 𝑑𝑡 =
Put 𝑠 = 6,
∞
∴ 0 𝑒 −6𝑡 𝑡 𝐽0 8𝑡 𝑑𝑡 =
29. Evaluate
∞
𝑡𝑒 −4𝑡 𝑓
0
−
𝑠
𝑠 2 +64
1
2
∴
∞
0
𝑠
3
6
36+64
3
100
=
3
500
1
𝑠 2 +1
1
3
𝑠 2 +9
3
6
1000
=
1
𝑠 2 +9
𝑒 −𝑠𝑡 𝑡 𝑓 3𝑡 𝑑𝑡 = 𝐿 𝑡 𝑓 3𝑡
= −1
= −1
𝑑
𝑒 −𝑠𝑡 𝑡 𝑓 3𝑡 𝑑𝑡 =
Put 𝑠 = 4,
∞
∴ 0 𝑒 −4𝑡 𝑡 𝑓 3𝑡 𝑑𝑡 =
𝐿 𝑓 3𝑡
𝑑𝑠
𝑑
1
𝑠 2 +9
2
𝑑𝑠
𝑑
𝑠 +9
𝑑𝑠
= −1
∞
0
=
3
2
=
= .
𝑠 +9
9
= −1
∴
6
=
3
2
3𝑡 𝑑𝑡 if 𝐿 𝑓 𝑡
+1
× 2𝑠
3
2
[N13/Biot/6M]
Solution:
By definition,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = 𝑡 𝑓 3𝑡
∞ −𝑠𝑡
𝑒 𝑡 𝑓 3𝑡 𝑑𝑡 = 𝐿 𝑡 𝑓 3𝑡
0
We have,
1
𝐿𝑓 𝑡 = 2
(given)
𝑠 +1
By change of scale property,
1
1
1
1
𝐿 𝑓 3𝑡 = .
= . 2
2
3
3
−2
𝑠 2 + 64
−
𝑠
𝑠 2 +9
2
3
2
=
= −1
S.E/Paper Solutions
3
3 𝑑
𝐿
𝑑𝑠 3
3
𝑑
𝑠
4
25
30. Find L.T. of 𝑡 3 𝑐𝑜𝑠𝑡
[N14/AutoMechCivil/5M]
Solution:
𝐿 𝑡 3 𝑐𝑜𝑠𝑡 = −1
𝑠2 + 9
𝑐𝑜𝑠𝑡
𝑑𝑠 3 𝑠 2 +1
20
3
2
=
3
−2
3
2
4
16+9
1
1
2
−
4
125
× 2𝑠
Crescent Academy…….………………….…..For Research in Education
= −1
= −1
= −1
=
=
=
=
=
=
𝑑2
=
=
𝑠 2 +1 1 −𝑠 2𝑠
𝑑𝑠 2
𝑠 2 +1
2
2
𝑠 +1−2𝑠 2
𝑑
𝑠 2 +1
1−𝑠 2
𝑑𝑠 2
𝑑2
𝑑𝑠 2 𝑠 2 +1
𝑠 2 −1
2
2
2
𝑑𝑠 2 𝑠 2 +1 2
𝑑 𝑠 2 +1 2 2𝑠 − 𝑠 2 −1 2 𝑠 2 +1 2𝑠
𝑠 2 +1 4
𝑠 2 +1 2𝑠 − 𝑠 2 −1 4𝑠
𝑑𝑠
𝑠 2 +1
𝑑
𝑠 2 +1
𝑑𝑠
4
2𝑠 3 +2𝑠−4𝑠 3 +4𝑠
𝑑
𝑑𝑠
𝑠 2 +1
𝑑 6𝑠−2𝑠 3
3
𝑑𝑠 𝑠 2 +1 3
𝑠 2 +1 3 6−6𝑠 2 − 6𝑠−2𝑠 3 3 𝑠 2 +1
𝑠 2 +1 2
=
=
𝑑2
6
6𝑠 2 −6𝑠 4 +6−6𝑠 2 −36𝑠 2 +12𝑠 4
𝑠 2 +1
4
6𝑠 4 −36𝑠 2 +6
𝑠 2 +1 4
6 𝑠 4 −6𝑠 2 +1
4
31. Find Laplace of 𝑡 5 𝑐𝑜𝑠𝑕𝑡
[N15/CompIT/5M]
Solution:
𝑒 𝑡 +𝑒 −𝑡
𝐿 𝑡 5 𝑐𝑜𝑠𝑕𝑡 = 𝐿 𝑡 5
=
=
=
S.E/Paper Solutions
1
2
1
2
𝑡 5
𝐿 𝑒 𝑡 + 𝑒 −𝑡 𝑡 5
5!
2 𝑠−1
5!
1
2
𝑠−1
+
6
6
+
2𝑠
𝑠 2 +1 6
6−6𝑠 2 − 6𝑠−2𝑠 3 6𝑠
𝑠 2 +1
𝑠 2 +1
𝑠 2 +1
2
5!
𝑠+1
1
𝑠+1
6
6
21
Crescent Academy…….………………….…..For Research in Education
Type IV: Division by t
1.
𝑠𝑖𝑛𝑎𝑡
Find Laplace transform of
[M18/AutoMechCivil/5M]
Solution:
𝐿
𝑠𝑖𝑛𝑎𝑡
∞
𝐿 𝑠𝑖𝑛𝑎𝑡
𝑠
∞ 𝑎
𝑑𝑠
𝑠 𝑠 2 +𝑎
∞
−1 𝑠
=
𝑡
=
𝑡
. Does Laplace transform of
𝑐𝑜𝑠𝑎𝑡
𝑡
exist?
𝑑𝑠
= tan
𝑎 𝑠
𝑠
= − tan−1
2
𝑎
𝑠
= cot −1
𝑎
𝑐𝑜𝑠𝑎𝑡
∞
= 𝑠 𝐿 𝑐𝑜𝑠𝑎𝑡 𝑑𝑠
𝑡
∞ 𝑠
= 𝑠 2 𝑑𝑠
𝑠 +𝑎
∞
1
2
2
𝜋
𝐿
=
2
1
log 𝑠 + 𝑎
𝑠
2
= 𝑙𝑜𝑔∞ − log 𝑠 + 𝑎2
2
= 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡
2.
𝑠𝑖𝑛 𝑕2𝑡
Find L.T. of
𝑡
[N17/Extc/4M][M18/Extc/4M]
Solution:
𝐿
𝑠𝑖𝑛 𝑕2𝑡
𝑡
∞
𝐿 𝑠𝑖𝑛𝑕2𝑡 𝑑𝑠
𝑠
∞ 2
𝑑𝑠
𝑠 𝑠 2 −4
1
𝑠−2 ∞
=
=
= 2×
log
2 2
1
1
𝑠+2 𝑠
𝑠−2
= log 1 − log
2
1
2
𝑠−2
2
𝑠+2
𝑠+2
= − log
1
= log
2
3.
Find L.T. of
4.
Find L.T. of
5.
Find L.T. of
6.
Find L.T. of
𝑠+2
𝑠−2
𝑒 −4𝑡 𝑠𝑖𝑛 3𝑡
𝑡
𝑒 −2𝑡 𝑐𝑜𝑠 2𝑡𝑠𝑖𝑛 3𝑡
𝑡
𝑐𝑜𝑠 𝑕2𝑡 𝑠𝑖𝑛 2𝑡
𝑡
𝑒 −2𝑡 𝑠𝑖𝑛 2𝑡 𝑐𝑜𝑠 𝑕𝑡
S.E/Paper Solutions
𝑡
Ans. cot −1
Ans.
Ans.
22
𝑠+4
3
−1 𝑠+2
−1
cot
+ cot
𝑠+2
2
5
1
𝑠−2
𝑠+2
Ans. cot −1
+ cot −1
2
2
2
1
𝑠+1
𝑠+3
𝜋 − tan−1
− tan−1
2
2
2
1
Crescent Academy…….………………….…..For Research in Education
7.
∞ 𝑒−
2𝑡 𝑠𝑖𝑛 𝑕𝑡𝑠𝑖𝑛𝑡
Solve 0
𝑑𝑡
𝑡
[M16/AutoMechCivil/6M] [M19/Elex/5M][M19/Elect/6M]
Solution:
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 =
𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑡
∞ −𝑠𝑡 𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑒
0
𝑡
∞ −𝑠𝑡 𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑒
0
𝑡
∞ −𝑠𝑡 𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑒
0
𝑡
𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑑𝑡 = 𝐿
𝑡
𝑒 𝑡 −𝑒 −𝑡
𝑑𝑡 = 𝐿
1
𝑑𝑡 = 𝐿
2
.
𝑠𝑖𝑛𝑡
2
𝑡
𝑡 𝑠𝑖𝑛𝑡
𝑒
−
𝑡
𝑒 −𝑡
𝑠𝑖𝑛𝑡
𝑡
We have,
∞
∞ 1
𝜋
𝐿 𝑠𝑖𝑛𝑡 𝑑𝑠 = 𝑠 2 𝑑𝑠 = [tan−1 𝑠]∞
= − tan−1
𝑠
𝑠
𝑡
𝑠 +1
2
∞ −𝑠𝑡 𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝜋
1 𝜋
−1
−1
∴ 0 𝑒
𝑑𝑡 =
− tan 𝑠 − 1 − + tan 𝑠 + 1
𝑡
2
2 2
∞ −𝑠𝑡 𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
1
−1
−1
𝑑𝑡 = tan 𝑠 + 1 − tan 𝑠 − 1
𝑒
0
2
𝑡
∞ −𝑠𝑡 𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑠+1 − 𝑠−1
1
𝑑𝑡 = tan−1
𝑒
0
1+ 𝑠+1 𝑠−1
2
𝑡
∞ −𝑠𝑡 𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
2
1
1
−1
−1 2
𝑒
𝑑𝑡
tan
=
tan
=
2
0
𝑡
1+𝑠 −1
2
𝑠2
2
𝐿
𝑠𝑖𝑛𝑡
=
Put 𝑠 = 2,
∞ − 2𝑡 𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑒
0
𝑡
8.
Prove that
2
1
𝑑𝑡 = tan−1
2
∞ 𝑠𝑖𝑛 2𝑡+𝑠𝑖𝑛 3𝑡
0
𝑡𝑒 𝑡
𝑑𝑡 =
2
2
1
1 𝜋
𝜋
2
2 4
8
= tan−1 (1) = . =
3𝜋
4
[M14/ChemBiot/6M][M18/Comp/6M][N18/Elex/5M]
Solution:
By definition of L.T.,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 =
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
𝑠𝑖𝑛 2𝑡+𝑠𝑖𝑛 3𝑡
𝑡
𝑠𝑖𝑛 2𝑡+𝑠𝑖𝑛 3𝑡
𝑡
𝑠𝑖𝑛 2𝑡+𝑠𝑖𝑛 3𝑡
𝑡
𝑠𝑖𝑛 2𝑡+𝑠𝑖𝑛 3𝑡
𝑡
𝑠𝑖𝑛 2𝑡+𝑠𝑖𝑛 3𝑡
𝑡
𝑠𝑖𝑛 2𝑡+𝑠𝑖𝑛 3𝑡
𝑡
𝑠𝑖𝑛 2𝑡+𝑠𝑖𝑛 3𝑡
S.E/Paper Solutions
𝑡
𝑑𝑡 = 𝐿
𝑑𝑡 =
𝑑𝑡 =
𝑑𝑡 =
𝑠𝑖𝑛 2𝑡+𝑠𝑖𝑛 3𝑡
𝑡
∞
𝐿 𝑠𝑖𝑛2𝑡 + 𝑠𝑖𝑛3𝑡 𝑑𝑠
𝑠
∞ 2
3
+
𝑑𝑠
2
2
𝑠 𝑠 +4
𝑠 +9
𝑠
𝑠 ∞
tan−1 + tan−1
2
3 𝑠
𝜋
𝜋
−1 𝑠
−1 𝑠
𝑑𝑡 = − tan
2
𝑑𝑡 = 𝜋 −
+ − tan
2
2
3
−1 𝑠
−1 𝑠
tan
+ tan
2
3
23
𝑠
Crescent Academy…….………………….…..For Research in Education
∞ −𝑠𝑡 𝑠𝑖𝑛 2𝑡+𝑠𝑖𝑛 3𝑡
𝑒
𝑑𝑡
0
𝑡
𝑠 𝑠
+
2 3
𝑠 𝑠
1−2 .3
−1
= 𝜋 − tan
Put 𝑠 = 1,
∞ −𝑡 𝑠𝑖𝑛 2𝑡+𝑠𝑖𝑛 3𝑡
𝑒
𝑑𝑡
0
𝑡
= 𝜋 − tan
−1
1 1
+
2 3
1
1−6
= 𝜋 − tan−1 1
𝜋
=𝜋−
=
9.
𝑠𝑖𝑛𝑡
3𝜋
4
4
Find L.T. of
𝑡
[M15/CompIT/5M]
Solution:
𝐿
𝑠𝑖𝑛𝑡
𝑡
=
∞
𝐿 𝑠𝑖𝑛𝑡
𝑠
∞ 1
𝑑𝑠
𝑠 𝑠 2 +1
tan−1 𝑠 ∞
𝑠
𝜋
−1
=
=
= − tan
2
= cot −1 𝑠
𝑑𝑠
𝑠
1
10. Find the L.T. of 𝑒 −𝑡 𝑠𝑖𝑛𝑡
𝑡
[N17/Comp/5M]
Solution:
𝐿
𝑠𝑖𝑛𝑡
𝑡
=
∞
𝐿 𝑠𝑖𝑛𝑡
𝑠
∞ 1
𝑑𝑠
𝑠 𝑠 2 +1
tan−1 𝑠 ∞
𝑠
𝜋
−1
𝑑𝑠
=
=
= − tan 𝑠
2
= cot −1 𝑠
𝑠𝑖𝑛𝑡
∴ 𝐿 𝑒 −𝑡
= cot −1 (𝑠 + 1)
𝑡
∞
11. Evaluate using L.T.: 0 𝑒 −2𝑡 𝑠𝑖𝑛𝑕𝑡
[M16/ElexExctElectBiomInst/5M]
Solution:
By definition of L.T.,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 =
𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑡
∞ −𝑠𝑡 𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑒
0
𝑡
S.E/Paper Solutions
𝑑𝑡 = 𝐿
𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑡
24
𝑠𝑖𝑛𝑡
𝑡
𝑑𝑡
Crescent Academy…….………………….…..For Research in Education
∞ −𝑠𝑡 𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑒
0
𝑡
∞ −𝑠𝑡 𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑒
0
𝑡
We
have,
𝑒 𝑡 −𝑒 −𝑡
𝑑𝑡 = 𝐿
1
𝑑𝑡 = 𝐿
2
𝑠𝑖𝑛𝑡
𝐿
=
𝑡
tan−1 𝑠
∞
𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
1
∴ 0 𝑒 −𝑠𝑡
𝑑𝑡 =
𝑡
∞ −𝑠𝑡 𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑒
0
𝑡
∞ −𝑠𝑡 𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑒
0
𝑡
∞ −𝑠𝑡 𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑒
0
𝑡
𝑑𝑡 =
2
𝑡
𝑡 𝑠𝑖𝑛𝑡
𝑒
−
𝑡
∞
𝐿 𝑠𝑖𝑛𝑡
𝑠
𝜋
2 2
1
tan
2
1
𝑠𝑖𝑛𝑡
.
𝑑𝑡 = tan
𝑠𝑖𝑛𝑡
𝑡
𝑑𝑠 =
∞ 1
𝑑𝑠
𝑠 𝑠 2 +1
= [tan−1 𝑠]∞
𝑠 =
𝜋
− tan−1 𝑠 − 1 − + tan−1 𝑠 + 1
2
−1
𝑑𝑡 = tan−1
2
1
𝑒 −𝑡
−1
2
−1
𝑠 + 1 − tan
𝑠−1
𝑠+1 − 𝑠−1
1+ 𝑠+1 𝑠−1
2
1
2
1+𝑠 2 −1
𝑠2
= tan−1
2
Put 𝑠 = 2,
∞ −2𝑡 𝑠𝑖𝑛 𝑕𝑡 𝑠𝑖𝑛𝑡
𝑒
0
𝑡
1
𝑑𝑡 = tan−1
2
2
2
2
1
1
2
2
= tan−1
∞ 𝑒 −2𝑡 𝑐𝑜𝑠 2𝑡𝑠𝑖𝑛 3𝑡
12. Evaluate 0
𝑑𝑡
𝑡
[N15/AutoMechCivil/6M]
Solution:
By definition of L.T.,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 =
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
𝑐𝑜𝑠 2𝑡 𝑠𝑖𝑛 3𝑡
𝑡
𝑐𝑜𝑠 2𝑡 𝑠𝑖𝑛 3𝑡
𝑡
𝑐𝑜𝑠 2𝑡 𝑠𝑖𝑛 3𝑡
𝑡
𝑐𝑜𝑠 2𝑡 𝑠𝑖𝑛 3𝑡
𝑡
𝑐𝑜𝑠 2𝑡 𝑠𝑖𝑛 3𝑡
𝑡
𝑐𝑜𝑠 2𝑡 𝑠𝑖𝑛 3𝑡
𝑡
𝑐𝑜𝑠 2𝑡 𝑠𝑖𝑛 3𝑡
𝑡
𝑐𝑜𝑠 2𝑡 𝑠𝑖𝑛 3𝑡
𝑡
𝑐𝑜𝑠 2𝑡 𝑠𝑖𝑛 3𝑡
𝑡
𝑐𝑜𝑠 2𝑡 𝑠𝑖𝑛 3𝑡
𝑑𝑡 = 𝐿
𝑑𝑡 =
𝑑𝑡 =
𝑑𝑡 =
𝑑𝑡 =
𝑑𝑡 =
𝑑𝑡 =
𝑑𝑡 =
𝑑𝑡 =
𝑡
∞ −𝑠𝑡 𝑐𝑜𝑠 2𝑡 𝑠𝑖𝑛 3𝑡
𝑒
𝑑𝑡
0
𝑡
=
𝑐𝑜𝑠 2𝑡 𝑠𝑖𝑛 3𝑡
𝑡
∞
𝐿 𝑐𝑜𝑠2𝑡 𝑠𝑖𝑛3𝑡 𝑑𝑠
𝑠
1 ∞
𝐿 2𝑐𝑜𝑠2𝑡 𝑠𝑖𝑛3𝑡 𝑑𝑠
2 𝑠
1 ∞
𝐿 𝑠𝑖𝑛5𝑡 − sin⁡
(−𝑡) 𝑑𝑠
2 𝑠
1 ∞
𝐿 𝑠𝑖𝑛5𝑡 + sin⁡
(𝑡) 𝑑𝑠
2 𝑠
1 ∞
5
1
+
𝑑𝑠
2 𝑠 𝑠 2 +25
𝑠 2 +1
∞
1
−1 𝑠
−1
tan
+ tan
2
1 𝜋
2
1
2
1
2
𝑠
5
𝑠
𝜋
−1 𝑠
− tan
+ − tan−1
2
5
2
−1 𝑠
𝜋 − tan
+ tan−1 𝑠
5
𝑠
+𝑠
−1 5
𝜋 − tan
𝑠
1−5 .𝑠
𝑠
Put 𝑠 = 2,
∞ −2𝑡 𝑐𝑜𝑠 2𝑡 𝑠𝑖𝑛 3𝑡
𝑒
𝑑𝑡
0
𝑡
S.E/Paper Solutions
=
1
2
−1
𝜋 − tan
25
2
+2
5
4
1−5
=
1
2
𝜋 − tan−1 12
𝜋
2
−
Crescent Academy…….………………….…..For Research in Education
𝑐𝑜𝑠𝑎𝑡 −𝑐𝑜𝑠𝑏𝑡
13. Find L.T. of
𝑡
𝑒 −𝑎𝑡 −𝑐𝑜𝑠𝑎𝑡
14. Find L.T. of
1
𝑠 2 +𝑏 2
2
𝑠 2 +𝑎 2
𝑠 2 +𝑎 2
Ans. log
Ans. log
𝑡
sin 2 2𝑡
𝑠+𝑎
15. Find L.T. of
𝑡
[M16/AutoMechCivil/5M]
Solution:
𝐿
sin 2 2 𝑡
𝑡
=
=
=
=
=
∞
𝐿 sin2 2𝑡 𝑑𝑠
𝑠
∞
1−𝑐𝑜𝑠 4𝑡
𝐿
𝑑𝑠
𝑠
2
1 ∞1
𝑠
−
𝑑𝑠
2
2 𝑠 𝑠
𝑠 +16
1
1
2
𝑙𝑜𝑔𝑠 − log 𝑠 + 16
2
1
2
𝑠2
log
4
1
𝑠 2 +16
𝑠 2 +16
= log
𝑠
𝑠
𝑠2
4
16. Find L.T. of 𝑒 −3𝑡
∞
∞
1−𝑐𝑜𝑠 3𝑡
1
𝑠 2 +6𝑠+18
2
𝑠 2 +6𝑠+9
Ans. log
𝑡
𝑠𝑖𝑛𝑡 𝑠𝑖𝑛 5𝑡
17. Find L.T. of
𝑡
[M19/Extc/4M]
Solution:
𝐿
𝑠𝑖𝑛𝑡𝑠𝑖𝑛 5𝑡
𝑡
1
2𝑠𝑖𝑛𝑡𝑠𝑖𝑛 5𝑡
2
1
𝑡
cos −4𝑡 −𝑐𝑜𝑠 6𝑡
2
1
𝑡
𝑐𝑜𝑠 4𝑡−𝑐𝑜𝑠 6𝑡
= 𝐿
= 𝐿
= 𝐿
=
=
=
=
=
2
1
2
1
2
1
𝑡
∞
𝐿 𝑐𝑜𝑠4𝑡 − 𝑐𝑜𝑠6𝑡
𝑠
∞
𝑠
𝑠
− 2
𝑑𝑠
𝑠 𝑠 2 +16
𝑠 +36
1
1
2
2 2
1
4
1
4
1
log 𝑠 + 16 − log 𝑠 2 + 36
log
𝑠 2 +16
4
18. Use L.T. to show that
2
∞
∞
𝑠
𝑠 2 +36 𝑠
𝑠 2 +16
0 − log
= log
S.E/Paper Solutions
𝑑𝑠
𝑠 2 +36
𝑠 2 +36
𝑠 2 +16
∞ −𝑡
𝑒
0
𝑐𝑜𝑠𝑎𝑡 −𝑐𝑜𝑠𝑏𝑡
𝑡
26
1
𝑏 2 +1
2
𝑎 2 +1
𝑑𝑡 = 𝑙𝑜𝑔
Crescent Academy…….………………….…..For Research in Education
𝑐𝑜𝑠𝑏𝑡 −𝑐𝑜𝑠𝑎𝑡
19. Find L.T. of
𝑡
[N13/ElexExtcElectBiomInst/4M]
Solution:
𝐿
𝑐𝑜𝑠𝑏𝑡 −𝑐𝑜𝑠𝑎𝑡
=
𝑡
=
=
=
=
∞
𝐿 𝑐𝑜𝑠𝑏𝑡 − 𝑐𝑜𝑠𝑎𝑡 𝑑𝑠
𝑠
∞
𝑠
𝑠
−
𝑑𝑠
2
2
2
𝑠 𝑠 +𝑏
𝑠 +𝑎 2
1
1
log 𝑠 2 + 𝑏2 − log 𝑠 2
2
2
∞
2
2
1
𝑠 +𝑏
2
1
2
1
log
2
20. Evaluate using L.T.:
∞
0
∞
𝑠
𝑠 2 +𝑎 2 𝑠
𝑠 2 +𝑏 2
0 − log
= log
+ 𝑎2
𝑠 2 +𝑎 2
𝑠 2 +𝑎 2
𝑠 2 +𝑏 2
𝑐𝑜𝑠 3𝑡−𝑐𝑜𝑠 2𝑡
𝑒 −𝑡
𝑡
𝑑𝑡
[N15/ElexExctElectBiomInst][N17/IT/6M]
Solution:
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 =
𝑐𝑜𝑠 3𝑡−𝑐𝑜𝑠 2𝑡
𝑡
∞ −𝑠𝑡 𝑐𝑜𝑠 3𝑡−𝑐𝑜𝑠 2𝑡
𝑑𝑡
𝑒
0
𝑡
𝑐𝑜𝑠 3𝑡−𝑐𝑜𝑠 2𝑡
=𝐿
=
=
=
𝑡
∞
𝐿 𝑐𝑜𝑠3𝑡 − 𝑐𝑜𝑠2𝑡
𝑠
∞ 𝑠
𝑠
− 2 𝑑𝑠
𝑠 𝑠 2 +9
𝑠 +4
1
1
2
2
1
𝑑𝑠
log 𝑠 + 9 − log 𝑠 2 + 4
2
= log
2
𝑠 2 +4
𝑠 2 +9
Put 𝑠 = 1,
∞ −𝑡
𝑒
0
𝑐𝑜𝑠 3𝑡−𝑐𝑜𝑠 2𝑡
𝑡
1
5
2
10
𝑑𝑡 = 𝑙𝑜𝑔
1
1
2
2
= log
∞ 𝑐𝑜𝑠 6𝑡−𝑐𝑜𝑠 4𝑡
21. Evaluate using L.T.: 0
𝑑𝑡
𝑡
[M14/ElexExtcElectBiomInst/5M]
Solution:
By definition of L.T.,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 =
𝑐𝑜𝑠 6𝑡−𝑐𝑜𝑠 4𝑡
𝑡
∞ −𝑠𝑡 𝑐𝑜𝑠 6𝑡−𝑐𝑜𝑠 4𝑡
𝑒
𝑑𝑡
0
𝑡
=𝐿
=
S.E/Paper Solutions
𝑐𝑜𝑠 6𝑡−𝑐𝑜𝑠 4𝑡
∞
𝐿
𝑠
𝑡
𝑐𝑜𝑠6𝑡 − 𝑐𝑜𝑠4𝑡 𝑑𝑠
27
∞
𝑠
Crescent Academy…….………………….…..For Research in Education
∞ 𝑠
𝑠
−
𝑑𝑠
2
2
𝑠 𝑠 +36
𝑠 +16
1
1
2
=
=
2
1
log 𝑠 + 36 − log 𝑠 2 + 16
2
𝑠 2 +16
= log
∞
𝑠
𝑠 2 +36
2
Put 𝑠 = 0,
∞ 𝑐𝑜𝑠 6𝑡−𝑐𝑜𝑠 4𝑡
𝑑𝑡
0
𝑡
1
16
2
36
= 𝑙𝑜𝑔
= 𝑙𝑜𝑔
∞ 𝑐𝑜𝑠𝑎𝑡 −𝑐𝑜𝑠𝑏𝑡
22. Evaluate using L.T.: 0
[M14/AutoMechCivil/6M]
Solution:
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 =
𝑡
4
6
= log
2
3
𝑑𝑡
𝑐𝑜𝑠𝑎𝑡 −𝑐𝑜𝑠𝑏𝑡
∞ −𝑠𝑡 𝑐𝑜𝑠𝑎𝑡 −𝑐𝑜𝑠𝑏𝑡
𝑒
0
𝑡
𝑡
𝑑𝑡 = 𝐿
=
=
=
𝑐𝑜𝑠𝑎𝑡 −𝑐𝑜𝑠𝑏𝑡
𝑡
∞
𝐿 𝑐𝑜𝑠𝑎𝑡 − 𝑐𝑜𝑠𝑏𝑡
𝑠
∞ 𝑠
𝑠
− 2 2 𝑑𝑠
𝑠 𝑠 2 +𝑎 2
𝑠 +𝑏
1
1
2
2
2
1
log 𝑠 + 𝑎
= log
2
𝑑𝑠
2
− log 𝑠 + 𝑏
2
2
𝑠 2 +𝑏 2
∞
𝑠
𝑠 2 +𝑎 2
Put 𝑠 = 0,
∞ 𝑐𝑜𝑠𝑎𝑡 −𝑐𝑜𝑠𝑏𝑡
0
𝑡
1
𝑏2
2
𝑎2
𝑑𝑡 = 𝑙𝑜𝑔
= 𝑙𝑜𝑔
𝑏
𝑎
∞ 𝑐𝑜𝑠 4𝑡−𝑐𝑜𝑠 3𝑡
𝑑𝑡
0
𝑡
∞ −3𝑡 𝑐𝑜𝑠 7𝑡−𝑐𝑜𝑠 11𝑡
𝑒
𝑑𝑡
0
𝑡
Ans. log
23. Evaluate using L.T.:
24. Evaluate
3
4
[M19/IT/6M]
Solution:
By definition of L.T.,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 =
𝑐𝑜𝑠 7𝑡−𝑐𝑜𝑠 11𝑡
𝑡
∞ −𝑠𝑡 𝑐𝑜𝑠 7𝑡−𝑐𝑜𝑠 11𝑡
𝑒
𝑑𝑡
0
𝑡
=𝐿
=
=
=
𝑐𝑜𝑠 7𝑡−𝑐𝑜𝑠 11𝑡
𝑡
∞
𝐿 𝑐𝑜𝑠7𝑡 − 𝑐𝑜𝑠11𝑡
𝑠
∞ 𝑠
𝑠
−
𝑑𝑠
2
2
𝑠 𝑠 +49
𝑠 +121
1
1
2
2
1
log 𝑠 + 49 − log 𝑠 2 + 121
= log
2
S.E/Paper Solutions
𝑑𝑠
2
𝑠 2 +121
𝑠 2 +49
28
∞
𝑠
Crescent Academy…….………………….…..For Research in Education
Put 𝑠 = 3,
∞ −3𝑡 𝑐𝑜𝑠 7𝑡−𝑐𝑜𝑠 11𝑡
𝑒
0
𝑡
1
130
2
58
𝑑𝑡 = log
1
65
2
29
= log
1−𝑐𝑜𝑠𝑡
25. Find L.T. of
[N18/IT/4M]
Solution:
𝐿
1−𝑐𝑜𝑠𝑡
𝑡
=
=
=
=
𝑡
∞
𝐿 1 − 𝑐𝑜𝑠𝑡 𝑑𝑠
𝑠
∞1
𝑠
− 2 𝑑𝑠
𝑠 𝑠
𝑠 +1
1
𝑙𝑜𝑔𝑠 − log 𝑠 2 +
2
∞
1
𝑠2
2
1
log
𝑠 2 +1
𝑠 2 +1
= log
1
∞
𝑠
𝑠
𝑠2
2
sin 2 𝑡
∞
26. Evaluate using L.T.: 0 𝑒 −𝑡
𝑑𝑡
𝑡
[N13/AutoMechCivil/6M][N14/AutoMechCivil/6M]
[M15/ChemBiot/6M][N17/Comp/6M][M18/Elect/6M]
[N18/AutoMechCivil/6M]
Solution:
By definition of L.T.,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 =
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
sin 2 𝑡
sin 2 𝑡
𝑡
sin 2 𝑡
𝑡
sin 2 𝑡
𝑡
si n 2 𝑡
𝑡
𝑑𝑡 =
𝑑𝑡 =
𝑑𝑡 =
𝑡
∞ −𝑠𝑡 sin 2 𝑡
𝑒
𝑑𝑡
0
𝑡
∞ −𝑠𝑡 sin 2 𝑡
𝑒
𝑑𝑡
0
𝑡
=
sin 2 𝑡
∞
𝐿 sin2
𝑠
𝑡
∞
1−𝑐𝑜𝑠 2𝑡
𝑑𝑠
𝐿
𝑠
2
𝑠
1 ∞1
− 2 𝑑𝑠
𝑠
2
𝑠
𝑠 +4
1
1
2
𝑑𝑡 = 𝐿
2
1
4
1
=
𝑙𝑜𝑔𝑠 − log 𝑠 + 4
2
log
= log
4
𝑠2
∞
𝑠 2 +4
𝑠 2 +4
𝑠
𝑠2
Put 𝑠 = 1,
∞ −𝑡 sin 2 𝑡
𝑒
𝑑𝑡
0
𝑡
1
= 𝑙𝑜𝑔5
4
27. Evaluate using L.T.:
[N18/Comp/4M]
S.E/Paper Solutions
𝑡 𝑑𝑠
∞ 𝑒 −𝑡 −𝑒 −2𝑡
0
𝑡
𝑑𝑡
29
∞
𝑠
Crescent Academy…….………………….…..For Research in Education
Solution:
By definition of L.T.,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 =
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
∞ −𝑠𝑡
𝑒
0
𝑒 −𝑡 −𝑒 −2𝑡
𝑡
𝑒 −𝑡 −𝑒 −2𝑡
𝑡
𝑒 −𝑡 −𝑒 −2𝑡
𝑡
𝑒 −𝑡 −𝑒 −2𝑡
𝑡
𝑒 −𝑡 −𝑒 −2𝑡
𝑡
𝑒 −𝑡 −𝑒 −2𝑡
𝑡
𝑒 −𝑡 −𝑒 −2𝑡
𝑑𝑡 = 𝐿
𝑑𝑡 =
=
𝑡
∞ 1
𝑠 𝑠+1
−
1
𝑠+2
∞
𝑠
𝐿 𝑒 −𝑡 − 𝑒 −2𝑡 𝑑𝑠
𝑑𝑠
𝑑𝑡 = log 𝑠 + 1 − log 𝑠 + 2
𝑑𝑡 = log
𝑑𝑡 = log
𝑠+1
∞
𝑠+2
𝑠+2
𝑠
∞
𝑠
𝑠+1
Put 𝑠 = 0,
∞ 𝑒 −𝑡 −𝑒 −2𝑡
0
𝑡
𝑑𝑡 = 𝑙𝑜𝑔2
∞ 𝑒 −𝑡 −𝑐𝑜𝑠𝑡
28. Evaluate 0
𝑡 𝑒 4𝑡
[M19/Comp/6M]
Solution:
∞
𝑑𝑡
𝑒 −𝑡 −𝑐𝑜𝑠𝑡
We have, 0 𝑒 −4𝑡 .
𝑡
By definition of L.T.,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 =
𝑑𝑡
𝑒 −𝑡 −𝑐𝑜𝑠𝑡
𝑡
∞ −𝑠𝑡 𝑒 −𝑡 −𝑐𝑜𝑠𝑡
𝑒
0
𝑡
𝑑𝑡 = 𝐿
=
=
=
𝑒 −𝑡 −𝑐𝑜𝑠𝑡
𝑡
∞
−𝑡
𝐿
𝑒
− 𝑐𝑜𝑠𝑡 𝑑𝑠
𝑠
∞ 1
𝑠
−
. 𝑑𝑠
2
𝑠 𝑠+1
𝑠 +1
1
log 𝑠 + 1 − log 𝑠 2
2
= log 𝑠 + 1 − log
= log
= log
∞
𝑠+1
𝑠 2 +1 𝑠
𝑠 2 +1
𝑠+1
Put 𝑠 = 4,
∞ −4𝑡 𝑒 −𝑡 −𝑐𝑜𝑠𝑡
𝑒 .
0
𝑡
S.E/Paper Solutions
𝑑𝑡 = log
17
5
30
+1
𝑠2 + 1
∞
𝑠
∞
𝑠
Crescent Academy…….………………….…..For Research in Education
Type V: Laplace Transform of Derivatives
1.
Find L.T. of
𝑑
𝑠𝑖𝑛𝑡
𝑑𝑡
𝑡
[N13/ElexExtcElectBiomInst/4M]
Solution:
𝑠𝑖𝑛𝑡
Let 𝑓 𝑡 =
𝑡
𝑠𝑖𝑛𝑡
∴ 𝑓 0 = lim𝑡→0
And,
𝐿𝑓 𝑡
=𝐿
∴𝐹 𝑠 =
=
=1
𝑡
𝑠𝑖𝑛𝑡
𝑡
∞
𝐿
𝑠𝑖𝑛𝑡
𝑠
∞ 1
𝑑𝑠
𝑠 𝑠 2 +1
tan−1 𝑠 ∞
𝑠
𝜋
−1
=
= − tan
2
= cot −1 𝑠
𝑑𝑠
𝑠
Now,
𝐿
2.
𝑑
𝑠𝑖𝑛𝑡
𝑑𝑡
𝑡
Find L.T. of
𝑑
=𝐿
𝑓 𝑡
𝑑𝑡
= 𝐿 𝑓′ 𝑡
=𝑠𝐹 𝑠 −𝑓 0
= 𝑠 . cot −1 𝑠 – 𝑓 0
= 𝑠 cot −1 𝑠 − 1
𝑑
1−𝑐𝑜𝑠 2𝑡
𝑑𝑡
𝑡
[N14/ElexExtcElectBiomInst/4M]
Solution:
1−𝑐𝑜𝑠 2𝑡
Let 𝑓 𝑡 =
𝑡
∴ 𝑓 0 = lim𝑡→0
= lim𝑡→0
= 2𝑠𝑖𝑛0
∴𝑓 0 =0
And,
𝐿𝑓 𝑡
=𝐿
∴𝐹 𝑠 =
=
1−𝑐𝑜𝑠 2𝑡 0
𝑡
2𝑠𝑖𝑛 2𝑡
0
1
1−𝑐𝑜𝑠 2𝑡
∞
𝐿
𝑠
∞1
𝑠 𝑠
𝑡
1 − 𝑐𝑜𝑠2𝑡 𝑑𝑠
−
𝑠
𝑠 2 +4
1
𝑑𝑠
= 𝑙𝑜𝑔𝑠 − log 𝑠 2 + 4
2
S.E/Paper Solutions
∞
𝑠
31
Crescent Academy…….………………….…..For Research in Education
=
1
log
2
1
= log
𝑠2
∞
𝑠 2 +4
𝑠 2 +4
𝑠
𝑠2
2
Now,
𝐿
𝑑
1−𝑐𝑜𝑠 2𝑡
𝑑𝑡
𝑡
=𝐿
𝑑
𝑓 𝑡
=𝐿 𝑓 𝑡
=𝑠𝐹 𝑠 −𝑓 0
𝑑𝑡
′
𝑠
= log
𝑠 2 +4
2
= 𝑠 log
𝑑
𝑠2
𝑠 2 +4
–𝑓 0
𝑠
𝑠𝑖𝑛 3𝑡𝑠𝑖𝑛𝑡
Find L.T. of
4.
Find L.T. of
𝑑𝑡
𝑡
[M14/ElexExtcElectBiomInst/4M][N17/IT/5M]
Solution:
𝑑𝑡
𝑡
𝑑 𝑠𝑖𝑛 3𝑡
𝑠𝑖𝑛 3𝑡
Let 𝑓 𝑡 =
𝑡
∴ 𝑓 0 = lim𝑡→0
= lim𝑡→0
= 3𝑐𝑜𝑠0
∴𝑓 0 =3
And,
𝐿𝑓 𝑡
=𝐿
𝑠𝑖𝑛 3𝑡 0
𝑡
0
3𝑐𝑜𝑠 3𝑡
1
𝑠𝑖𝑛 3𝑡
𝑡
∞
𝐿 𝑠𝑖𝑛3𝑡
𝑠
∞ 3
𝑑𝑠
𝑠 𝑠 2 +9
∞
−1 𝑠
∴𝐹 𝑠 =
=
𝑑𝑠
= tan
=
3 𝑠
𝑠
− tan−1
2
3
−1 𝑠
𝜋
= cot
3
Now,
𝐿
𝑑
𝑠𝑖𝑛 3𝑡
𝑑𝑡
𝑡
=𝐿
𝑑
𝑑𝑡
′
𝑓 𝑡
=𝐿 𝑓 𝑡
=𝑠𝐹 𝑠 −𝑓 0
𝑠
= 𝑠 . cot −1 – 𝑓 0
=
S.E/Paper Solutions
𝑠
𝑠 2 +16
2
𝑠 2 +4
Ans. log
3.
3
−1 𝑠
𝑠𝑐𝑜𝑡
3
−3
32
Crescent Academy…….………………….…..For Research in Education
5.
𝑑
𝑠𝑖𝑛 4𝑡
Find L.T. of
𝑑𝑡
𝑡
[N17/Elect/5M]
Solution:
𝑠𝑖𝑛 4𝑡
Let 𝑓 𝑡 =
𝑡
∴ 𝑓 0 = lim𝑡→0
= lim𝑡→0
= 4𝑐𝑜𝑠0
∴𝑓 0 =4
And,
𝐿𝑓 𝑡
=𝐿
𝑠𝑖𝑛 4𝑡 0
𝑡
0
4𝑐𝑜𝑠 4𝑡
1
𝑠𝑖𝑛 4𝑡
𝑡
∞
𝐿 𝑠𝑖𝑛4𝑡
𝑠
∞ 4
𝑑𝑠
𝑠 𝑠 2 +16
∞
−1 𝑠
∴𝐹 𝑠 =
=
𝑑𝑠
= tan
=
4 𝑠
𝑠
− tan−1
2
4
−1 𝑠
𝜋
= cot
4
Now,
𝐿
𝑑
𝑠𝑖𝑛 4𝑡
𝑑𝑡
𝑡
𝑑
=𝐿
𝑑𝑡
′
𝑓 𝑡
=𝐿 𝑓 𝑡
=𝑠𝐹 𝑠 −𝑓 0
𝑠
= 𝑠 . cot −1 – 𝑓 0
=
6.
Find L.T. of
4
−1 𝑠
𝑠𝑐𝑜𝑡
4
𝑑
sin 2 𝑡
𝑑𝑡
𝑡
−4
[M19/IT/5M]
Solution:
Let 𝑓 𝑡 =
sin 2 𝑡
=
𝑡
∴ 𝑓 0 = lim𝑡→0
= lim𝑡→0
= 𝑠𝑖𝑛0
∴𝑓 0 =0
And,
𝐿𝑓 𝑡
=𝐿
S.E/Paper Solutions
1−𝑐𝑜𝑠 2𝑡
2𝑡
1−𝑐𝑜𝑠 2𝑡 0
2𝑡
2𝑠𝑖𝑛 2𝑡
0
2
1−𝑐𝑜𝑠 2𝑡
2𝑡
33
Crescent Academy…….………………….…..For Research in Education
∴𝐹 𝑠 =
=
=
=
∞
𝐿
2 𝑠
1 ∞1
2 𝑠 𝑠
1
1
2
1
4
1
1 − 𝑐𝑜𝑠2𝑡 𝑑𝑠
𝑠
−
𝑠 2 +4
𝑑𝑠
1
𝑙𝑜𝑔𝑠 − log 𝑠 2 + 4
2
log
= log
𝑠2
∞
𝑠 2 +4
𝑠 2 +4
𝑠
∞
𝑠
𝑠2
4
Now,
𝐿
𝑑
sin 2 𝑡
𝑑𝑡
𝑡
=𝐿
𝑑
𝑑𝑡
𝑓 𝑡
= 𝐿 𝑓′ 𝑡
=𝑠𝐹 𝑠 −𝑓 0
𝑠
𝑠 2 +4
4
𝑠2
= log
𝑠
𝑠 2 +4
4
𝑠2
– 0 = log
Type VI: Laplace Transform of Integrals
𝑡
𝑢𝑐𝑜𝑠𝑕𝑢𝑑𝑢
0
𝑡 −2𝑡
𝑒
cos 2 𝑡 𝑑𝑡
0
𝑡
𝑢𝑒 −2𝑢 𝑠𝑖𝑛3𝑢𝑑𝑢
0
𝑡 𝑡 𝑡
𝑡 𝑠𝑖𝑛𝑡 𝑑𝑡 3
0 0 0
1.
Find L.T. of
2.
Find L.T. of
3.
Find L.T. of
4.
Find L.T. of
[M18/IT/4M]
Solution:
𝐿
𝑡 𝑡 𝑡
0 0 0
𝑡𝑠𝑖𝑛𝑡 𝑑𝑡
3
=
=
1
𝑠3
1
𝑠3
=−
=−
=
5.
Ans.
Ans.
𝐿 𝑡𝑠𝑖𝑛𝑡
𝑑
. −1
𝑑𝑠
1
1 𝑑
𝐿 𝑠𝑖𝑛𝑡
𝑠 3 𝑑𝑠 𝑠 2 +1
1 𝑠 2 +1 [0]−1[2𝑠]
𝑠3
2
𝑠 2 +1
𝑠 2 𝑠 2 +1
2
2
1
𝑠
𝐿 𝑡. 𝑒 −3𝑡 . 𝑠𝑖𝑛𝑡
𝑠 +1
S.E/Paper Solutions
34
2
1
𝑠+2
+ .
2 𝑠
2𝑠 𝑠+2
3 2𝑠+4
1
𝑠
𝑡
𝑡
𝑠 𝑠 2 −1
1
Ans. .
Find L.T. of 0 𝑢. 𝑒 −3𝑢 . 𝑠𝑖𝑛 𝑢 𝑑𝑢
[N18/AutoMechCivil/5M]
Solution:
𝐿 0 𝑢. 𝑒 −3𝑢 . 𝑠𝑖𝑛𝑢 𝑑𝑢 =
We have,
1
𝐿 𝑠𝑖𝑛𝑡 = 2
𝑠 2 +1
𝑠 2 +4𝑠+13
2
𝑠 2 +4𝑠+8
Crescent Academy…….………………….…..For Research in Education
𝑑
𝐿 𝑡𝑠𝑖𝑛𝑡 = −1
= −1
1
𝑑𝑠 𝑠 2 +1
𝑠 2 +1 [0]−1[2𝑠]
= −1
∴ 𝐿{𝑡𝑠𝑖𝑛𝑡} =
𝐿 𝑠𝑖𝑛𝑡
𝑑𝑠
𝑑
𝑠 2 +1
2𝑠
𝑠 2 +1
2
2 𝑠+3
∴ 𝐿 𝑒 −3𝑡 𝑡 𝑠𝑖𝑛𝑡 =
6.
𝑑𝑢 =
2 𝑠+3
=
𝑠 2 +6𝑠+10 2
2 𝑠+3
𝑠
𝑠 2 +6𝑠+10 2
𝑡
Find L.T. of 0 𝑢. 𝑒 −3𝑢 . cos 2 2 𝑢 𝑑𝑢
[M16/ElexExctElectBiomInst/4M]
Solution:
𝐿 cos 2 2𝑡 = 𝐿
1+𝑐𝑜𝑠 4𝑡
2
2
𝐿 𝑡 cos 2𝑡 = −1
= −
= −
=
2
2
1
1
1
2
𝑠2
1
+
+
+
2
𝑠
1
+
𝑠 2 +16 1 −𝑠 2𝑠
𝑠 2 +16 2
2
−𝑠 +16
2
𝑠+3 2 −16
+
2
2
2
𝑠+3 2 +16
𝑠 2 +6𝑠−7
2
𝑠 2 +6𝑠+25
1 1
1
2
+
= .
𝑠 2
𝑠+3
+
2
𝑠 2 +6𝑠−7
𝑠 2 +6𝑠+25
𝑡
Find L.T. of 0 𝑢. 𝑒 −3𝑢 . 𝑠𝑖𝑛 4𝑢 𝑑𝑢
[N13/ElexExtcElectBiomInst/5M]
Solution:
𝑡
𝐿 0 𝑢. 𝑒 −3𝑢 . 𝑠𝑖𝑛4𝑢 𝑑𝑢 =
We have,
4
𝐿 𝑠𝑖𝑛4𝑡 = 2
𝐿 𝑡𝑠𝑖𝑛4𝑡 = −1
= −1
= −4
S.E/Paper Solutions
𝑑
𝑑𝑠
𝑑
𝑠
𝑠 2 +16
𝑠
𝑠 2 +16
1
𝑠+3
𝑡
𝑢. 𝑒 −3𝑢 . cos 2 2 𝑢 𝑑𝑢
0
𝑠 +16
2 𝑠
+
𝑠 2 +16
𝑠2
𝑠 2 +16
2
𝑠 −16
2 𝑠+3
1
1
2
1 1
𝐿 cos 2𝑡
−
2
𝐿 1 + 𝑐𝑜𝑠4𝑡 =
2
−
=
𝐿
𝑑
𝑑𝑠 2 𝑠
1
1
𝐿 𝑒 −3𝑡 𝑡 cos 2 2𝑡 =
1
=
𝑑𝑠
𝑑 1 1
= −1
7.
2 +1 2
𝑠+3
𝑡
𝑢. 𝑒 −3𝑢 . 𝑠𝑖𝑛𝑢
0
∴𝐿
2
1
𝑠
𝐿 𝑡. 𝑒 −3𝑡 . 𝑠𝑖𝑛4𝑡
𝐿 𝑠𝑖𝑛4𝑡
4
𝑑𝑠
𝑑
𝑠 2 +16
𝑑𝑠
𝑠 2 +16
1
35
2
Crescent Academy…….………………….…..For Research in Education
𝑠 2 +16 [0]−1[2𝑠]
= −4
∴ 𝐿{𝑡𝑠𝑖𝑛4𝑡} =
𝑠 2 +16
∴ 𝐿 𝑒 −3𝑡 𝑡 𝑠𝑖𝑛4𝑡 =
∴𝐿
8.
9.
10.
𝑠 2 +16
8𝑠
2
2
8 𝑠+3
𝑠+3
𝑡
−3𝑢
𝑢. 𝑒
. 𝑠𝑖𝑛4𝑢 𝑑𝑢
0
2 +16 2
=
=
8 𝑠+3
𝑠 2 +6𝑠+25 2
8 𝑠+3
𝑠 𝑠 2 +6𝑠+25
𝑡 −1 −𝑢
𝑢 𝑒 𝑠𝑖𝑛𝑢 𝑑𝑢
0
𝑡 1−𝑒 −𝑎𝑢
Find L.T. of 0
𝑑𝑢
𝑢
∞
𝑡 𝑠𝑖𝑛𝑢
Evaluate using L.T.: 0 𝑒 −𝑡 0
𝑢
2
1
Ans. cot −1 𝑠 + 1
Find L.T. of
𝑠
1
𝑠+𝑎
𝑠
𝑠
Ans. log
𝑑𝑢 𝑑𝑡
[N14/ChemBiot/6M][N15/ChemBiot/6M]
Solution:
By definition of L.T.,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
𝑡 𝑠𝑖𝑛𝑢
𝑑𝑢
0 𝑢
∞ −𝑠𝑡 𝑡 𝑠𝑖𝑛𝑢
𝑒
𝑑𝑢 𝑑𝑡
0 𝑢
0
Put 𝑓 𝑡 =
=𝐿
=
=
=
=
=
=
1
𝐿
𝑡 𝑠𝑖𝑛𝑢
0 𝑢
𝑠𝑖𝑛𝑡
𝑑𝑢
𝑡
∞
𝐿 𝑠𝑖𝑛𝑡 𝑑𝑠
𝑠 𝑠
1 ∞ 1
𝑑𝑠
𝑠 𝑠 𝑠 2 +1
1
tan−1 𝑠 ∞
𝑠
𝑠
1 𝜋
− tan−1 𝑠
𝑠 2
1
cot −1 𝑠
𝑠
𝑠
1
Put 𝑠 = 1,
∞ −𝑡 𝑡 𝑠𝑖𝑛𝑢
𝜋
1
𝑒
𝑑𝑢 𝑑𝑡 = cot −1 1 =
0
0
𝑢
11. Find L.T. of
12. Find L.T. of
13. Find L.T. of
1
4
𝑡 𝑠𝑖𝑛𝑢
𝑑𝑢
0 𝑢
𝑡 𝑢 𝑠𝑖𝑛 4𝑢
𝑒
𝑑𝑢
0
𝑢
𝑡
𝑒 −2𝑡 0 𝑡𝑐𝑜𝑠3𝑡𝑑𝑡
𝑡
𝑐𝑜𝑠𝑕𝑡 0 𝑒 𝑥 𝑠𝑖𝑛𝑕𝑥𝑑𝑥
1
Ans. cot −1 𝑠
𝑠
1
Ans. cot −1
𝑠
Ans.
14. Find L.T. of
[N15/ElexExtcElectBiomInst/4M]
Solution:
1
𝐿 𝑠𝑖𝑛𝑕𝑡 = 2
𝑠 −1
S.E/Paper Solutions
36
1
𝑠+2
.
𝑠−1
4
𝑠 2 +4𝑠−5
𝑠 2 +4𝑠+13
2
Crescent Academy…….………………….…..For Research in Education
𝐿 𝑒 𝑡 𝑠𝑖𝑛𝑕𝑡 =
𝐿
𝐿
1
2 −1
=
1
𝑠−1
𝑡 𝑥
1
1
𝑒 𝑠𝑖𝑛𝑕𝑥𝑑𝑥 = .
0
𝑠 𝑠 𝑠−2
𝑡 𝑥
𝑐𝑜𝑠𝑕𝑡 0 𝑒 𝑠𝑖𝑛𝑕𝑥𝑑𝑥 = 𝐿
1
=
2
1
=
=
=
1
𝑠 𝑠−2
1
𝑠 2 𝑠−2
𝑡
𝑒 +𝑒 −𝑡
𝐿
= 𝐿
∞
=
𝑠 2 −2𝑠
2
1
𝑡 𝑥
𝑒 𝑠𝑖𝑛𝑕𝑥𝑑𝑥
0
2
𝑡
𝑒 𝑡 + 𝑒 −𝑡 0 𝑒 𝑥 𝑠𝑖𝑛𝑕𝑥𝑑𝑥
𝑡
𝑡
𝑒 𝑡 0 𝑒 𝑥 𝑠𝑖𝑛𝑕𝑥𝑑𝑥 + 𝑒 −𝑡 0 𝑒 𝑥 𝑠𝑖𝑛𝑕𝑥𝑑𝑥
1
1
2
1
𝑠−1
2
𝑠−1
2
𝑠−1−2
1
2
+
+
𝑠−3
2
𝑠+1
1
𝑠+1
2
𝑠+1−2
𝑠−1
𝑡
15. Evaluate using L.T.: 0 𝑒 −𝑡 0 𝑢4 𝑠𝑖𝑛𝑕𝑢𝑐𝑜𝑠𝑕𝑢𝑑𝑢 𝑑𝑡
[N18/IT/4M][N18/Elect/6M]
Solution:
By definition of L.T.,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
𝑡 4
𝑢 𝑠𝑖𝑛𝑕𝑢𝑐𝑜𝑠𝑕𝑢𝑑𝑢
0
∞ −𝑠𝑡 𝑡 4
𝑒
𝑢 𝑠𝑖𝑛𝑕𝑢𝑐𝑜𝑠𝑕𝑢𝑑𝑢 𝑑𝑡
0
0
Put 𝑓 𝑡 =
=𝐿
=
1
𝑠
1
𝑡 4
𝑢 𝑠𝑖𝑛𝑕𝑢𝑐𝑜𝑠𝑕𝑢𝑑𝑢
0
4
𝐿 𝑡 𝑠𝑖𝑛𝑕𝑡 𝑐𝑜𝑠𝑕𝑡
= 𝐿 𝑡4
=
=
=
𝑠
1
4𝑠
1
4𝑠
1
4𝑠
𝑒 𝑡 −𝑒 −𝑡
4
𝐿 𝑡 𝑒
2
2𝑡
𝑒 𝑡 +𝑒 −𝑡
2
−𝑒
−2𝑡
𝐿 𝑒 2𝑡 𝑡 4 − 𝑒 −2𝑡 𝑡 4
4!
𝑠−2
5
−
4!
𝑠+2
5
Put 𝑠 = 1,
∞ −𝑡
𝑒
0
𝑡
0
𝑢4 𝑠𝑖𝑛𝑕𝑢𝑐𝑜𝑠𝑕𝑢𝑑𝑢 𝑑𝑡 =
𝑡 𝑠𝑖𝑛 𝑢
16. Find L.T. of 𝑒 −𝑡 0
𝑑𝑢
𝑢
[M15/ElexExtcElectBiomInst/4M]
Solution:
We have,
1
𝐿 𝑠𝑖𝑛𝑡 = 2
𝐿
𝑠𝑖𝑛𝑡
𝑡
=
=
=
𝑠 +1
∞
𝐿 𝑠𝑖𝑛𝑡 𝑑𝑠
𝑠
∞ 1
𝑑𝑠
𝑠 𝑠 2 +1
tan−1 𝑠 ∞
𝑠
S.E/Paper Solutions
37
1 4!
4 −1
−
4!
35
=−
488
81
Crescent Academy…….………………….…..For Research in Education
𝜋
=
𝐿
𝐿
2
𝑡 𝑠𝑖𝑛𝑢
𝑑𝑢
0 𝑢
𝑡 𝑠𝑖𝑛𝑢
𝑒 −𝑡 0
𝑢
− tan−1 𝑠 = cot −1 𝑠
=
1
𝑠𝑖𝑛𝑡
𝐿
𝑠
𝑑𝑢 =
1
1
= cot −1 𝑠
𝑡
𝑠+1
cot
𝑠
−1
(𝑠 + 1)
𝑡
17. Find L.T. of 𝑒 −3𝑡 0 𝑢𝑠𝑖𝑛3𝑢𝑑𝑢
[N14/ElexExtcElectBiomInst/4M]
Solution:
We have,
3
𝐿 𝑠𝑖𝑛3𝑡 = 2
𝑠 +9
𝐿 𝑡𝑠𝑖𝑛3𝑡 = −1
= −1
= −3
= −3
∴ 𝐿{𝑡𝑠𝑖𝑛3𝑡} =
∴𝐿
∴𝐿
𝑑
1
𝑑𝑠 𝑠 2 +9
𝑠 2 +9 [0]−1[2𝑠]
𝑠 2 +9
6𝑠
𝑠 2 +9
𝐿 𝑡𝑠𝑖𝑛4𝑡 = −1
𝑡
𝑡
0
= −1
= −4
= −4
∴ 𝐿{𝑡𝑠𝑖𝑛4𝑡} =
𝑑
𝑑𝑠
𝑑
2
2
𝑡
1
𝑢𝑠𝑖𝑛3𝑢𝑑𝑢 =
0
𝑠
−3𝑡 𝑡
𝑒
𝑢𝑠𝑖𝑛3𝑢𝑑𝑢
0
𝑠 +16
∴𝐿
3
𝑠 2 +9
𝑑𝑠
𝑑
18. Find the L.T. of 𝑒 −3𝑡
[N18/Comp/4M]
Solution:
We have,
4
𝐿 𝑠𝑖𝑛4𝑡 = 2
∴𝐿
𝐿 𝑠𝑖𝑛3𝑡
𝑑𝑠
𝑑
𝐿 𝑡𝑠𝑖𝑛3𝑡 =
=
6
𝑠+3
2 +9 2
1
𝑠
.
=
6𝑠
𝑠 2 +9 2
=
6
𝑠 2 +9 2
6
𝑠 2 +6𝑠+18 2
𝑠𝑖𝑛4𝑡 𝑑𝑡
𝐿 𝑠𝑖𝑛4𝑡
4
𝑑𝑠 𝑠 2 +16
1
𝑑
𝑑𝑠 𝑠 2 +16
𝑠 2 +16 [0]−1[2𝑠]
8𝑠
𝑠 2 +16
2
𝑠 2 +16 2
𝑡
1
1
8𝑠
8
𝑡𝑠𝑖𝑛4𝑡𝑑𝑡
=
𝐿
𝑡𝑠𝑖𝑛4𝑡
=
.
=
2
2
2
0
𝑠
𝑠
𝑠 +16
𝑠 +16
𝑡
8
8
𝑒 −3𝑡 0 𝑡 𝑠𝑖𝑛4𝑡 𝑑𝑡 =
= 2
𝑠+3 2 +16 2
𝑠 +6𝑠+25 2
S.E/Paper Solutions
38
2
Crescent Academy…….………………….…..For Research in Education
𝑡
𝑢
0
19. Find the L.T. of 𝑒 −𝑡
[M17/CompIT/5M]
Solution:
We have,
𝑠
𝐿 𝑐𝑜𝑠2𝑡 = 2
𝑠 +4
𝐿 𝑡𝑐𝑜𝑠2𝑡 = −1
= −1
∴𝐿
𝑡
𝑢
0
∴ 𝐿 𝑒 −𝑡
𝐿 𝑐𝑜𝑠2𝑡
𝑑𝑠
𝑑
𝑠
𝑑𝑠 𝑠 2 +4
𝑠 2 +4 1 −𝑠 2𝑠
= −1
∴ 𝐿{𝑡𝑐𝑜𝑠2𝑡} =
𝑑
𝑠 2 +4
𝑠2 − 4
𝑠 2 +4
2
2
𝑐𝑜𝑠2𝑢𝑑𝑢 =
𝑡
𝑢
0
𝑐𝑜𝑠2𝑢 𝑑𝑢
1
1
𝐿 𝑡𝑐𝑜𝑠2𝑡 =
𝑠
𝑐𝑜𝑠2𝑢𝑑𝑢 =
𝑠+1
2 −4
𝑠
(𝑠+1) 𝑠+1 2 +4
𝑠2 − 4
.
𝑠 2 +4
2
=
2
𝑠 2 +2𝑠−3
(𝑠+1) 𝑠 2 +2𝑠+5
𝑡
2
20. Find L.T. of 𝑒 −4𝑡 0 𝑢 𝑠𝑖𝑛3𝑢𝑑𝑢
[M15/AutoMechCivil/6M][N16/AutoMechCivil/4M][N17/Elex/5M]
[M18/Elex/6M]
Solution:
We have,
3
𝐿 𝑠𝑖𝑛3𝑡 = 2
𝑠 +9
𝑑
𝐿 𝑡𝑠𝑖𝑛3𝑡 = −1
= −1
𝑑𝑠 𝑠 2 +9
𝑠 2 +9 [0]−1[2𝑠]
= −3
∴𝐿
∴𝐿
3
𝑑𝑠 𝑠 2 +9
𝑑
1
= −3
∴ 𝐿{𝑡𝑠𝑖𝑛3𝑡} =
𝐿 𝑠𝑖𝑛3𝑡
𝑑𝑠
𝑑
𝑠 2 +9
6𝑠
𝑠 2 +9
2
𝑡
1
𝑢𝑠𝑖𝑛3𝑢𝑑𝑢
=
0
𝑠
𝑡
𝑒 −4𝑡 0 𝑢𝑠𝑖𝑛3𝑢𝑑𝑢
21. Find L.T. of 𝑒 −2𝑡
[N18/Extc/6M]
Solution:
We have,
𝑠
𝐿 𝑐𝑜𝑠4𝑡 = 2
𝑡
𝑢
0
2
𝐿 𝑡𝑠𝑖𝑛3𝑡 =
=
6
𝑠+4
2 +9 2
𝑒 3𝑢 𝑐𝑜𝑠4𝑢 𝑑𝑢
𝑠 +16
S.E/Paper Solutions
39
1
𝑠
.
=
6𝑠
𝑠 2 +9 2
=
6
𝑠 2 +9 2
6
𝑠 2 +8𝑠+25 2
Crescent Academy…….………………….…..For Research in Education
𝑑
𝐿 𝑡𝑐𝑜𝑠4𝑡 = −1
= −1
𝑡
𝑢
0
𝑠 2 +16
𝑠 2 − 16
𝑠−3 2 +16
1
𝑒 3𝑢 𝑐𝑜𝑠4𝑢 𝑑𝑢 =
𝑡
𝑢
0
∴ 𝐿 𝑒 −2𝑡
2
𝑠 2 +16 2
𝑠−3 2 −16
∴ 𝐿 𝑡𝑒 3𝑡 𝑐𝑜𝑠4𝑡 =
∴𝐿
𝑠
𝑑𝑠 𝑠 2 +16
𝑠 2 +16 1 −𝑠 2𝑠
= −1
∴ 𝐿{𝑡𝑐𝑜𝑠4𝑡} =
𝐿 𝑐𝑜𝑠4𝑡
𝑑𝑠
𝑑
2
𝑠 2 −6𝑠−7
=
𝑠 2 −6𝑠+25
3𝑡
2
𝐿 𝑡𝑒 𝑐𝑜𝑠4𝑡 =
𝑠
1
𝑒 3𝑢 𝑐𝑜𝑠4𝑢 𝑑𝑢 =
=
𝑠+2
1
𝑠+2
1
=
𝑠+2
𝑠
2 −6
𝑠+2
.
1
𝑠 2 −6𝑠−7
.
𝑠 2 −6𝑠+25
𝑠+2 −7
2
𝑠+2 2 −6 𝑠+2 +25 2
𝑠 2 +4𝑠+4−6𝑠−12−7
.
𝑠 2 +4𝑠+4−6𝑠−12+25
𝑠 2 −2𝑠−15
.
𝑠 2 −2𝑠+17
2
2
𝑡
∞
22. Evaluate 0 𝑒 −4𝑡 0 𝑢𝑠𝑖𝑛𝑕𝑢 2 𝑐𝑜𝑠𝑕5𝑢 𝑒 3𝑢 𝑑𝑢 𝑑𝑡
[N17/Elect/8M]
Solution:
By definition of L.T.,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
𝑡 𝑠𝑖𝑛𝑢
0 𝑢
Put 𝑓 𝑡 =
∞ −𝑠𝑡 𝑡
𝑒
0
0
=
1
𝑠
1
= .
𝑡
0
𝑢𝑠𝑖𝑛𝑕2 𝑢 2 𝑐𝑜𝑠𝑕5𝑢. 𝑒 3𝑢 𝑑𝑢 𝑑𝑡 = 𝐿
𝑢𝑠𝑖𝑛𝑕2 𝑢 2 𝑐𝑜𝑠𝑕5𝑢. 𝑒 3𝑢 𝑑𝑢
𝐿 𝑡 2 sinh4 𝑡 𝑐𝑜𝑠𝑕5𝑡 𝑒 3𝑡
= 𝐿 𝑡2
𝑠
1
𝑑𝑢
1
𝑠 32
𝑒 𝑡 −𝑒 −𝑡
4
𝑒 5𝑡 +𝑒 −5𝑡
2
𝑒 3𝑡
2
𝐿 𝑡 2 𝑒 4𝑡 − 4𝑒 2𝑡 + 6 − 4𝑒 −2𝑡 + 𝑒 −4𝑡 𝑒 8𝑡 + 𝑒 −2𝑡
1
= 32𝑠 𝐿 𝑡 2 𝑒12𝑡 + 𝑒 2𝑡 − 4𝑒10𝑡 − 4 + 6𝑒 8𝑡 + 6𝑒 −2𝑡 − 4𝑒 6𝑡 − 4𝑒 −4𝑡 + 𝑒 4𝑡 + 𝑒 −6𝑡
=
1
32𝑠
2
𝑠−12 3
+
2
𝑠−2 3
−
4.2
𝑠−10 3
−
4.2
𝑠3
+
6.2
𝑠−8 3
+
6.2
𝑠+2 3
−
4.2
𝑠−6 3
−
4.2
𝑠+4 3
Put 𝑠 = 4,
∞ −𝑠𝑡 𝑡
𝑒
0
0
𝑢𝑠𝑖𝑛𝑕2 𝑢 2 𝑐𝑜𝑠𝑕5𝑢. 𝑒 3𝑢 𝑑𝑢 𝑑𝑡 = ∞
23. Find Laplace Transform of 𝑓 𝑡 = 𝑡
[M19/AutoMechCivil/5M]
Solution:
We have,
4
𝐿 𝑠𝑖𝑛4𝑡 = 2
𝑠 +16
S.E/Paper Solutions
40
𝑡 −2𝑢
𝑒
𝑠𝑖𝑛4𝑢
0
𝑑𝑢
+
2
𝑠−4 3
+
2
𝑠+6 3
Crescent Academy…….………………….…..For Research in Education
4
𝐿 𝑒 −2𝑡 𝑠𝑖𝑛4𝑡 =
𝐿
𝐿
2 +16
=
4
𝑠 2 +4𝑠+20
𝑠+2
𝑡 −2𝑢
1
4
𝑒
𝑠𝑖𝑛4𝑢 𝑑𝑢 = . 2
0
𝑠 𝑠 +4𝑠+20
𝑡 −2𝑢
𝑑
4
𝑡 0𝑒
𝑠𝑖𝑛4𝑢 𝑑𝑢 = −1
3
𝑑𝑠 𝑠 +4𝑠 2 +20𝑠
𝑠 3 +4𝑠 2 +20𝑠 0 − 4 3𝑠 2 +8𝑠+20
= −1
=
4
𝑠 3 +4𝑠 2 +20
3𝑠 2 +8𝑠+20
𝑠 3 +4𝑠 2 +20𝑠
2
2
Type VII: Heaviside Unit step Function& Delta Function
1.
2.
Find the Laplace transform of 𝑡 2 𝐻 𝑡 − 3
Find the Laplace transform of𝑠𝑖𝑛𝑡. 𝐻 𝑡 −
𝜋
2
−𝐻 𝑡−
𝜋
𝑠 +1
1
𝑠 2 +1
+ 𝑒 −𝜋𝑠
−𝑠
𝑠 2 +1
+
6
𝑠2
+
2
𝑠
1
𝑠 2 +1
Find the L.T. of 𝑒 −𝑡 𝑐𝑜𝑠𝑡 𝐻 𝑡 − 𝜋
[N14/ChemBiot/6M][M17/ElexExtcElectBiomInst/4M]
Solution:
𝐿 𝑒 −𝑡 𝑐𝑜𝑠𝑡 𝐻 𝑡 − 𝜋 = 𝑒 −𝜋𝑠 𝐿 𝑒 − 𝑡+𝜋 cos 𝑡 + 𝜋
= 𝑒 −𝜋𝑠 𝐿 𝑒 −𝑡 . 𝑒 −𝜋 𝑐𝑜𝑠𝑡 𝑐𝑜𝑠𝜋 − 𝑠𝑖𝑛𝑡 𝑠𝑖𝑛𝜋
= 𝑒 −𝜋𝑠 . 𝑒 −𝜋 𝐿 𝑒 −𝑡 – 𝑐𝑜𝑠𝑡
= −𝑒 −𝜋𝑠 . 𝑒 −𝜋 𝐿 𝑒 −𝑡 𝑐𝑜𝑠𝑡
𝑠+1
= −𝑒 −𝜋𝑠 . 𝑒 −𝜋
2
= −𝑒
5.
+
9
𝑠
𝜋
1
Ans. 𝑒 . 2 − 𝑒 −3 2 𝑠 .
𝑠 +1
𝑠
Find the L.T. of 𝑠𝑖𝑛𝑡 𝐻 𝑡 + 𝑐𝑜𝑠𝑡 − 𝑠𝑖𝑛𝑡 𝐻 𝑡 − 𝜋
[M19/IT/4M]
Solution:
𝐿 𝑠𝑖𝑛𝑡 𝐻 𝑡 + 𝑐𝑜𝑠𝑡 − 𝑠𝑖𝑛𝑡 𝐻 𝑡 − 𝜋
= 𝑒 −0𝑠 𝐿 sin 𝑡 + 0 + 𝑒 −𝜋𝑠 𝐿 cos 𝑡 + 𝜋 − sin 𝑡 + 𝜋
= 𝐿 𝑠𝑖𝑛𝑡 + 𝑒 −𝜋𝑠 𝐿{𝑐𝑜𝑠𝑡 𝑐𝑜𝑠𝜋 − 𝑠𝑖𝑛𝑡 𝑠𝑖𝑛𝜋 − 𝑠𝑖𝑛𝑡 𝑐𝑜𝑠𝜋 − 𝑐𝑜𝑠𝑡 𝑠𝑖𝑛𝜋}
1
= 2 + 𝑒 −𝜋𝑠 𝐿 −𝑐𝑜𝑠𝑡 + 𝑠𝑖𝑛𝑡
=
4.
𝑠3
3𝜋
−2 𝑠
3.
2
Ans. 𝑒 −3𝑠
−𝜋𝑠
.𝑒
−𝜋
.
𝑠+1 +1
𝑠+1
𝑠 2 +2𝑠+2
Find the L.T. of 𝑒 −𝑡 𝑠𝑖𝑛𝑡 𝐻 𝑡 − 𝜋
[M19/Comp/6M]
Solution:
𝐿 𝑒 −𝑡 𝑠𝑖𝑛𝑡 𝐻 𝑡 − 𝜋 = 𝑒 −𝜋𝑠 𝐿 𝑒 − 𝑡+𝜋 sin 𝑡 + 𝜋
= 𝑒 −𝜋𝑠 𝐿 𝑒 −𝑡 . 𝑒 −𝜋 𝑠𝑖𝑛𝑡 𝑐𝑜𝑠𝜋 + 𝑐𝑜𝑠𝑡 𝑠𝑖𝑛𝜋
S.E/Paper Solutions
41
Crescent Academy…….………………….…..For Research in Education
= 𝑒 −𝜋𝑠 . 𝑒 −𝜋 𝐿 𝑒 −𝑡 – 𝑠𝑖𝑛𝑡
= −𝑒 −𝜋𝑠 . 𝑒 −𝜋 𝐿 𝑒 −𝑡 𝑠𝑖𝑛𝑡
1
= −𝑒 −𝜋𝑠 . 𝑒 −𝜋
2
= −𝑒 −𝜋𝑠 . 𝑒 −𝜋 .
6.
𝑠+1 +1
1
𝑠 2 +2𝑠+2
Find L.T. of 𝑡𝑒 −2𝑡 𝐻 𝑡 − 1
[N17/IT/4M]
Solution:
𝐿 𝑡𝑒 −2𝑡 𝐻 𝑡 − 1
= 𝑒 −𝑠 𝐿 𝑒 −2 𝑡+1 𝑡 + 1
= 𝑒 −𝑠 𝐿 𝑒 −2𝑡 . 𝑒 −2 𝑡 + 1
= 𝑒 −𝑠 . 𝑒 −2 𝐿 𝑒 −2𝑡 𝑡 + 𝑒 −2𝑡
1
1
= 𝑒 −𝑠−2
+
2
𝑠+2
𝑠+2
7.
Find the Laplace transform of: 𝑡. 𝐻 𝑡 − 4 + 𝑡 2 𝛿 𝑡 − 4
8.
Ans. 2 1 + 4𝑠 + 16𝑠 2
𝑠
Find the Laplace transform of: 𝑡 2 𝐻 𝑡 − 2 − 𝑐𝑜𝑠𝑕𝑡 𝛿 𝑡 − 4
𝑒 −4𝑠
2
Ans. 𝑒 −2𝑠
9.
𝑠3
+
4
𝑠2
4
+
𝑠
− 𝑒 −4𝑠 𝑐𝑜𝑠𝑕4
Find the Laplace transform of: 1 + 2𝑡 − 3𝑡 2 + 4𝑡 3 𝐻 𝑡 − 2
[M15/ChemBiot/6M]
Solution:
𝐿 1 + 2t − 3t 2 + 4t 3 H t − 2
= 𝑒 −2𝑠 𝐿 1 + 2 𝑡 + 2 − 3 𝑡 + 2 2 + 4 𝑡 + 2 3
= 𝑒 −2𝑠 𝐿 1 + 2𝑡 + 4 − 3𝑡 2 − 12𝑡 − 12 + 4𝑡 3 + 24𝑡 2 + 48𝑡 + 32
= 𝑒 −2𝑠 𝐿 4𝑡 3 + 21𝑡 2 + 38𝑡 + 25
4.3!
21.2!
38
25
= 𝑒 −2𝑠 4 + 3 + 2 +
=
𝑠
−2𝑠 24
𝑒
𝑠4
+
𝑠
42
38
𝑠3
𝑠2
+
𝑠
+
25
𝑠
𝑠
10. Find the Laplace transform of: 2 − 2𝑡 + 3𝑡 2 − 𝑡 3 𝐻 𝑡 − 3
Ans. 𝑒 −3𝑠 −
2
3
6
𝑠4
−
12
𝑠3
−
11
𝑠2
−
4
𝑠
11. Find the Laplace transform of: 1 + 3𝑡 − 𝑡 + 𝑡 𝐻(𝑡 − 4)
6
22
43
Ans. 𝑒 −4𝑠 4 + 3 + 2 +
𝑠
𝑠
𝑠
12. Find the Laplace transform of: 1 + 2𝑡 − 𝑡 2 + 𝑡 3 𝐻(𝑡 − 4)
[N18/Elex/4M]
Solution:
𝐿 1 + 2t − t 2 + t 3 H t − 4
S.E/Paper Solutions
42
61
𝑠
Crescent Academy…….………………….…..For Research in Education
= 𝑒 −4𝑠 𝐿 1 + 2 𝑡 + 4 − 𝑡 + 4 2 + 𝑡 + 4 3
= 𝑒 −4𝑠 𝐿 1 + 2𝑡 + 8 − 𝑡 2 − 8𝑡 − 16 + 𝑡 3 + 12𝑡 2 + 48𝑡 + 64
= 𝑒 −4𝑠 𝐿 𝑡 3 + 11𝑡 2 + 42𝑡 + 57
3!
11.2!
42
57
= 𝑒 −4𝑠 4 + 3 + 2 +
=𝑒
−4𝑠
𝑠
6
𝑠4
+
𝑠
22
𝑠3
+
42
𝑠2
𝑠
+
57
𝑠
𝑠
13. Find the Laplace transform of: 𝑡 3 + 2𝑡 2 − 3𝑡 + 1 𝐻 𝑡 − 1
Ans. 𝑒 −𝑠
6
𝑠4
+
14. Find the Laplace transform of: 𝑡 2 𝐻 𝑡 − 2 + 𝑡 3 𝛿 𝑡 − 3
Ans. 𝑒 −2𝑠
15. Evaluate the following:
∞
0
𝑠3
+
4
𝑠2
+
𝑠3
4
𝑠
+
4
𝑠2
+
1
𝑠
+ 𝑒 −3𝑠 . 27
𝑒 −2𝑡 1 + 2𝑡 − 3𝑡 2 + 4𝑡 3 𝐻 𝑡 − 1 𝑑𝑡
Ans.
∞
0
2
10
−𝑡
2
3
31
4𝑒 2
16. Evaluate the following:
𝑒 1 + 2𝑡 − 𝑡 + 𝑡 𝐻 𝑡 − 1 𝑑𝑡
[N17/Comp/6M]
Solution:
By definition of L.T.,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = 1 + 2𝑡 − 𝑡 2 + 𝑡 3 𝐻 𝑡 − 1
∞ −𝑠𝑡
𝑒
1 + 2𝑡 − 𝑡 2 + 𝑡 3 𝐻 𝑡 − 1 𝑑𝑡 = 𝐿 1 + 2𝑡 − 𝑡 2 + 𝑡 3 𝐻 𝑡 − 1
0
= 𝑒 −𝑠 𝐿 (1 + 2 𝑡 + 1 − 𝑡 + 1 2 + 𝑡 + 1 3
= 𝑒 −𝑠 𝐿 1 + 2𝑡 + 2 − 𝑡 2 − 2𝑡 − 1 + 𝑡 3 + 3𝑡 2 + 3𝑡 + 1
= 𝑒 −𝑠 𝐿 𝑡 3 + 2𝑡 2 + 3𝑡 + 3
2.2!
3
3
3!
= 𝑒 −𝑠 4 + 3 + 2 +
𝑠
𝑠
𝑠
𝑠
Put 𝑠 = 1,
16
1 + 2𝑡 − 𝑡 2 + 𝑡 3 𝐻 𝑡 − 1 𝑑𝑡 = 𝑒 −1 6 + 4 + 3 + 3 =
∞ −𝑡
𝑒
0
17. Evaluate the following:
𝑒
∞
0
𝑒 −2𝑡 1 − 𝑡 + 𝑡 2 𝐻 𝑡 − 3 𝑑𝑡
Ans.
∞
0
5
𝑒6
18. Evaluate the following:
𝑒 −𝑡 1 + 3𝑡 + 𝑡 2 𝐻 𝑡 − 2 𝑑𝑡
[M18/Comp/6M]
Solution:
By definition of L.T.,
∞ −𝑠𝑡
𝑒 𝑓 𝑡 𝑑𝑡 = 𝐿 𝑓 𝑡
0
Put 𝑓 𝑡 = 1 + 3𝑡 + 𝑡 2 𝐻 𝑡 − 2
∞ −𝑠𝑡
𝑒
1 + 3𝑡 + 𝑡 2 𝐻 𝑡 − 2 𝑑𝑡 = 𝐿 1 + 3𝑡 + 𝑡 2 𝐻 𝑡 − 2
0
S.E/Paper Solutions
43
Crescent Academy…….………………….…..For Research in Education
= 𝑒 −2𝑠 𝐿 (1 + 3 𝑡 + 2 + 𝑡 + 2 2
= 𝑒 −2𝑠 𝐿 1 + 3𝑡 + 6 + 𝑡 2 + 4𝑡 + 4
= 𝑒 −2𝑠 𝐿 𝑡 2 + 7𝑡 + 11
2!
7
11
= 𝑒 −2𝑠 3 + 2 +
𝑠
𝑠
𝑠
Put 𝑠 = 1,
∞ −𝑡
𝑒 1 + 3𝑡 + 𝑡 2 𝐻 𝑡 − 2 𝑑𝑡 = 𝑒 −2 2 + 7 + 11 = 20𝑒 −2
0
S.E/Paper Solutions
44
Crescent Academy…….………………….…..For Research in Education
Type VIII: L.T. of Periodic Functions
1.
1 𝑓𝑜𝑟 0 ≤ 𝑡 < 𝑎
and 𝑓(𝑡) is periodic with period
−1 𝑓𝑜𝑟 𝑎 < 𝑡 < 2𝑎
Find L.T. of 𝑓 𝑡 =
2a
[N18/IT/6M]
Solution:
We have, 𝑓(𝑡) is periodic with period 2a,
1
2𝑎 −𝑠𝑡
𝐿𝑓 𝑡 =
𝑒 𝑓 𝑡 𝑑𝑡
−2𝑎𝑠 0
𝐿𝑓 𝑡
𝑎 −𝑠𝑡
2𝑎
𝑒 . 1𝑑𝑡 + 𝑎
1−𝑒 −2𝑎𝑠 0
𝑎
2𝑎
𝑒 −𝑠𝑡
𝑒 −𝑠𝑡
1
=
=
=
=
=
−
−𝑠 0
1−𝑒 −2𝑎𝑠
1
𝑒 −𝑎𝑠
𝑒0
−
=
−𝑠 𝑎
𝑒 −2𝑎𝑠
−
1−𝑒 −2𝑎𝑠
1
𝑠
1−𝑒 −𝑎𝑠
1−𝑒 −2𝑎𝑠
.
1−𝑒 −𝑎𝑠
𝑠
1−𝑒 −𝑎𝑠 1+𝑒 −𝑎𝑠
1−𝑒 −𝑎𝑠
1 1−𝑒 −𝑎𝑠
𝑠
= .
1+𝑒 −𝑎𝑠
𝑠
−𝑠
2
𝑠
1
1+𝑒 −𝑎𝑠
2
1
= . tanh
𝑠
𝐸 𝑓𝑜𝑟 0 ≤ 𝑡 <
Find L.T. of 𝑓 𝑡 =
𝑒 −𝑠𝑡 . −1𝑑𝑡
𝑒 −𝑎𝑠
+
1−𝑒 −2𝑎𝑠 −𝑠
−𝑠
−𝑠
1
1−2𝑒 −𝑎𝑠 +𝑒 −2𝑎𝑠
=
2.
1−𝑒
1
𝑝
−𝐸
𝑎𝑠
2
𝐿𝑓 𝑡
=
=
=
=
=
𝑝
2
0
1−𝑒 −𝑝𝑠
2
𝑓𝑜𝑟 < 𝑡 < 𝑝
1
𝐸
1−𝑒 −𝑝𝑠
𝐸
1−𝑒 −𝑝𝑠
𝑒
−
𝑒 −𝑠𝑡
𝑝
2
−𝑠 0
𝑝𝑠
2
−
−𝑠
𝑒0
−𝑠
− 𝐸
−
𝐸
1−𝑒 −𝑝𝑠
𝑠
1−𝑒 −𝑝𝑠
S.E/Paper Solutions
1−𝑒
−
𝑒 −𝑠𝑡 . −𝐸𝑑𝑡
𝑒 −𝑠𝑡
𝑝
−𝑠
𝑝
2
𝑒 −𝑝𝑠
𝑝𝑠
−
1−2𝑒 2 +𝑒 −𝑝𝑠
𝐸
𝑝
𝑝
2
−𝑠
+
𝑝𝑠 2
2
𝑠
45
𝑒
1+𝑒 −𝜃
and 𝑓 𝑡 + 𝑝 = 𝑓(𝑡)
2
𝑒 −𝑠𝑡 . 𝐸𝑑𝑡 +
1−𝑒 −𝜃
𝑝
[N14/CompIT/6M]
Solution:
We have, 𝑓(𝑡) is periodic with period p,
1
𝑝 −𝑠𝑡
𝐿𝑓 𝑡 =
𝑒 𝑓 𝑡 𝑑𝑡
−𝑝𝑠 0
1−𝑒
1
since
−
𝑝𝑠
2
−𝑠
= tanh
𝜃
2
Crescent Academy…….………………….…..For Research in Education
𝐸
=
=
𝑝𝑠
−
1−𝑒 2
𝐸 1−𝑒
−
𝑠 1+𝑒
−
𝑠
𝑠
𝑝𝑠
2
−
𝐸
𝑝𝑠
𝑠
since
4
1−𝑒 −𝜃
1+𝑒 −𝜃
= tanh
𝑡 0<𝑡<1
Find L.T. of 𝑓 𝑡 =
and 𝑓 𝑡 + 2 = 𝑓 𝑡
0 1<𝑡<2
[M18/IT/5M][N18/AutoMechCivil/6M][N18/Comp/6M]
[M19/AutoMechCivil/6M]
Solution:
We have, 𝑓(𝑡) is periodic with period 2,
1
2 −𝑠𝑡
𝐿𝑓 𝑡 =
𝑒 𝑓 𝑡 𝑑𝑡
−2𝑠 0
𝐿𝑓 𝑡
=
=
=
=
4.
.
𝑝𝑠
2
𝑝𝑠 2
2
𝑝𝑠
2
= . tanh
3.
−
−
𝑝𝑠
2
𝑝𝑠
−
1+𝑒 2
𝐸 1−𝑒
= .
1+𝑒
1−𝑒
1−𝑒
1
1 −𝑠𝑡
𝑒 . 𝑡𝑑𝑡
0
1−𝑒 −2𝑠
1
1−𝑒 −2𝑠
1
𝑒 −𝑠𝑡
𝑡.
−𝑠
𝑒 −𝑠
−
2
2 −𝑠𝑡
𝑒 . 0𝑑𝑡
1
−𝑠𝑡 1
+
− 1 .
𝑒 −𝑠
𝜃
𝑒
𝑠2
1
−0+
𝑠2
1−𝑒 −2𝑠 −𝑠
1
1−𝑠𝑒 −𝑠 −𝑒 −𝑠
0
𝑠2
𝑠2
1−𝑒 −2𝑠
𝑡
𝜋−𝑡
Find L.T. of 𝑓 𝑡 =
0<𝑡<𝜋
and 𝑓 𝑡 = 𝑓(𝑡 + 2𝜋)
𝜋 < 𝑡 < 2𝜋
Solution:
We have, 𝑓(𝑡) is periodic with period 2𝜋,
1
2𝜋 −𝑠𝑡
𝐿𝑓 𝑡 =
𝑒 𝑓 𝑡 𝑑𝑡
−2𝜋𝑠 0
𝐿𝑓 𝑡
=
=
=
=
=
=
=
1−𝑒
1
𝜋
0
1−𝑒 −2𝜋𝑠
1
1−𝑒 −2𝜋𝑠
1
𝑡.
𝜋
𝑒 −𝑠𝑡
− 1 .
−𝑠
𝑒 −𝜋𝑠
−
𝑒 −𝜋𝑠
1−𝑒 −2𝜋𝑠
−𝑠
𝑠2
1
𝜋(𝑒 −2𝜋𝑠 −𝑒 −𝜋𝑠 )
1−𝑒 −2𝜋𝑠
1
1−𝑒 −2𝜋𝑠
1−𝑒 −𝜋𝑠
𝑠
1−𝑒 −𝜋𝑠 2
𝑠2
−
2𝜋
𝜋
−𝑠𝑡 𝜋
𝑒 −𝑠𝑡 . 𝑡𝑑𝑡 +
−
𝑒
𝑠2
−0+
+
+
𝜋−𝑡 .
1
+ −𝜋 .
2
𝑠
𝑒 −2𝜋𝑠 −2𝑒 −𝜋𝑠 +1
𝜋𝑒 −𝜋𝑠
𝑠2
−𝜋𝑠
1−𝑒
𝑠
𝜋𝑒 −𝜋𝑠
𝑠 2 1+𝑒 −𝜋𝑠
𝑠 1+𝑒 −𝑠𝜋
−𝜋𝑠
−𝑠𝜋
1−𝑒
−𝜋𝑠𝑒
S.E/Paper Solutions
0
𝑒 −𝑠𝑡 . 𝜋 − 𝑡 𝑑𝑡
𝑠 2 1+𝑒 −𝜋𝑠
46
𝑒 −𝑠𝑡
−𝑠
𝑒 −2𝜋𝑠
−𝑠
− −1 .
+
𝑒 −2𝜋𝑠
𝑠2
𝑒 −𝑠𝑡
2𝜋
𝑠2
𝜋
−0−
𝑒 −𝜋𝑠
𝑠2
Crescent Academy…….………………….…..For Research in Education
5.
𝐾𝑡
Find L.T. of 𝑓 𝑡 = for 0 < 𝑡 < 𝑇 & 𝑓 𝑡 + 𝑇 = 𝑓 𝑡
𝑇
[N16/CompIT/6M]
Solution:
We have, 𝑓(𝑡) is periodic with period T,
𝑇 −𝑠𝑡
1
𝑒 𝑓 𝑡 𝑑𝑡
𝐿𝑓 𝑡 =
−𝑇𝑠 0
𝐿𝑓 𝑡
=
=
=
=
6.
𝑇 −𝑠𝑡 𝐾𝑡
𝑒 .
0
𝑇
1−𝑒 −𝑇𝑠
1
.
𝐾
1−𝑒 −𝑇𝑠 𝑇
1
𝐾
1−𝑒 −𝑇𝑠
.
𝑒 −𝑠𝑡
𝑡.
𝑇.
−𝑠
𝑒 −𝑇𝑠
𝑑𝑡
− 1 .
−
𝑒 −𝑇𝑠
𝐾
−𝑠
𝑠2
−𝑇𝑠
1−𝑠𝑇𝑒
−𝑒 −𝑇𝑠
𝑇 1−𝑒 −𝑇𝑠
𝑠2
𝑇
𝑒 −𝑠𝑡
𝑇
𝑠2
0
1
−0+
𝑠2
2𝑡
Find the Laplace Transform of 𝑓 𝑡 = , 0 ≤ 𝑡 ≤ 3, 𝑓 𝑡 + 3 = 𝑓 𝑡
3
[M17/CompIT/6M]
Solution:
We have, 𝑓(𝑡) is periodic with period 3,
3 −𝑠𝑡
1
𝑒 𝑓 𝑡 𝑑𝑡
𝐿𝑓 𝑡 =
−3𝑠 0
𝐿𝑓 𝑡
=
=
=
=
7.
1−𝑒
1
1−𝑒
1
3 −𝑠𝑡 2𝑡
𝑒 .
0
3
1−𝑒 −3𝑠
1
.
2
1−𝑒 −3𝑠 3
2
1
.
𝑡.
3.
𝑒 −𝑠𝑡
−𝑠
𝑒 −3𝑠
𝑑𝑡
− 1 .
−
𝑒 −3𝑠
2
−𝑠
𝑠2
1−3𝑠𝑒 −3𝑠 −𝑒 −3𝑠
3 1−𝑒 −3𝑠
𝑠2
1−𝑒 −3𝑠
3
𝑒 −𝑠𝑡
3
𝑠2
0
−0+
1
𝑠2
3𝑡
0<𝑡<2
where 𝑓 𝑡 has period 4. (i) Draw graph of
6
2<𝑡<4
𝑓 𝑡 (ii) Find 𝐿 𝑓 𝑡
Solution:
We have, 𝑓(𝑡) is periodic with period 4,
1
4 −𝑠𝑡
𝐿𝑓 𝑡 =
𝑒 𝑓 𝑡 𝑑𝑡
−4𝑠 0
If 𝑓 𝑡 =
𝐿𝑓 𝑡
=
=
=
=
=
1−𝑒
1
2 −𝑠𝑡
𝑒 . 3𝑡𝑑𝑡
0
1−𝑒 −4𝑠
1
1−𝑒 −4𝑠
1
3𝑡.
6
𝑒 −𝑠𝑡
−𝑠
𝑒 −2𝑠
−3
1−𝑒 −4𝑠
−𝑠
1
3 1−𝑒 −2𝑠
1−𝑒 −4𝑠
𝑠2
3 1−𝑒 −2𝑠 −6𝑠𝑒 −4𝑠
S.E/Paper Solutions
− 3 .
𝑒 −2𝑠
−
4 −𝑠𝑡
𝑒 . 6𝑑𝑡
2
2
4
𝑒 −𝑠𝑡
6𝑒 −𝑠𝑡
+
𝑠2
0
−0+
𝑠2
6𝑒 −4𝑠
𝑠
𝑠 2 1−𝑒 −4𝑠
47
+
3
−𝑠 2
6𝑒 −4𝑠
𝑠
−𝑠
+
2
−
6𝑒 −2𝑠
−𝑠
Crescent Academy…….………………….…..For Research in Education
𝑡
8.
Find L.T. of 𝑓 𝑡 =
0<𝑡<𝑎
𝑎
2𝑎−𝑡
𝑎 < 𝑡 < 2𝑎
𝑎
where 𝑓 𝑡 = 𝑓(𝑡 + 2𝑎)
Solution:
We have, 𝑓(𝑡) is periodic with period 2𝑎,
1
2𝑎 −𝑠𝑡
𝐿𝑓 𝑡 =
𝑒 𝑓 𝑡 𝑑𝑡
−2𝑎𝑠 0
𝐿𝑓 𝑡
=
=
=
=
=
1−𝑒
1
1−𝑒 −2𝑎𝑠
1
𝑎 1−𝑒 −2𝑎𝑠
1
𝑎 1−𝑒 −2𝑎𝑠
1
𝑎
0
𝑡
𝑎
𝑡.
𝑎
𝑒 −𝑠𝑡
− 1 .
−𝑠
𝑒 −𝑎𝑠
−
𝑒 −𝑎𝑠
𝑒 −𝑠𝑡 .
𝑒 −𝑠𝑡
𝑎
𝑠2
0
−0+
−𝑠
𝑠2
−𝑎𝑠
−2𝑎𝑠
1−2𝑒
+𝑒
𝑎 1−𝑒 −2𝑎𝑠
1
2𝑎
𝑎
𝑒 −𝑠𝑡 . 𝑑𝑡 +
2𝑎−𝑡
𝑎
𝑑𝑡
+ 2𝑎 − 𝑡 .
1
𝑠2
+0+
𝑒 −𝑠𝑡
𝑒 −2𝑎𝑠
𝑠2
−𝑠
− −1 .
− 𝑎.
𝑒 −𝑎𝑠
−
−𝑠
𝑒 −𝑠𝑡
2𝑎
𝑠2
𝑎
𝑒 −𝑎𝑠
𝑠2
𝑠2
−𝑎𝑠
2
1−𝑒
𝑎 1−𝑒 −2𝑎𝑠
𝑠2
−𝑎𝑠
2
1−𝑒
1
= .
=
=
9.
𝑎 𝑠 2 1−𝑒 −𝑎𝑠 1+𝑒 −𝑎𝑠
1 1−𝑒 −𝑎𝑠
.
𝑎𝑠 2 1+𝑒 −𝑎𝑠
1
𝑎𝑠
𝑎𝑠 2
. tanh
2
𝜋−𝑡 2
Find L.T. of 𝑓 𝑡 =
; 0 < 𝑡 < 2𝜋 and 𝑓 𝑡 = 𝑓 𝑡 + 2𝜋
2
Solution:
We have, 𝑓(𝑡) is periodic with period 2𝜋,
1
2𝜋 −𝑠𝑡
𝐿𝑓 𝑡 =
𝑒 𝑓 𝑡 𝑑𝑡
−2𝜋𝑠 0
𝐿𝑓 𝑡
=
=
=
=
1−𝑒
1
1−𝑒 −2𝜋𝑠
2𝜋
0
1
𝜋−𝑡
4 1−𝑒 −2𝜋𝑠
1
4 1−𝑒 −2𝜋𝑠
𝑒 −𝑠𝑡 .
𝜋2
𝑒 −2𝜋𝑠
−𝑠
𝜋−𝑡 2
−𝑠𝑡
2 𝑒
.
−𝑠
− 2𝜋
1
𝜋 2 1−𝑒 −2𝜋𝑠
4 1−𝑒 −2𝜋𝑠
𝑠
𝑎 𝑠𝑖𝑛𝑝𝑡
10. Find L.T. of 𝑓 𝑡 =
0
2
𝑑𝑡
− 2 𝜋 − 𝑡 −1 .
𝑒 −2𝜋𝑠
𝑠2
−
+ 2.
𝑠2
0<𝑡<
𝜋
𝑝
<𝑡<
𝑝
48
−𝑠 3
2𝜋 1+𝑒 −2𝜋𝑠
Solution:
2𝜋
We have, 𝑓(𝑡) is periodic with period ,
S.E/Paper Solutions
𝑒 −2𝜋𝑠
𝑒 −𝑠𝑡
𝑠2
− 𝜋2.
+
1
−𝑠
+ 2.
− 2𝜋.
2 1−𝑒 −2𝜋𝑠
𝑒 −𝑠𝑡
2𝜋
−𝑠 3 0
1
𝑠2
− 2.
𝑠3
𝜋
𝑝
2𝜋 and
𝑝
𝑓 𝑡 =𝑓 𝑡+
2𝜋
𝑝
1
−𝑠 3
Crescent Academy…….………………….…..For Research in Education
𝐿𝑓 𝑡
=
1−𝑒
𝐿𝑓 𝑡
1−𝑒
2𝜋
𝑠
𝑝
−
1−𝑒
−
𝜋
𝑝
2𝜋
𝑠
𝑝
𝑒 −𝑠𝑡
2𝜋
𝑠
𝑝
−𝑠 2 +𝑝 2
𝑒
2𝜋
− 𝑠
1−𝑒 𝑝
−
−
𝑠𝜋
𝑝
𝑠 2 +𝑝 2
𝑎𝑝
1−𝑒
𝑒 −𝑠𝑡 . 𝑎𝑠𝑖𝑛𝑝𝑡𝑑𝑡 +
0
𝑎
=
𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡
0
𝑎
=
=
−
1
=
=
2𝜋
𝑝
1
1+𝑒
2𝜋
𝑠
𝑝
𝑎𝑝
𝑒 −𝑠𝑡 . 0𝑑𝑡
𝜋
𝑝
−𝑠. 𝑠𝑖𝑛𝑝𝑡 − 𝑝. 𝑐𝑜𝑠𝑝𝑡
0
1
. 𝑝 −
𝑠 2 +𝑝 2
. (−𝑝)
𝑠𝜋
𝑝
𝑠 2 +𝑝 2
1+𝑒
𝑠 2 +𝑝 2
−
2𝜋
𝑝
𝜋
𝑝
.
1−𝑒
−
−
𝑠𝜋
𝑝
2𝜋
𝑠
𝑝
𝑠𝑖𝑛2𝑡
0<𝑡<
𝜋
2
and 𝑓 𝑡 = 𝑓 𝑡 + 𝜋
0
<𝑡<𝜋
2
[N16/AutoMechCivil/6M]
Solution:
We have, 𝑓(𝑡) is periodic with period 𝜋,
𝜋 −𝑠𝑡
1
𝑒 𝑓 𝑡 𝑑𝑡
𝐿𝑓 𝑡 =
−𝜋𝑠 0
11. Find L.T. of 𝑓 𝑡 =
𝐿𝑓 𝑡
=
=
=
=
=
𝜋
1−𝑒
1
𝜋
2
0
1−𝑒 −𝜋𝑠
1
𝑒 −𝑠𝑡
1−𝑒 −𝜋𝑠
−𝑠 2 +22
1
𝑒
−
𝑠𝜋
2
1−𝑒 −𝜋𝑠
𝑠 2 +4
2
1+𝑒
1−𝑒 −𝜋𝑠
2
𝑠 2 +4
.
𝜋
𝑒 −𝑠𝑡 . 𝑠𝑖𝑛2𝑡𝑑𝑡 +
−
𝜋
2
−𝑠. 𝑠𝑖𝑛2𝑡 − 2. 𝑐𝑜𝑠2𝑡
. 2 −
−
1
. (−2)
𝑠 2 +4
𝑠𝜋
2
𝑠𝜋
2
1−𝑒 −𝜋𝑠
𝑠𝑖𝑛𝜔𝑡
12. Find L.T. of 𝑓 𝑡 =
0
0<𝑡<
𝜋
𝜔
<𝑡<
𝜋
𝜔
2𝜋
𝜔
Solution:
2𝜋
We have, 𝑓(𝑡) is periodic with period ,
𝜔
S.E/Paper Solutions
𝜋
2
0
𝑠 2 +4
1+𝑒
𝑒 −𝑠𝑡 . 0𝑑𝑡
49
Crescent Academy…….………………….…..For Research in Education
𝐿𝑓 𝑡
=
1−𝑒
𝐿𝑓 𝑡
1−𝑒
2𝜋
𝑠
𝜔
−
1−𝑒
−
𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡
0
𝜋
𝜔
2𝜋
𝑠
𝜔
0
𝑒
−𝑠𝑡
𝑒 −𝑠𝑡
1
=
=
−
1
=
=
2𝜋
𝑝
1
2𝜋
𝑠
𝜔
−𝑠 2 +𝜔 2
1
𝑒
2𝜋
− 𝑠
1−𝑒 𝜔
−
𝑠𝜋
𝜔
𝜔
1+𝑒
2𝜋
− 𝑠
1−𝑒 𝜔
−
𝑠𝜋
𝜔
𝑠𝑖𝑛7𝑡
𝜋
𝜔
0
1
𝑠 2 +𝜔 2
𝜔
𝑠 2 +𝜔 2
. (−𝜔)
1+𝑒
.
1−𝑒
0<𝑡<
𝜋
2
𝑒 −𝑠𝑡 . 0𝑑𝑡
−𝑠. 𝑠𝑖𝑛𝜔𝑡 − 𝜔. 𝑐𝑜𝑠𝜔𝑡
=
𝑠 2 +𝜔 2
13. Find L.T. of 𝑓 𝑡 =
. 𝑠𝑖𝑛𝜔𝑡𝑑𝑡 +
. 𝜔 −
𝑠 2 +𝜔 2
2𝜋
𝜔
𝜋
𝜔
−
𝑠𝜋
𝜔
2𝜋
𝑠
𝜔
𝜋
2
and 𝑓 𝑡 = 𝑓 𝑡 + 𝜋
<𝑡<𝜋
2
−
[N17/Elect/6M]
Solution:
We have, 𝑓(𝑡) is periodic with period 𝜋,
𝜋 −𝑠𝑡
1
𝑒 𝑓 𝑡 𝑑𝑡
𝐿𝑓 𝑡 =
−𝜋𝑠 0
𝐿𝑓 𝑡
=
=
=
=
1−𝑒
1
𝜋
2
0
1−𝑒 −𝜋𝑠
1
𝑒 −𝑠𝑡
1−𝑒 −𝜋𝑠
−𝑠 2 +72
1
𝑒
−
𝑠𝜋
2
1−𝑒 −𝜋𝑠 𝑠 2 +49
1
1−𝑒 −𝜋𝑠
𝑠
𝜋
𝑒 −𝑠𝑡 . 𝑠𝑖𝑛7𝑡𝑑𝑡 +
𝑠 2 +49
+
𝑒 −𝑠𝑡 . 2𝑑𝑡
𝜋
2
−𝑠. 𝑠𝑖𝑛7𝑡 − 7. 𝑐𝑜𝑠7𝑡
1
𝑠 2 +49
7
𝑠 2 +49
−
. −7 + 2
2𝑒 −𝑠𝜋
𝑠
+
𝑒 −𝑠𝜋
−𝑠
=
1−𝑒 −2𝜋𝑠
1
= 1−𝑒 −2𝜋𝑠
=
=
1
𝜋
0
𝑒 −𝑠𝑡 . 𝑠𝑖𝑛𝑡𝑑𝑡 +
. −𝑠. 𝑠𝑖𝑛𝑡 − 1. 𝑐𝑜𝑠𝑡
𝑠 2 +1
. 1 −
1
1−𝑒 −2𝜋𝑠 𝑠 2 +1
𝑠 2 +1
1
𝑒 −𝑠𝜋 +1−𝑠.𝑒 −2𝜋𝑠 −𝑠
1−𝑒 −2𝜋𝑠
S.E/Paper Solutions
2𝜋
𝜋
−𝑠
𝜋
2
𝑠 2 +1
50
0
−2
𝑒
−𝑠
𝜋
2
−𝑠
𝑠𝑖𝑛𝑡
𝑐𝑜𝑠𝑡
0<𝑡<𝜋
𝜋 < 𝑡 < 2𝜋
𝑒 −𝑠𝑡 . 𝑐𝑜𝑠𝑡𝑑𝑡
𝜋
𝑒 −𝑠𝑡
𝑒 −𝑠𝜋
𝜋
𝑠
Solution:
We have, 𝑓(𝑡) is periodic with period 2𝜋,
1
2𝜋 −𝑠𝑡
𝐿𝑓 𝑡 =
𝑒 𝑓 𝑡 𝑑𝑡
−2𝜋𝑠 0
𝐿𝑓 𝑡
𝑒 −𝑠𝑡
𝜋
−𝑠
2𝑒 2
14. Find the Laplace Transforms of f(t), where 𝑓 𝑡 =
1−𝑒
1
+2
0
. 𝑠 −
𝜋
−𝑠
𝑒 2
𝜋
2
+
. −1 +
2𝜋
𝑒 −𝑠𝑡
. −𝑠. 𝑐𝑜𝑠𝑡 + 1. 𝑠𝑖𝑛𝑡
𝑠 2 +1
𝑒 −2𝜋𝑠
𝑠 2 +1
. −𝑠 −
1
𝑠 2 +1
. (𝑠)
𝜋
Crescent Academy…….………………….…..For Research in Education
15. Find the Laplace Transforms of f(t), where 𝑓 𝑡 = 𝑠𝑖𝑛𝑝𝑡 ; 𝑡 ≥ 0
[N15/CompIT/6M]
Solution:
We note that,
𝑓 𝑡+
𝜋
= 𝑠𝑖𝑛𝑝 𝑡 +
𝑝
𝜋
= sin(𝑝𝑡 + 𝜋) = 𝑠𝑖𝑛𝑝𝑡
𝑝
∴ 𝑓(𝑡) is periodic with period
𝐿𝑓 𝑡
𝐿𝑓 𝑡
=
=
1
𝜋
− 𝑠
1−𝑒 𝑝
0
𝜋
− 𝑠
𝑝
𝑒
𝜋
− 𝑠
1−𝑒 𝑝
−
𝑠𝜋
𝑝
𝑠 2 +𝑝 2
1+𝑒
1
1−𝑒
𝜋
− 𝑠
𝑝
1
−
𝜋
𝑝
0
. 𝑝 −
1
𝑠 2 +𝑝 2
. (−𝑝)
𝑠𝜋
𝑝
𝑠 2 +𝑝 2
1+𝑒
𝑠 2 +𝑝 2
−𝑠. 𝑠𝑖𝑛𝑝𝑡 − 𝑝. 𝑐𝑜𝑠𝑝𝑡
−𝑠 2 +𝑝 2
1
=
𝑒 −𝑠𝑡 . 𝑠𝑖𝑛𝑝𝑡𝑑𝑡
𝑒 −𝑠𝑡
1
1−𝑒
=
𝜋
𝑝
𝜋
− 𝑠
1−𝑒 𝑝
𝑝
𝑒 −𝑠𝑡 𝑓 𝑡 𝑑𝑡
0
1
=
=
𝜋
𝑝
𝜋
.
−
𝑠𝜋
𝑝
=
𝑠𝜋
−
1−𝑒 𝑝
1
𝑠 2 +𝑝 2
. coth
𝜋𝑠
2𝑝
16. Find the Laplace Transforms of f(t), where𝑓 𝑡 = 𝑠𝑖𝑛𝑡
[N18/Elect/5M]
Solution:
We note that,
𝑓 𝑡 + 𝜋 = 𝑠𝑖𝑛 𝑡 + 𝜋 = 𝑠𝑖𝑛𝑡
∴ 𝑓(𝑡) is periodic with period 𝜋
1
𝜋 −𝑠𝑡
𝐿𝑓 𝑡 =
𝑒 𝑓 𝑡 𝑑𝑡
−𝜋𝑠 0
𝐿𝑓 𝑡
=
=
=
=
=
1−𝑒
1
1−𝑒 −𝜋𝑠
1
𝜋 −𝑠𝑡
𝑒 . 𝑠𝑖𝑛𝑡𝑑𝑡
0
𝑒 −𝑠𝑡
1−𝑒 −𝜋𝑠 −𝑠 2 +1
1
𝑒 −𝑠𝜋
. 1 −
1−𝑒 −𝜋𝑠 𝑠 2 +1
1+𝑒 −𝜋𝑠
1
1−𝑒 −𝜋𝑠 𝑠 2 +1
1
1+𝑒 −𝑠𝜋
𝑠 2 +1
.
1−𝑒 −𝑠𝜋
=
𝜋
−𝑠. 𝑠𝑖𝑛𝑡 − 1. 𝑐𝑜𝑠𝑡
1
𝑠 2 +1
1
𝑠 2 +1
0
. (−1)
. coth
𝜋𝑠
2
17. Find the Laplace Transforms of f(t), where𝑓 𝑡 = 𝑐𝑜𝑠𝑡
Solution:
S.E/Paper Solutions
51
Crescent Academy…….………………….…..For Research in Education
We note that,
𝑓 𝑡 + 𝜋 = 𝑐𝑜𝑠 𝑡 + 𝜋 = 𝑐𝑜𝑠𝑡
∴ 𝑓(𝑡) is periodic with period 𝜋
1
𝜋 −𝑠𝑡
𝐿𝑓 𝑡 =
𝑒 𝑓 𝑡 𝑑𝑡
−𝜋𝑠 0
𝐿𝑓 𝑡
=
=
=
=
=
=
1−𝑒
1
1−𝑒 −𝜋𝑠
1
𝜋 −𝑠𝑡
𝑒 . 𝑐𝑜𝑠𝑡𝑑𝑡
0
𝑒 −𝑠𝑡
1−𝑒 −𝜋𝑠 −𝑠 2 +1
1
𝑒 −𝑠𝜋
−𝑠. 𝑐𝑜𝑠𝑡 + 1. 𝑠𝑖𝑛𝑡
. 𝑠 −
1−𝑒 −𝜋𝑠 𝑠 2 +1
1+𝑒 −𝜋𝑠
𝑠
0
1
𝑠 2 +1
. (−𝑠)
1−𝑒 −𝜋𝑠 𝑠 2 +1
𝑠
1+𝑒 −𝑠𝜋
𝑠 2 +1
𝑠
𝑠 2 +1
S.E/Paper Solutions
.
1−𝑒 −𝑠𝜋
𝜋𝑠
. coth
𝜋
2
52
Crescent Academy…….………………….…..For Research in Education
Type IX: Miscellaneous Solved Problems
1.
Find the Laplace Transforms of sin 𝑡
[N13/AutoMechCivil/5M][N13/Chem/5M][M18/Elex/5M]
Solution:
𝑥3
We have, 𝑠𝑖𝑛𝑥 = 𝑥 −
sin 𝑡 = 𝑡 −
𝑡2
𝐿 sin 𝑡 = 𝐿 𝑡
𝐿 sin 𝑡 =
𝐿 sin 𝑡 =
𝐿 sin 𝑡 =
𝐿 sin 𝑡 =
3
2
3
𝑠2
Γ
𝐿 sin 𝑡 =
−𝐿
𝜋
𝜋
3
5
𝑡2
𝑡2
5
2
5
6𝑠 2
31 1
. Γ
22 2
5
6𝑠 2
3
2
3
1−
2𝑠 2
𝜋
1
+𝐿
+
+
+
4𝑠
1
− ⋯ … ..
− ⋯ … ..
7
1
2
7
120𝑠 2
53
.
22
120𝑠 2
− ⋯ … ..
− ⋯…..
1
− ⋯…..
32𝑠 2
1 2
4𝑠
+
4𝑠
120
120𝑠 2
531
. . Γ
222
6𝑠
1
7
2
Γ
+
1−
1−
3
2𝑠 2
− ⋯ ….
6
Γ
−
− ⋯ ….
− ⋯ … ..
120
−
1 1
Γ
2 2
3
𝑠2
1 1
Γ
2 2
3
𝑠2
𝐿 sin 𝑡 =
5
1
2
5!
5!
𝑡2
+
6
5
𝑡
+
3!
3
1
2
3
𝑡
sin 𝑡 = 𝑡 −
3!
𝑥5
+
− ⋯…..
2!
−
3 . 𝑒 4𝑠
2𝑠 2
2.
Find the Laplace Transforms of
𝑠𝑖𝑛 𝑡
𝑡
Solution:
We have, 𝑠𝑖𝑛𝑥 = 𝑥 −
𝑡
sin 𝑡 = 𝑡 −
1
sin 𝑡 = 𝑡 2 −
sin 𝑡
𝐿
𝐿
𝐿
3
𝑡2
6
𝑡
sin 𝑡
𝑡
sin 𝑡
𝑡
sin 𝑡
𝑡
3!
+
5
𝑡2
120
6
120
1
𝑠
1
6𝑠 2
𝑠
6𝑠 2
= −
S.E/Paper Solutions
1
𝑥5
5
5!
− ⋯ ….
− ⋯ ….
5!
− ⋯ … ..
− ⋯ … ….
= 𝐿 1 −𝐿
1
+
𝑡
+
𝑡2
= −
3!
3
𝑡
= 1− +
𝑥3
+
+
𝑡
+𝐿
6
2!
120𝑠 3
1
60𝑠 3
𝑡2
120
−⋯…
− ⋯ ….
53
−⋯…
Crescent Academy…….………………….…..For Research in Education
3.
cos 𝑡
Find the Laplace Transforms of
𝑡
[M14/ChemBiot/5M][N16/ChemBiot/5M]
Solution:
𝑥2
We have, 𝑐𝑜𝑠𝑥 = 1 −
cos 𝑡 = 1 −
+
2!
𝑡
𝑡2
cos 𝑡 = 1 − +
2
cos 𝑡
𝑡
𝐿
𝐿
𝐿
𝐿
𝐿
𝐿
𝐿
4.
1
=𝑡
cos 𝑡
𝑡
cos 𝑡
𝑡
cos 𝑡
𝑡
cos 𝑡
𝑡
cos 𝑡
𝑡
cos 𝑡
𝑡
cos 𝑡
𝑡
−2
−
24
1
𝑡2
2
2!
𝑡
2
𝑡
3
− ⋯ … ….
1
=𝐿 𝑡
=
=
=
=
=
=
1
2
1
𝑠2
1
Γ 2
1
𝑠2
1
Γ 2
1
𝑠2
1
Γ 2
1
𝑠2
1
Γ 2
1
𝑠2
Γ
𝜋
𝑠
−𝐿
3
2
3
2𝑠 2
1 1
Γ
2 2
3
2𝑠 2
1
2
Γ
−
−
− ⋯ ….
− ⋯ … ..
24
−2
− ⋯ ….
4!
4
4!
𝑡2
+
𝑥4
+
1−
1−
3
𝑡2
+𝐿
2
+
1−
1
𝑡2
Γ
5
2
5
1
2
5
24𝑠 2
31
.
22
24𝑠 2
+
2𝑠
1
1
+
4𝑠
1
− ⋯ ….
− ⋯…..
− ⋯…..
32𝑠 2
1 2
4𝑠
+
4𝑠
−⋯…
− ⋯ … ..
24𝑠 2
31
. Γ
22
+
24
2!
− ⋯…..
1
𝑒 −4𝑠
Find the Laplace Transforms of 𝑒𝑟𝑓 𝑡
Solution:
2 𝑥 −𝑢 2
By definition, 𝑒𝑟𝑓𝑥 =
𝑒 𝑑𝑢
0
𝜋
𝑒𝑟𝑓 𝑡 =
𝑒𝑟𝑓 𝑡 =
𝑒𝑟𝑓 𝑡 =
𝑒𝑟𝑓 𝑡 =
𝑒𝑟𝑓 𝑡 =
2
𝑡
𝜋 0
2
𝑡
𝜋 0
2
𝜋
2
𝜋
2
𝜋
S.E/Paper Solutions
𝑒
−𝑢 2
1 − 𝑢2 +
𝑢−
𝑢3
3
𝑡 −
+
𝑡
𝑡−
1
2
𝑑𝑢
3
𝑢5
10
5
𝑡2
3
+
10
𝑢6
3!
− ⋯…
+
3
2!
−
+ ⋯ . . 𝑑𝑢
𝑡
3
𝑡2
𝑢4
𝑡
10
5
0
−⋯……
−⋯…
54
Crescent Academy…….………………….…..For Research in Education
𝐿 𝑒𝑟𝑓 𝑡 =
𝐿 𝑒𝑟𝑓 𝑡 =
𝐿 𝑒𝑟𝑓 𝑡 =
𝐿 𝑒𝑟𝑓 𝑡 =
𝐿 𝑒𝑟𝑓 𝑡 =
𝐿 𝑒𝑟𝑓 𝑡 =
𝐿 𝑒𝑟𝑓 𝑡 =
2
𝐿 𝑡
𝜋
3
2
3
𝑠2
Γ
2
𝜋
2
𝜋
2
𝜋
2
𝜋
1
.
.
3
𝐿 𝑒𝑟𝑓 𝑡 =
𝐿 𝑒𝑟𝑓 𝑡 =
𝐿 𝑒𝑟𝑓 𝑡 =
5.
1 1
Γ
2 2
3
𝑠2
1 1
Γ
2 2
3
𝑠2
5
𝑡2
1−
1
+𝐿
3
+
1−
3
2𝑠 2
3
𝑡2
5
2
5
3𝑠 2
31 1
. Γ
22 2
5
3𝑠 2
3
2
Γ
−
𝜋
3𝑠
7
2
Γ
+
+
+
− ⋯…
7
1
2
7
10𝑠 2
53
.
22
10𝑠 2
3
− ⋯….
−⋯…
2 4 𝑠2
1
1 −2
− ⋯……
− ⋯…..
1− . + . .
2 𝑠
−⋯…
10
10𝑠 2
531
. . Γ
222
2𝑠
8𝑠 2
1 1
1 3 1
𝑠2
1
1+
3
𝑠
1
1 𝑠+1 −2
3
𝑠2
𝑠
1
𝑠
1
2
3
𝑠 2 𝑠+1
𝑠
1
𝑠 𝑠
1
.
𝑠+1
𝑠 𝑠+1
Find the Laplace Transforms of erf𝑐 𝑡
Solution:
𝑒𝑟𝑓𝑥 + 𝑒𝑟𝑓𝑐 𝑥 = 1
erf𝑐 𝑡 = 1 − erf 𝑡
𝐿 erf𝑐 𝑡 = 𝐿 1 − 𝐿 erf 𝑡
1
1
𝑠
𝑠 𝑠+1
𝐿 erf𝑐 𝑡 = −
6.
−𝐿
−
𝑠2
𝐿 𝑒𝑟𝑓 𝑡 =
1
2
Find the Laplace Transforms of 𝐽0 𝑡 𝑤𝑕𝑒𝑟𝑒𝐽0 𝑡 =
Solution:
𝐽0 𝑡 =
𝐽0 𝑡 =
𝑟 𝑡 2𝑟
∞ −1
0 𝑟! 2 2
−1 0 𝑡 0
−1
0!
2
𝐽0 𝑡 = 1 −
S.E/Paper Solutions
𝑡2
4
2
+
+
𝑡4
64
1!
1
2
𝑡 2
2
+
− ⋯ … ..
55
−1
2
𝑡 4
2!
2
2
+ ⋯…
𝑟
∞ −1
0 𝑟! 2
𝑡 2𝑟
2
Crescent Academy…….………………….…..For Research in Education
𝐿 𝐽0 𝑡
=𝐿 1 −𝐿
𝐿 𝐽0 𝑡
= −
𝐿 𝐽0 𝑡
=
𝐿 𝐽0 𝑡
=
𝐿 𝐽0 𝑡
=
𝐿 𝐽0 𝑡
=
𝐿 𝐽0 𝑡
=
𝐿 𝐽0 𝑡
=
𝐿 𝐽0 𝑡
=
𝐿 𝐽0 𝑡
1
2!
𝑠
1
4𝑠 3
𝑠
1
𝑠
1
𝑠
1
𝑠
1−
1−
1 𝑠 2 +1
𝑠
𝑠2
1
𝑠2
4
4!
+
+𝐿
4𝑠 2
1
+
+
2𝑠 2
1 1
64𝑠 4
3
64
− ⋯….
8𝑠 4
1 3 1
2 4 𝑠4
− ⋯…..
𝑠2
1
−2
1
2
𝑠 𝑠 2 +1
1
𝑠
𝑠
=
S.E/Paper Solutions
1
− ⋯ ….
− ⋯….
+ . .
2 𝑠2
1
1 −2
𝑡4
− ⋯ ….
64𝑠 5
2
24
1− .
1+
𝑡2
𝑠 2 +1
𝑠 2 +1
56