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Exam soln LT chappter

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LAPLACE TRANSFORM
Type I: Basic
1.
2.
Find the Laplace Transforms of 4𝑑 2 + 𝑠𝑖𝑛3𝑑 + 𝑒 2𝑑
Ans.
𝑠3
+
3
𝑠 2 +9
+
1
𝑠−2
Find the Laplace Transforms of 1 + 𝑠𝑖𝑛𝑑
[M18/Extc/5M]
Solution:
𝐿 1 + 𝑠𝑖𝑛𝑑
𝑑
= 𝐿 cos + sin
2
=
𝑠
1
𝑠 2 +4
+
1
=
=
3.
8
1
2
1
𝑠 2 +4
𝑑
πœƒ
πœƒ
2
2
since 1 + π‘ π‘–π‘›πœƒ = cos + sin
2
𝑠+2
1
𝑠 2 +4
4𝑠+2
4𝑠 2 +1
Find the LT of 𝑠𝑖𝑛2𝑑 − π‘π‘œπ‘ 2𝑑 2
[M19/Extc/5M]
Solution:
𝐿 𝑠𝑖𝑛2𝑑 − π‘π‘œπ‘ 2𝑑 2
= 𝐿{sin2 2𝑑 − 2𝑠𝑖𝑛2𝑑 π‘π‘œπ‘ 2𝑑 + cos 2 2𝑑}
= 𝐿 1 − 𝑠𝑖𝑛4𝑑
since, 2π‘ π‘–π‘›πœƒπ‘π‘œπ‘ πœƒ = 𝑠𝑖𝑛2πœƒ
=
1
𝑠
−
4
𝑠 2 +16
2
4.
Find the Laplace Transforms of sin2 𝑑
Ans.
5.
Find the Laplace Transforms of cos 2 3𝑑
Ans.
6.
7.
Evaluate using L.T.: 0 𝑒 −2𝑑 sin2 2𝑑 𝑑𝑑
Ans.
5
2
−2𝑑
2
Find the Laplace Transforms of 𝑑 − 𝑒
+ cosh 3𝑑
∞
Ans.
3
8.
Find the Laplace Transforms of sin 𝑑
Ans.
9.
Find the Laplace Transforms of cos 3 2𝑑
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 −2𝑑 sin3 𝑑 𝑑𝑑
[M17/ElexExtcElectBiomInst/5M]
Solution:
By definition,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = sin3 𝑑
∞ −𝑠𝑑
𝑒 sin3 𝑑 𝑑𝑑 = 𝐿 sin3 𝑑
0
= 𝐿
3𝑠𝑖𝑛𝑑 −𝑠𝑖𝑛 3𝑑
4
1
= 𝐿 3𝑠𝑖𝑛𝑑 − 𝑠𝑖𝑛3𝑑
4
1
1
3
= 3. 2 − 2
4
𝑠 +1
𝑠 +9
Put 𝑠 = 2
∞ −2𝑑
𝑒
0
sin3 𝑑 𝑑𝑑 =
1
3
4 4+1
−
3
=
4+9
6
65
11. Find the Laplace Transforms of 𝑠𝑖𝑛3π‘‘π‘π‘œπ‘ 4𝑑
Ans.
12. Find the Laplace Transforms of 𝑠𝑖𝑛𝑑𝑠𝑖𝑛5𝑑
Ans.
3𝑠 2 −21
𝑠 2 +49 𝑠 2 +1
10𝑠
𝑠 2 +16 𝑠 2 +36
13. Find the Laplace Transforms of π‘π‘œπ‘ π‘‘ π‘π‘œπ‘ 2𝑑 π‘π‘œπ‘ 3𝑑
[N15/ChemBiot/5M][M16/ChemBiot/5M]
Solution:
𝐿{π‘π‘œπ‘ 2𝑑 π‘π‘œπ‘ π‘‘ π‘π‘œπ‘ 3𝑑}
1
= 𝐿 2π‘π‘œπ‘ 2𝑑 π‘π‘œπ‘ π‘‘ π‘π‘œπ‘ 3𝑑
2
1
= 𝐿 π‘π‘œπ‘ 3𝑑 + π‘π‘œπ‘ π‘‘ π‘π‘œπ‘ 3𝑑
2
1
= 𝐿{cos 2 3𝑑 + π‘π‘œπ‘ 3𝑑 π‘π‘œπ‘ π‘‘}
2
1 1
= . 𝐿 2 cos 2 3𝑑 + 2π‘π‘œπ‘ 3𝑑 π‘π‘œπ‘ π‘‘
=
=
2 2
1
4
1
4
𝐿 1 + π‘π‘œπ‘ 6𝑑 + π‘π‘œπ‘ 4𝑑 + π‘π‘œπ‘ 2𝑑
1
𝑠
+
𝑠
𝑠 2 +36
+
𝑠
𝑠 2 +16
+
𝑠
𝑠 2 +4
∞
1
14. If 0 𝑒 −2𝑑 sin 𝑑 + 𝛼 cos 𝑑 − 𝛼 𝑑𝑑 = , then find α
4
[M16/CompIT/5M]
Solution:
By definition,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = sin 𝑑 + 𝛼 cos 𝑑 − 𝛼
∞ −𝑠𝑑
𝑒 sin 𝑑 + 𝛼 cos 𝑑 − 𝛼 𝑑𝑑 = 𝐿 sin 𝑑 + 𝛼 cos 𝑑 − 𝛼
0
S.E/Paper Solutions
2
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=
=
∴
∞
0
1
𝐿 2 sin 𝑑 + 𝛼 cos 𝑑 − 𝛼
2
1
𝐿 sin 2𝑑 + sin 2𝛼
2
𝑒 −𝑠𝑑 sin 𝑑 + 𝛼 cos 𝑑 − 𝛼 𝑑𝑑 =
1
2
2
𝑠 2 +4
+
Put 𝑠 = 2,
∞ −2𝑑
𝑒
0
sin 𝑑 + 𝛼 cos 𝑑 − 𝛼 𝑑𝑑 =
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œ‹
1
2
6
πœ‹
12
15. Find the Laplace Transform of 𝑠𝑖𝑛2π‘‘π‘π‘œπ‘ 3𝑑
[M17/AutoMechCivil/5M]
Solution:
𝐿{𝑠𝑖𝑛2𝑑 π‘π‘œπ‘ 3𝑑}
1
= 𝐿 2𝑠𝑖𝑛2𝑑 π‘π‘œπ‘ 3𝑑
2
1
= 𝐿 𝑠𝑖𝑛5𝑑 + sin(−𝑑)
2
1
= 𝐿{𝑠𝑖𝑛5𝑑 − 𝑠𝑖𝑛𝑑}
=
2
1
5
2 𝑠 2 +25
−
1
𝑠 2 +1
16. Find the Laplace Transform of 𝑠𝑖𝑛2𝑑𝑠𝑖𝑛𝑑𝑠𝑖𝑛3𝑑
[N13/Biot/5M]
Solution:
𝐿{𝑠𝑖𝑛2𝑑 𝑠𝑖𝑛𝑑 𝑠𝑖𝑛3𝑑}
1
= 𝐿 2𝑠𝑖𝑛2𝑑 𝑠𝑖𝑛𝑑 𝑠𝑖𝑛3𝑑
2
1
= 𝐿 π‘π‘œπ‘ π‘‘ − π‘π‘œπ‘ 3𝑑 𝑠𝑖𝑛3𝑑
2
1
= 𝐿{π‘π‘œπ‘ π‘‘π‘ π‘–π‘›3𝑑 − π‘π‘œπ‘ 3𝑑 𝑠𝑖𝑛3𝑑}
2
1 1
= . 𝐿 2π‘π‘œπ‘ π‘‘ 𝑠𝑖𝑛3𝑑 − 2π‘π‘œπ‘ 3𝑑 𝑠𝑖𝑛3𝑑
=
2 2
1
4
𝐿 𝑠𝑖𝑛4𝑑 − sin −2𝑑 − 𝑠𝑖𝑛6𝑑
S.E/Paper Solutions
3
𝑠𝑖𝑛 2𝛼
𝑠
Crescent Academy…….………………….…..For Research in Education
=
=
1
4
1
𝐿 𝑠𝑖𝑛4𝑑 + 𝑠𝑖𝑛2𝑑 − 𝑠𝑖𝑛6𝑑
4
4 𝑠 2 +16
+
2
𝑠 2 +4
−
6
𝑠 2 +36
∞
17. Evaluate 0 𝑒 𝑑 𝑠𝑖𝑛2π‘‘π‘π‘œπ‘ 3𝑑𝑑𝑑
[N17/AutoMechCivil/6M]
Solution:
By definition,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = 𝑠𝑖𝑛2π‘‘π‘π‘œπ‘ 3𝑑
∞ −𝑠𝑑
𝑒 𝑠𝑖𝑛2π‘‘π‘π‘œπ‘ 3𝑑𝑑𝑑 = 𝐿 𝑠𝑖𝑛2π‘‘π‘π‘œπ‘ 3𝑑
0
=
1
2
1
𝐿 2𝑠𝑖𝑛2𝑑 π‘π‘œπ‘ 3𝑑
= 𝐿 𝑠𝑖𝑛5𝑑 + sin(−𝑑)
2
1
= 𝐿{𝑠𝑖𝑛5𝑑 − 𝑠𝑖𝑛𝑑}
=
2
1
5
2
𝑠 2 +25
−
1
𝑠 2 +1
Put 𝑠 = −1
∞ 𝑑
𝑒 𝑠𝑖𝑛2π‘‘π‘π‘œπ‘ 3𝑑𝑑𝑑
0
=
1
5
2 1+25
−
1
1+1
=−
∞
2
13
3
18. If 0 𝑒 −2𝑑 sin 𝑑 + 𝛼 cos 𝑑 − 𝛼 𝑑𝑑 = , then find α
8
[N14/CompIT/6M]
Solution:
By definition,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = sin 𝑑 + 𝛼 cos 𝑑 − 𝛼
∞ −𝑠𝑑
𝑒 sin 𝑑 + 𝛼 cos 𝑑 − 𝛼 𝑑𝑑 = 𝐿 sin 𝑑 + 𝛼 cos 𝑑 − 𝛼
0
=
=
∴
∞
0
1
2
1
2
𝐿 2 sin 𝑑 + 𝛼 cos 𝑑 − 𝛼
𝐿 sin 2𝑑 + sin 2𝛼
1
𝑒 −𝑠𝑑 sin 𝑑 + 𝛼 cos 𝑑 − 𝛼 𝑑𝑑 =
2
Put 𝑠 = 2,
∞ −2𝑑
𝑒
0
sin 𝑑 + 𝛼 cos 𝑑 − 𝛼 𝑑𝑑 =
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
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1
𝑠𝑖𝑛 2𝛼
=
4
4
𝑠𝑖𝑛2𝛼 = 1
πœ‹
2𝛼 =
𝛼=
19. Obtain 𝐿 𝑓 3𝑑
if 𝐿 𝑓 𝑑
=
πœ‹
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 𝑑 =
then find 𝐿 𝑠𝑖𝑛2 𝑑
21. Evaluate using L.T.:
S.E/Paper Solutions
Ans.
∞
π‘’π‘Ÿπ‘“
0
2 𝑑 𝑒 −5𝑑 𝑑𝑑
5
Ans.
πœ‹
𝑠 𝑠
2
15
.𝑒
1
−𝑠
πœ‹
2𝑠 𝑠
1
.𝑒
−4𝑠
,
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Type II: First Shifting Property
1.
State the first shifting theorem for Laplace Transforms. Use the theorem
to obtain 𝐿 𝑒 −𝑑 sin2 𝑑
2.
3.
Find L.T. of 𝑒
4𝑑
Ans.
3
sin 𝑑
Ans.
1
1
−
2 𝑠+1
3
1
4
𝑠+1
𝑠 2 +2𝑠+5
𝑠 2 −8𝑠+17
−
−2𝑑
Find L.T. of 𝑒
1 − 𝑠𝑖𝑛𝑑
[N17/IT/4M]
Solution:
𝐿 𝑒 −2𝑑 1 − 𝑠𝑖𝑛𝑑
𝑑
= 𝐿 𝑒 −2𝑑 cos − sin
=𝐿 𝑒
𝑑
2
cos − 𝑒
2
𝑠+2
=
=
−2𝑑
−
1
4
𝑠+2 2 +
1
2
𝑑
2
−2𝑑
1
2
sin
𝑑
𝑑
2
2
since, 1 − 𝑠𝑖𝑛𝑑 = cos − sin
𝑑
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 𝑑 π‘π‘œπ‘ π‘‘
2
Ans.
𝑠𝑖𝑛𝑑 2
Find L.T. of
Ans.
𝑒𝑑
π‘π‘œπ‘  2𝑑 𝑠𝑖𝑛𝑑
Find L.T. of
𝑒𝑑
[M15/ChemBiot/5M][M18/AutoMechCivil/5M]
Solution:
𝐿
π‘π‘œπ‘  2𝑑 𝑠𝑖𝑛𝑑
𝑒𝑑
−𝑑
= 𝐿 𝑒 π‘π‘œπ‘ 2𝑑 𝑠𝑖𝑛𝑑
1
= 𝐿 𝑒 −𝑑 2π‘π‘œπ‘ 2𝑑 𝑠𝑖𝑛𝑑
=
2
1
2
1
𝐿 𝑒 −𝑑 𝑠𝑖𝑛3𝑑 − 𝑠𝑖𝑛𝑑
= 𝐿 𝑒 −𝑑 𝑠𝑖𝑛3𝑑 − 𝑒 −𝑑 𝑠𝑖𝑛𝑑
=
=
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 π‘ π‘–π‘›π‘‘π‘π‘œπ‘ 2π‘‘π‘π‘œπ‘ π‘•π‘‘
[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 𝑠𝑖𝑛𝑑 π‘π‘œπ‘ 2𝑑 π‘π‘œπ‘ π‘•π‘‘
𝑒 𝑑 +𝑒 −𝑑
= 𝐿 𝑠𝑖𝑛𝑑 π‘π‘œπ‘ 2𝑑
2
𝑑
1
= 𝐿 𝑠𝑖𝑛𝑑 π‘π‘œπ‘ 2𝑑 𝑒 + 𝑒 −𝑑
=
=
=
=
=
2
1
4
1
4
1
4
1
4
1
𝐿 2𝑠𝑖𝑛𝑑 π‘π‘œπ‘ 2𝑑 𝑒 𝑑 + 𝑒 −𝑑
𝑒 𝑑 + 𝑒 −𝑑
𝐿 𝑠𝑖𝑛3𝑑 + sin −𝑑
𝐿 𝑒 𝑑 𝑠𝑖𝑛3𝑑 + 𝑒 −𝑑 𝑠𝑖𝑛3𝑑 − 𝑒 𝑑 𝑠𝑖𝑛𝑑 − 𝑒 −𝑑 𝑠𝑖𝑛𝑑
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 𝑑 𝑠𝑖𝑛2𝑑 𝑠𝑖𝑛3𝑑
9.
Find L.T. of 𝑠𝑖𝑛2π‘‘π‘π‘œπ‘ π‘‘π‘π‘œπ‘ π‘•2𝑑
[N18/IT/5M]
Solution:
𝐿 𝑠𝑖𝑛2𝑑 π‘π‘œπ‘ π‘‘ π‘π‘œπ‘ π‘•2𝑑
1
= 𝐿 𝑠𝑖𝑛2𝑑 π‘π‘œπ‘ π‘‘ 𝑒
=
=
=
=
=
4
1
4
1
4
1
4
1
4
2
2𝑑
=
2
1
1
𝑠−1
2
𝑠 2 −2𝑠+2
𝑒 2𝑑 + 𝑒 −2𝑑
𝐿 𝑠𝑖𝑛3𝑑 + sin 𝑑
𝐿 𝑒 2𝑑 𝑠𝑖𝑛3𝑑 + 𝑒 −2𝑑 𝑠𝑖𝑛3𝑑 + 𝑒 2𝑑 𝑠𝑖𝑛𝑑 + 𝑒 −2𝑑 𝑠𝑖𝑛𝑑
3
𝑠−2
3
2 +9
𝑠 2 −4𝑠+13
+
+
3
2 +9
𝑠+2
+
3
𝑠 2 +4𝑠+13
𝑒 𝑑 −𝑒 −𝑑
1
=
𝑠 2 +2𝑠+2
+ 𝑒 −2𝑑
+
1
𝑠−2
2 +1
1
𝑠 2 −4𝑠+5
2
𝑠𝑖𝑛𝑑
= 𝐿 𝑒 −3𝑑 − 𝑒 −5𝑑 𝑠𝑖𝑛𝑑
2
1
𝑠+1 2 +1
1
𝐿 2𝑠𝑖𝑛2𝑑 π‘π‘œπ‘ π‘‘ 𝑒 2𝑑 + 𝑒 −2𝑑
= 𝐿 𝑒 −4𝑑
=
−
1
Ans.
10. Find L.T. of 𝑒 −4𝑑 𝑠𝑖𝑛𝑕𝑑 𝑠𝑖𝑛𝑑
[N14/ChemBiot/5M]
Solution:
𝐿 𝑒 −4𝑑 𝑠𝑖𝑛𝑕𝑑 𝑠𝑖𝑛𝑑
2
1
−
𝑒 2𝑑 +𝑒 −2𝑑
= 𝐿 𝑠𝑖𝑛2𝑑 π‘π‘œπ‘ π‘‘
2
1
2 +1
𝐿 𝑒 −3𝑑 𝑠𝑖𝑛𝑑 − 𝑒 −5𝑑 𝑠𝑖𝑛𝑑
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
2π‘Ž 2 𝑠
11. Find L.T. of π‘ π‘–π‘›π‘Žπ‘‘ π‘ π‘–π‘›π‘•π‘Žπ‘‘
Ans. 4 4
𝑠 +4π‘Ž
12. State and prove first shifting property of Laplace transform and also find
𝐿 𝑑𝑒 3𝑑 𝑠𝑖𝑛𝑕𝑑
Ans.
13. If Laplace Transform of erf
[N18/Comp/5M]
Solution:
We have, 𝐿 erf 𝑑
𝑑 =
4
2
𝑠−1
∞
𝑠−1+4
1
𝑠
+1
4
𝑠−1
3𝑑
𝑑
=
2
𝑠 𝑠+4
2
=
𝑠+3
3
14. Show that 0 𝑒 −𝑑 . sin . sinh
𝑑𝑑 =
2
2
2
−𝑑
15. Find the Laplace Transform of 𝑑𝑒 π‘π‘œπ‘ π‘•2𝑑
[M15/AutoMechCivil/5M]
Solution:
𝐿 𝑑𝑒 −𝑑 π‘π‘œπ‘ π‘•2𝑑
𝑒 2𝑑 +𝑒 −2𝑑
= 𝐿 𝑑 𝑒 −𝑑
=
=
=
1
2
1
2
1
2
2
−3𝑑
𝑑
𝐿 𝑑 𝑒 +𝑒
𝐿 𝑒 𝑑 𝑑 + 𝑒 −3𝑑 𝑑
1
𝑠−1
2
1
+
𝑠+3
2
16. Prove that 𝐿 π‘π‘œπ‘ 2𝑑. π‘π‘œπ‘ π‘•2𝑑 =
17. Prove that 𝐿{π‘π‘œπ‘ π‘Žπ‘‘. π‘π‘œπ‘ π‘•π‘Žπ‘‘} =
𝑑
18. Find L.T. of sinh sin
2
[N17/Extc/5M]
Solution:
𝑑
3𝑑
2
2
𝐿 sinh sin
𝑑
=𝐿
1
𝑒 2 −𝑒
−
. sin
2
𝑑
2
= 𝐿 𝑒 sin
2
𝑑
2
S.E/Paper Solutions
3𝑑
2
−𝑒
𝑠3
𝑠 4 +64
𝑠3
𝑠 4 +4π‘Ž 4
3𝑑
2
3𝑑
2
𝑑
−2
sin
3𝑑
2
8
2
1
𝑠−4
𝑑
2
−
1
𝑠−2
then find 𝐿 𝑒 . erf 2 𝑑
𝑠 𝑠+1
4
=
𝑠 𝑠+1
1
=
By change of scale property,
1
𝐿 erf 2 𝑑 = 𝐿 π‘’π‘Ÿπ‘“ 4𝑑 = . 𝑠
Now,
𝐿 𝑒 𝑑 erf 2 𝑑
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 −5𝑑 . erf 𝑑
[M19/Elect/5M]
Solution:
We have,
1
𝐿 erf 𝑑 =
𝑠 𝑠+1
∴𝐿 𝑒
−5𝑑
erf 𝑑 =
1
=
𝑠+5+1
𝑠+5
1
(𝑠+5) 𝑠+6
∞
20. Evaluate 0 𝑒 −2𝑑 . 𝑑 5 π‘π‘œπ‘ π‘•π‘‘ 𝑑𝑑
[N16/ElexExtcElectBiomInst/5M]
Solution:
By definition,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = 𝑑 5 π‘π‘œπ‘ π‘•π‘‘
∞ −𝑠𝑑 5
𝑒 𝑑 π‘π‘œπ‘ π‘•π‘‘π‘‘π‘‘ = 𝐿 𝑑 5 π‘π‘œπ‘ π‘•π‘‘
0
𝑒 𝑑 +𝑒 −𝑑
= 𝐿 𝑑5
=
=
1
2
1
2
𝐿 𝑒 𝑑 + 𝑒 −𝑑 𝑑 5
2
𝑑 5
5!
𝑠−1
6
+
5!
𝑠+1
6
Put 𝑠 = 2,
∞ −2𝑑 5
𝑒 t π‘π‘œπ‘ π‘•π‘‘π‘‘π‘‘
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 π‘‘π‘ π‘–π‘›π‘Žπ‘‘
Ans.
2.
Find L.T. of π‘‘π‘π‘œπ‘ π‘Žπ‘‘
Ans.
3.
3
= −
= −
= −
= −
=
=
5.
𝑠 2 +π‘Ž 2
2
Find L.T. of 𝑑 sin 𝑑
[N15/AutoMechCivil/5M]
Solution:
𝑑
𝐿 𝑑𝑠𝑖𝑛3 𝑑 = −1
𝐿 sin3 𝑑
𝑑𝑠
𝑑
= −1
4.
2π‘Žπ‘ 
𝑠 2 +π‘Ž 2 2
𝑠 2 −π‘Ž 2
𝑑𝑠
𝑑
1
4
𝐿 3𝑠𝑖𝑛𝑑 − 𝑠𝑖𝑛3𝑑
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
3𝑑
3𝑠𝑖𝑛𝑑 −𝑠𝑖𝑛 3𝑑
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 𝑑𝑒 𝑠𝑖𝑛𝑑
Ans.
−3𝑑
2𝑠−6
𝑠 2 −6𝑠+10
Find L.T. of 𝑑𝑒 π‘π‘œπ‘ 2π‘‘π‘π‘œπ‘ 3𝑑
[M14/ElexExtcElectBiomInst/4M]
Solution:
We have,
1
𝐿 π‘π‘œπ‘ 2𝑑 π‘π‘œπ‘ 3𝑑 = 𝐿 2π‘π‘œπ‘ 2𝑑 π‘π‘œπ‘ 3𝑑
=
=
2
1
2
1
2
𝐿 π‘π‘œπ‘ 5𝑑 + cos −𝑑
𝑠
𝑠 2 +25
𝑑
𝐿 𝑑 π‘π‘œπ‘ 2𝑑 π‘π‘œπ‘ 3𝑑 = −1
= −1
=−
=−
=
1
2
1
2
1
2
𝑑𝑠
𝑑
𝑠
𝑠 2 +1
𝐿 π‘π‘œπ‘ 2𝑑 π‘π‘œπ‘ 3𝑑
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 −3𝑑 𝑑 π‘π‘œπ‘ 2𝑑 π‘π‘œπ‘ 3𝑑 =
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 −3𝑑 . π‘‘π‘π‘œπ‘ π‘‘π‘‘π‘‘
[M15/ElexExtcElectBiomInst/5M]
Solution:
By definition,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = 𝑑 π‘π‘œπ‘ π‘‘
∞ −𝑠𝑑
𝑒 𝑑 π‘π‘œπ‘ π‘‘π‘‘π‘‘ = 𝐿 𝑑 π‘π‘œπ‘ π‘‘
0
∞ −𝑠𝑑
𝑒 𝑑
0
π‘π‘œπ‘ π‘‘π‘‘π‘‘ = 𝐿 𝑑 π‘π‘œπ‘ π‘‘
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 π‘π‘œπ‘ π‘‘
𝑠 2 +1
𝑠 2 −1
𝑠 2 +1
2
2
2
2
Put 𝑠 = 3,
∴
7.
∞
0
𝑒 −3𝑑 𝑑 π‘π‘œπ‘ π‘‘π‘‘π‘‘ =
Evaluate using L.T.:
32 −1
32 +1 2
=
∞
𝑠𝑖𝑛𝑑 2
𝑑
0
𝑒𝑑
8
100
=
2
25
𝑑𝑑
[N13/Chem/6M]
Solution:
We have,
∞
𝑑. 𝑒 −2𝑑 sin2 𝑑 𝑑𝑑 =?
0
By definition,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = 𝑑 sin2 𝑑
∞ −𝑠𝑑
𝑒 𝑑 sin2 𝑑 𝑑𝑑 = 𝐿 𝑑 sin2 𝑑
0
∞ −𝑠𝑑
𝑒 𝑑 sin2
0
1−π‘π‘œπ‘  2𝑑
𝑑 𝑑𝑑 = 𝐿 𝑑
2
1
= 𝐿 𝑑 − π‘‘π‘π‘œπ‘ 2𝑑
=
=
S.E/Paper Solutions
2
1 1
𝑑
2 𝑠
1 1
𝑑𝑠
𝑑
− −1
2
2
𝑠2
− −1
11
𝑑𝑠
𝐿 π‘π‘œπ‘ 2𝑑
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
−2𝑑
𝑑 sin2 𝑑 𝑑𝑑 =
1 1
−0 =
2 4
1
8
Find L.T. of 𝑑𝑒 3𝑑 π‘π‘œπ‘ π‘‘
Ans.
Find Laplace Transform of 𝑑𝑒
[N17/AutoMechCivil/5M]
Solution:
𝑑
𝐿 𝑑 𝑠𝑖𝑛𝑑 = −1
𝐿 𝑠𝑖𝑛𝑑
𝑑𝑠
𝑑
= −1
=
−1[2𝑠]
𝑠 2 +1
2𝑠
𝑠 2 +1
𝑠𝑖𝑛𝑑
1
𝑑𝑠 𝑠 2 +1
𝑠 2 +1 0
= −1
−3𝑑
2
2
2 𝑠+3
∴ 𝐿 𝑒 −3𝑑 𝑑 𝑠𝑖𝑛𝑑 =
𝑠+3 2 +1
2
=
2 𝑠+3
𝑠 2 +6𝑠+10
2
10. Find L.T. of 𝑑𝑒 3𝑑 𝑠𝑖𝑛4𝑑
[N16/CompIT/5M][M18/IT/4M]
Solution:
𝐿 𝑑𝑒 3𝑑 𝑠𝑖𝑛4𝑑 =?
We have,
4
𝐿 𝑠𝑖𝑛4𝑑 = 2
𝑠 +16
𝐿 𝑑 𝑠𝑖𝑛4𝑑 = −1
= −1
= −4
=
𝐿 𝑠𝑖𝑛4𝑑
4
𝑠 2 +16
𝑑𝑠
𝑠 2 +16 0 −1[2𝑠]
𝑠 2 +16
8𝑠
𝑠 2 +16
∴ 𝐿 𝑒 3𝑑 𝑑 𝑠𝑖𝑛4𝑑 =
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 −2𝑑 𝑑 π‘π‘œπ‘ π‘‘
[M18/Comp/5M]
Solution:
𝑑
𝐿 π‘π‘œπ‘ π‘‘
𝐿 𝑑 π‘π‘œπ‘ π‘‘ = −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 −2𝑑 𝑑 π‘π‘œπ‘ π‘‘ =
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 −𝑑 𝑑 π‘π‘œπ‘ 2𝑑 𝑠𝑖𝑛4𝑑
[M18/Elect/5M]
Solution:
𝐿 𝑑𝑒 −𝑑 π‘π‘œπ‘ 2𝑑 𝑠𝑖𝑛4𝑑 =?
We have,
1
𝐿 π‘π‘œπ‘ 2𝑑 𝑠𝑖𝑛4𝑑 = 𝐿 2π‘π‘œπ‘ 2𝑑 𝑠𝑖𝑛4𝑑
=
=
2
1
2
1
2
𝐿 𝑠𝑖𝑛6𝑑 + sin −2𝑑
6
𝑑
𝐿 𝑑 π‘π‘œπ‘ 2𝑑 𝑠𝑖𝑛4𝑑 = −1
𝑑𝑠
𝑑
= −1
=−
=−
=
−
𝑠 2 +36
𝐿 π‘π‘œπ‘ 2𝑑 𝑠𝑖𝑛4𝑑
1
6
𝑠 2 +36
12𝑠
−
𝑠 2 +36
2
6𝑠
𝑠 2 +36 2
∴ 𝐿 𝑑𝑒 −𝑑 π‘π‘œπ‘ 2𝑑 𝑠𝑖𝑛4𝑑 =
=
−
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 𝑑 𝑑 𝑠𝑖𝑛2𝑑 π‘π‘œπ‘ π‘‘
[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 𝑑 𝑑 𝑠𝑖𝑛2𝑑 π‘π‘œπ‘ π‘‘ =?
We have,
1
𝐿 𝑠𝑖𝑛2𝑑 π‘π‘œπ‘ π‘‘ = 𝐿 2𝑠𝑖𝑛2𝑑 π‘π‘œπ‘ π‘‘
2
=
=
1
𝐿 𝑠𝑖𝑛3𝑑 + sint
2
1
3
𝑠 2 +9
2
𝐿 𝑑 π‘π‘œπ‘ 2𝑑 𝑠𝑖𝑛4𝑑 = −1
=−
=
2
3𝑠
𝑑
𝐿 𝑠𝑖𝑛2𝑑 π‘π‘œπ‘ π‘‘
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
∞
𝑠𝑖𝑛𝑑𝑑𝑑 = 𝐿 𝑑 𝑠𝑖𝑛𝑑
𝑑
= −1
𝑑𝑠
𝑑
= −1
=
Put 𝑠 = 3,
∞
∴ 0 𝑒 −3𝑑 𝑑 𝑠𝑖𝑛𝑑𝑑𝑑 =
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
−2𝑑
𝐿 𝑠𝑖𝑛𝑑
𝑑𝑠 𝑠 2 +1
𝑠 2 +1 [0]−1[2𝑠]
= −1
=
2
2 𝑠−1
14. Evaluate using L.T.: 0 𝑒 −3𝑑 𝑑 𝑠𝑖𝑛𝑑 𝑑𝑑
[M17/AutoMechCivil/6M]
Solution:
By definition,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = 𝑑 𝑠𝑖𝑛𝑑
∞ −𝑠𝑑
𝑒 𝑑 𝑠𝑖𝑛𝑑𝑑𝑑 = 𝐿 𝑑 𝑠𝑖𝑛𝑑
0
∞ −𝑠𝑑
𝑒 𝑑
0
𝑠 2 +1
2𝑠
𝑠 2 +1 2
2 +9 2
𝑠−1
+
2
2
3 𝑠−1
∴ 𝐿 𝑒 𝑑 𝑑 𝑠𝑖𝑛2𝑑 π‘π‘œπ‘ π‘‘ =
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
=
𝑑 𝑠𝑖𝑛𝑑 𝑑𝑑
14
3
50
2
2
Crescent Academy…….………………….…..For Research in Education
∞ −𝑠𝑑
𝑒 𝑓
0
𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
Put 𝑓 𝑑 = 𝑑 𝑠𝑖𝑛𝑑
∞ −𝑠𝑑
𝑒 𝑑 𝑠𝑖𝑛𝑑𝑑𝑑 = 𝐿 𝑑 𝑠𝑖𝑛𝑑
0
∞ −𝑠𝑑
𝑒 𝑑
0
𝑠𝑖𝑛𝑑𝑑𝑑 = 𝐿 𝑑 𝑠𝑖𝑛𝑑
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 −2𝑑 𝑑 𝑠𝑖𝑛𝑑𝑑𝑑 =
𝐿 𝑠𝑖𝑛𝑑
𝑑𝑠
𝑑
22 +1 2
4
=
𝑑
25
2
16. Evaluate using L.T.:
𝑒 𝑑 cos 𝑑 𝑑𝑑
[M19/AutoMechCivil/6M]
Solution:
We have,
∞ 𝑑
𝑒 𝑑 cos 2 𝑑 𝑑𝑑 =?
0
By definition,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = 𝑑 cos 2 𝑑
∞ −𝑠𝑑
𝑒 𝑑 cos 2 𝑑 𝑑𝑑 = 𝐿 𝑑 cos 2 𝑑
0
∞ −𝑠𝑑
𝑒 𝑑 cos 2
0
1+π‘π‘œπ‘  2𝑑
𝑑 𝑑𝑑 = 𝐿 𝑑
2
1
= 𝐿 𝑑 + π‘‘π‘π‘œπ‘ 2𝑑
=
=
=
=
=
=
2
1 1
𝑠2
2
1 1
2
𝑠2
+ −1
+ −1
𝑑
𝑑𝑠
𝑑
𝑑𝑠
𝐿 π‘π‘œπ‘ 2𝑑
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 𝑑 𝑑 cos 2 𝑑 𝑑𝑑 =
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 𝑑 1 + 𝑠𝑖𝑛𝑑
[M15/ElexExtcElectBiomInst/4M][N15/ElexExtcElectBiomInst/4M]
[M16/ElexExctElectBiomInst/4M][N17/AutoMechCivil/6M]
[N17/Extc/4M][M18/Extc/4M][M19/Extc/4M]
Solution:
𝑑
𝐿 𝑑 1 + 𝑠𝑖𝑛𝑑 = −1
𝐿 1 + 𝑠𝑖𝑛𝑑
= −1
= −1
= −1
= −1
𝑑𝑠
𝑑
𝑑𝑠
𝑑
𝑑
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
𝑑 1 + 𝑠𝑖𝑛𝑑𝑑𝑑
Ans. −4
19. Find L.T. of 𝑑 1 + 𝑠𝑖𝑛2𝑑
Ans.
20. Find L.T. of 𝑑 1 − 𝑠𝑖𝑛𝑑
Ans.
∞
0
21. Evaluate using L.T.:
𝑒 −𝑑 . 𝑑 1 + 𝑠𝑖𝑛𝑑𝑑𝑑
[N16/AutoMechCivil/5M]
Solution:
By definition,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = 𝑑 1 + 𝑠𝑖𝑛𝑑
∞ −𝑠𝑑
𝑒 𝑑 1 + 𝑠𝑖𝑛𝑑𝑑𝑑 = 𝐿 𝑑 1 + 𝑠𝑖𝑛𝑑
0
= −1
= −1
S.E/Paper Solutions
𝑑
𝑑𝑠
𝑑
𝑑𝑠
16
𝐿
1 + 𝑠𝑖𝑛𝑑
𝑑
𝑑
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 −𝑑 . 𝑑 1 + 𝑠𝑖𝑛𝑑𝑑𝑑 =
25
22. Find L.T. of 𝑑 𝑒 3𝑑 erf 5 𝑑
[M19/IT/4M]
Solution:
We have,
1
𝐿 erf 𝑑 =
𝑠 𝑠+1
By change of scale property,
1
1
𝐿 erf 5 𝑑 = . 𝑠 𝑠 =
25
25
25
Now,
𝐿 𝑑 erf 5 𝑑 = −1
= −1
= −1
= (−5)
= −5
𝑑𝑠
𝑑
𝐿 erf 5 𝑑
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 3𝑑 𝑑 erf 5 𝑑 = .
2
𝑠−3
1
23. Given 𝐿 erf 𝑑 =
𝑠 𝑠+1
2
5
3
2
𝑠−3+25
2
∞
𝑑
0
, evaluate
3𝑠+41
= .
𝑠−3
2
𝑠+22
3
2
𝑒 −𝑑 erf 𝑑 𝑑𝑑
[M14/CompIT/6M]
Solution:
By definition,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = 𝑑 erf 𝑑
∞ −𝑠𝑑
𝑒 𝑑 erf 𝑑 𝑑𝑑 = 𝐿 𝑑 erf 𝑑
0
𝑑
= −1
= −1
= −1
𝑑
1
𝑑𝑠
𝑠 3 +𝑠 2
𝑑
𝑠3 + 𝑠2
𝑑𝑠
= −1
𝑑 𝑑𝑑 =
25. Prove that
=
1
−
3𝑠 2 +2𝑠
2 𝑠 3 +𝑠 2
Put 𝑠 = 1,
∞
∴ 0 𝑒 −𝑑 𝑑 erf 𝑑 𝑑𝑑 =
24. If 𝐿 𝐽0 𝑑
1
𝑑𝑠 𝑠 𝑠+1
= −1
∞ −𝑠𝑑
𝑒 𝑑 erf
0
𝐿 erf 𝑑
𝑑𝑠
𝑑
5
2 2
3
2
1
2
3
−2
× 3𝑠 2 + 2𝑠
3
2
=
5
4 2
, find 𝐿 𝑑𝐽0 3𝑑
Ans.
4𝑑 𝑑𝑑 =
3
125
[M18/IT/6M]
Solution:
By definition,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = 𝑑 𝐽0 4𝑑
∞ −𝑠𝑑
𝑒 𝑑 𝐽0 4𝑑 𝑑𝑑 = 𝐿 𝑑 𝐽0 4𝑑
0
We have,
S.E/Paper Solutions
1
−2
𝑠3 + 𝑠2
𝑠 2 +1
∞
𝑑𝑒 −3𝑑 𝐽0
0
1
2
18
where 𝐿 𝐽0 𝑑
𝑠
𝑠 2 +9
=
3
2
1
1+𝑠 2
Crescent Academy…….………………….…..For Research in Education
𝐿 𝐽0 𝑑
1
=
(given)
𝑠 2 +1
By change of scale property,
1
1
1
= .
𝐿 𝐽0 4𝑑 = .
2
4
∴
∞
0
𝑒
−𝑠𝑑
𝑠
4
+1
1
4
𝑠2 +16
16
.
𝑠 2 +16
=
1
𝑠 2 +16
𝑑 𝐽0 4𝑑 𝑑𝑑 = 𝐿 𝑑 𝐽0 4𝑑
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 −3𝑑 𝑑 𝐽0 4𝑑 𝑑𝑑 =
1
𝑑𝑠
𝑑
= −1
𝑒 −𝑠𝑑 𝑑 𝐽0 4𝑑 𝑑𝑑 =
𝐿 𝐽0 4𝑑
𝑑𝑠
𝑑
= −1
∞
0
1
16
16
= −1
∴
=
3
2
3
=
25
3
2
=
3
125
3𝑠+8
26. Find L.T. of 𝑑 erf 2 𝑑
Ans.
27. Find L.T. of π‘‘π‘’π‘Ÿπ‘“ 3 𝑑
Ans. .
28. Show that
∞
0
𝑑𝑒 −6𝑑 𝐽0 8𝑑 𝑑𝑑 =
3
500
where 𝐿 𝐽0 𝑑
[N14/ElexExtcElectBiomInst/6M]
Solution:
By definition,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = 𝑑 𝐽0 8𝑑
∞ −𝑠𝑑
𝑒 𝑑 𝐽0 8𝑑 𝑑𝑑 = 𝐿 𝑑 𝐽0 8𝑑
0
We have,
1
(given)
𝐿 𝐽0 𝑑 = 2
𝑠 +1
By change of scale property,
1
1
1
1
1
𝐿 𝐽0 8𝑑 = .
= . 2 = .
2
8
∴
∞
0
𝑠
8
+1
8
8
𝑠 +64
64
8
𝑠 2 +64
𝑒 −𝑠𝑑 𝑑 𝐽0 8𝑑 𝑑𝑑 = 𝐿 𝑑 𝐽0 8𝑑
= −1
= −1
= −1
S.E/Paper Solutions
𝑑
𝑑𝑠
𝑑
𝑑𝑠
𝑑
𝑑𝑠
𝐿 𝐽0 8𝑑
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 −𝑠𝑑 𝑑 𝐽0 8𝑑 𝑑𝑑 =
Put 𝑠 = 6,
∞
∴ 0 𝑒 −6𝑑 𝑑 𝐽0 8𝑑 𝑑𝑑 =
29. Evaluate
∞
𝑑𝑒 −4𝑑 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 −𝑠𝑑 𝑑 𝑓 3𝑑 𝑑𝑑 = 𝐿 𝑑 𝑓 3𝑑
= −1
= −1
𝑑
𝑒 −𝑠𝑑 𝑑 𝑓 3𝑑 𝑑𝑑 =
Put 𝑠 = 4,
∞
∴ 0 𝑒 −4𝑑 𝑑 𝑓 3𝑑 𝑑𝑑 =
𝐿 𝑓 3𝑑
𝑑𝑠
𝑑
1
𝑠 2 +9
2
𝑑𝑠
𝑑
𝑠 +9
𝑑𝑠
= −1
∞
0
=
3
2
=
= .
𝑠 +9
9
= −1
∴
6
=
3
2
3𝑑 𝑑𝑑 if 𝐿 𝑓 𝑑
+1
× 2𝑠
3
2
[N13/Biot/6M]
Solution:
By definition,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = 𝑑 𝑓 3𝑑
∞ −𝑠𝑑
𝑒 𝑑 𝑓 3𝑑 𝑑𝑑 = 𝐿 𝑑 𝑓 3𝑑
0
We have,
1
𝐿𝑓 𝑑 = 2
(given)
𝑠 +1
By change of scale property,
1
1
1
1
𝐿 𝑓 3𝑑 = .
= . 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 𝑑 3 π‘π‘œπ‘ π‘‘
[N14/AutoMechCivil/5M]
Solution:
𝐿 𝑑 3 π‘π‘œπ‘ π‘‘ = −1
𝑠2 + 9
π‘π‘œπ‘ π‘‘
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 𝑑 5 π‘π‘œπ‘ π‘•π‘‘
[N15/CompIT/5M]
Solution:
𝑒 𝑑 +𝑒 −𝑑
𝐿 𝑑 5 π‘π‘œπ‘ π‘•π‘‘ = 𝐿 𝑑 5
=
=
=
S.E/Paper Solutions
1
2
1
2
𝑑 5
𝐿 𝑒 𝑑 + 𝑒 −𝑑 𝑑 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.
π‘ π‘–π‘›π‘Žπ‘‘
Find Laplace transform of
[M18/AutoMechCivil/5M]
Solution:
𝐿
π‘ π‘–π‘›π‘Žπ‘‘
∞
𝐿 π‘ π‘–π‘›π‘Žπ‘‘
𝑠
∞ π‘Ž
𝑑𝑠
𝑠 𝑠 2 +π‘Ž
∞
−1 𝑠
=
𝑑
=
𝑑
. Does Laplace transform of
π‘π‘œπ‘ π‘Žπ‘‘
𝑑
exist?
𝑑𝑠
= tan
π‘Ž 𝑠
𝑠
= − tan−1
2
π‘Ž
𝑠
= cot −1
π‘Ž
π‘π‘œπ‘ π‘Žπ‘‘
∞
= 𝑠 𝐿 π‘π‘œπ‘ π‘Žπ‘‘ 𝑑𝑠
𝑑
∞ 𝑠
= 𝑠 2 𝑑𝑠
𝑠 +π‘Ž
∞
1
2
2
πœ‹
𝐿
=
2
1
log 𝑠 + π‘Ž
𝑠
2
= π‘™π‘œπ‘”∞ − log 𝑠 + π‘Ž2
2
= π‘‘π‘œπ‘’π‘  π‘›π‘œπ‘‘ 𝑒π‘₯𝑖𝑠𝑑
2.
𝑠𝑖𝑛 𝑕2𝑑
Find L.T. of
𝑑
[N17/Extc/4M][M18/Extc/4M]
Solution:
𝐿
𝑠𝑖𝑛 𝑕2𝑑
𝑑
∞
𝐿 𝑠𝑖𝑛𝑕2𝑑 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 −4𝑑 𝑠𝑖𝑛 3𝑑
𝑑
𝑒 −2𝑑 π‘π‘œπ‘  2𝑑𝑠𝑖𝑛 3𝑑
𝑑
π‘π‘œπ‘  𝑕2𝑑 𝑠𝑖𝑛 2𝑑
𝑑
𝑒 −2𝑑 𝑠𝑖𝑛 2𝑑 π‘π‘œπ‘  𝑕𝑑
S.E/Paper Solutions
𝑑
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œ‹ − tan−1
− tan−1
2
2
2
1
Crescent Academy…….………………….…..For Research in Education
7.
∞ 𝑒−
2𝑑 𝑠𝑖𝑛 𝑕𝑑𝑠𝑖𝑛𝑑
Solve 0
𝑑𝑑
𝑑
[M16/AutoMechCivil/6M] [M19/Elex/5M][M19/Elect/6M]
Solution:
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 =
𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑑
∞ −𝑠𝑑 𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑒
0
𝑑
∞ −𝑠𝑑 𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑒
0
𝑑
∞ −𝑠𝑑 𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑒
0
𝑑
𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑑𝑑 = 𝐿
𝑑
𝑒 𝑑 −𝑒 −𝑑
𝑑𝑑 = 𝐿
1
𝑑𝑑 = 𝐿
2
.
𝑠𝑖𝑛𝑑
2
𝑑
𝑑 𝑠𝑖𝑛𝑑
𝑒
−
𝑑
𝑒 −𝑑
𝑠𝑖𝑛𝑑
𝑑
We have,
∞
∞ 1
πœ‹
𝐿 𝑠𝑖𝑛𝑑 𝑑𝑠 = 𝑠 2 𝑑𝑠 = [tan−1 𝑠]∞
= − tan−1
𝑠
𝑠
𝑑
𝑠 +1
2
∞ −𝑠𝑑 𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
πœ‹
1 πœ‹
−1
−1
∴ 0 𝑒
𝑑𝑑 =
− tan 𝑠 − 1 − + tan 𝑠 + 1
𝑑
2
2 2
∞ −𝑠𝑑 𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
1
−1
−1
𝑑𝑑 = tan 𝑠 + 1 − tan 𝑠 − 1
𝑒
0
2
𝑑
∞ −𝑠𝑑 𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑠+1 − 𝑠−1
1
𝑑𝑑 = tan−1
𝑒
0
1+ 𝑠+1 𝑠−1
2
𝑑
∞ −𝑠𝑑 𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
2
1
1
−1
−1 2
𝑒
𝑑𝑑
tan
=
tan
=
2
0
𝑑
1+𝑠 −1
2
𝑠2
2
𝐿
𝑠𝑖𝑛𝑑
=
Put 𝑠 = 2,
∞ − 2𝑑 𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑒
0
𝑑
8.
Prove that
2
1
𝑑𝑑 = tan−1
2
∞ 𝑠𝑖𝑛 2𝑑+𝑠𝑖𝑛 3𝑑
0
𝑑𝑒 𝑑
𝑑𝑑 =
2
2
1
1 πœ‹
πœ‹
2
2 4
8
= tan−1 (1) = . =
3πœ‹
4
[M14/ChemBiot/6M][M18/Comp/6M][N18/Elex/5M]
Solution:
By definition of L.T.,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 =
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
𝑠𝑖𝑛 2𝑑+𝑠𝑖𝑛 3𝑑
𝑑
𝑠𝑖𝑛 2𝑑+𝑠𝑖𝑛 3𝑑
𝑑
𝑠𝑖𝑛 2𝑑+𝑠𝑖𝑛 3𝑑
𝑑
𝑠𝑖𝑛 2𝑑+𝑠𝑖𝑛 3𝑑
𝑑
𝑠𝑖𝑛 2𝑑+𝑠𝑖𝑛 3𝑑
𝑑
𝑠𝑖𝑛 2𝑑+𝑠𝑖𝑛 3𝑑
𝑑
𝑠𝑖𝑛 2𝑑+𝑠𝑖𝑛 3𝑑
S.E/Paper Solutions
𝑑
𝑑𝑑 = 𝐿
𝑑𝑑 =
𝑑𝑑 =
𝑑𝑑 =
𝑠𝑖𝑛 2𝑑+𝑠𝑖𝑛 3𝑑
𝑑
∞
𝐿 𝑠𝑖𝑛2𝑑 + 𝑠𝑖𝑛3𝑑 𝑑𝑠
𝑠
∞ 2
3
+
𝑑𝑠
2
2
𝑠 𝑠 +4
𝑠 +9
𝑠
𝑠 ∞
tan−1 + tan−1
2
3 𝑠
πœ‹
πœ‹
−1 𝑠
−1 𝑠
𝑑𝑑 = − tan
2
𝑑𝑑 = πœ‹ −
+ − tan
2
2
3
−1 𝑠
−1 𝑠
tan
+ tan
2
3
23
𝑠
Crescent Academy…….………………….…..For Research in Education
∞ −𝑠𝑑 𝑠𝑖𝑛 2𝑑+𝑠𝑖𝑛 3𝑑
𝑒
𝑑𝑑
0
𝑑
𝑠 𝑠
+
2 3
𝑠 𝑠
1−2 .3
−1
= πœ‹ − tan
Put 𝑠 = 1,
∞ −𝑑 𝑠𝑖𝑛 2𝑑+𝑠𝑖𝑛 3𝑑
𝑒
𝑑𝑑
0
𝑑
= πœ‹ − tan
−1
1 1
+
2 3
1
1−6
= πœ‹ − tan−1 1
πœ‹
=πœ‹−
=
9.
𝑠𝑖𝑛𝑑
3πœ‹
4
4
Find L.T. of
𝑑
[M15/CompIT/5M]
Solution:
𝐿
𝑠𝑖𝑛𝑑
𝑑
=
∞
𝐿 𝑠𝑖𝑛𝑑
𝑠
∞ 1
𝑑𝑠
𝑠 𝑠 2 +1
tan−1 𝑠 ∞
𝑠
πœ‹
−1
=
=
= − tan
2
= cot −1 𝑠
𝑑𝑠
𝑠
1
10. Find the L.T. of 𝑒 −𝑑 𝑠𝑖𝑛𝑑
𝑑
[N17/Comp/5M]
Solution:
𝐿
𝑠𝑖𝑛𝑑
𝑑
=
∞
𝐿 𝑠𝑖𝑛𝑑
𝑠
∞ 1
𝑑𝑠
𝑠 𝑠 2 +1
tan−1 𝑠 ∞
𝑠
πœ‹
−1
𝑑𝑠
=
=
= − tan 𝑠
2
= cot −1 𝑠
𝑠𝑖𝑛𝑑
∴ 𝐿 𝑒 −𝑑
= cot −1 (𝑠 + 1)
𝑑
∞
11. Evaluate using L.T.: 0 𝑒 −2𝑑 𝑠𝑖𝑛𝑕𝑑
[M16/ElexExctElectBiomInst/5M]
Solution:
By definition of L.T.,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 =
𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑑
∞ −𝑠𝑑 𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑒
0
𝑑
S.E/Paper Solutions
𝑑𝑑 = 𝐿
𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑑
24
𝑠𝑖𝑛𝑑
𝑑
𝑑𝑑
Crescent Academy…….………………….…..For Research in Education
∞ −𝑠𝑑 𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑒
0
𝑑
∞ −𝑠𝑑 𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑒
0
𝑑
We
have,
𝑒 𝑑 −𝑒 −𝑑
𝑑𝑑 = 𝐿
1
𝑑𝑑 = 𝐿
2
𝑠𝑖𝑛𝑑
𝐿
=
𝑑
tan−1 𝑠
∞
𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
1
∴ 0 𝑒 −𝑠𝑑
𝑑𝑑 =
𝑑
∞ −𝑠𝑑 𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑒
0
𝑑
∞ −𝑠𝑑 𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑒
0
𝑑
∞ −𝑠𝑑 𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑒
0
𝑑
𝑑𝑑 =
2
𝑑
𝑑 𝑠𝑖𝑛𝑑
𝑒
−
𝑑
∞
𝐿 𝑠𝑖𝑛𝑑
𝑠
πœ‹
2 2
1
tan
2
1
𝑠𝑖𝑛𝑑
.
𝑑𝑑 = tan
𝑠𝑖𝑛𝑑
𝑑
𝑑𝑠 =
∞ 1
𝑑𝑠
𝑠 𝑠 2 +1
= [tan−1 𝑠]∞
𝑠 =
πœ‹
− tan−1 𝑠 − 1 − + tan−1 𝑠 + 1
2
−1
𝑑𝑑 = tan−1
2
1
𝑒 −𝑑
−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,
∞ −2𝑑 𝑠𝑖𝑛 𝑕𝑑 𝑠𝑖𝑛𝑑
𝑒
0
𝑑
1
𝑑𝑑 = tan−1
2
2
2
2
1
1
2
2
= tan−1
∞ 𝑒 −2𝑑 π‘π‘œπ‘  2𝑑𝑠𝑖𝑛 3𝑑
12. Evaluate 0
𝑑𝑑
𝑑
[N15/AutoMechCivil/6M]
Solution:
By definition of L.T.,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 =
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
π‘π‘œπ‘  2𝑑 𝑠𝑖𝑛 3𝑑
𝑑
π‘π‘œπ‘  2𝑑 𝑠𝑖𝑛 3𝑑
𝑑
π‘π‘œπ‘  2𝑑 𝑠𝑖𝑛 3𝑑
𝑑
π‘π‘œπ‘  2𝑑 𝑠𝑖𝑛 3𝑑
𝑑
π‘π‘œπ‘  2𝑑 𝑠𝑖𝑛 3𝑑
𝑑
π‘π‘œπ‘  2𝑑 𝑠𝑖𝑛 3𝑑
𝑑
π‘π‘œπ‘  2𝑑 𝑠𝑖𝑛 3𝑑
𝑑
π‘π‘œπ‘  2𝑑 𝑠𝑖𝑛 3𝑑
𝑑
π‘π‘œπ‘  2𝑑 𝑠𝑖𝑛 3𝑑
𝑑
π‘π‘œπ‘  2𝑑 𝑠𝑖𝑛 3𝑑
𝑑𝑑 = 𝐿
𝑑𝑑 =
𝑑𝑑 =
𝑑𝑑 =
𝑑𝑑 =
𝑑𝑑 =
𝑑𝑑 =
𝑑𝑑 =
𝑑𝑑 =
𝑑
∞ −𝑠𝑑 π‘π‘œπ‘  2𝑑 𝑠𝑖𝑛 3𝑑
𝑒
𝑑𝑑
0
𝑑
=
π‘π‘œπ‘  2𝑑 𝑠𝑖𝑛 3𝑑
𝑑
∞
𝐿 π‘π‘œπ‘ 2𝑑 𝑠𝑖𝑛3𝑑 𝑑𝑠
𝑠
1 ∞
𝐿 2π‘π‘œπ‘ 2𝑑 𝑠𝑖𝑛3𝑑 𝑑𝑠
2 𝑠
1 ∞
𝐿 𝑠𝑖𝑛5𝑑 − sin⁑
(−𝑑) 𝑑𝑠
2 𝑠
1 ∞
𝐿 𝑠𝑖𝑛5𝑑 + sin⁑
(𝑑) 𝑑𝑠
2 𝑠
1 ∞
5
1
+
𝑑𝑠
2 𝑠 𝑠 2 +25
𝑠 2 +1
∞
1
−1 𝑠
−1
tan
+ tan
2
1 πœ‹
2
1
2
1
2
𝑠
5
𝑠
πœ‹
−1 𝑠
− tan
+ − tan−1
2
5
2
−1 𝑠
πœ‹ − tan
+ tan−1 𝑠
5
𝑠
+𝑠
−1 5
πœ‹ − tan
𝑠
1−5 .𝑠
𝑠
Put 𝑠 = 2,
∞ −2𝑑 π‘π‘œπ‘  2𝑑 𝑠𝑖𝑛 3𝑑
𝑒
𝑑𝑑
0
𝑑
S.E/Paper Solutions
=
1
2
−1
πœ‹ − tan
25
2
+2
5
4
1−5
=
1
2
πœ‹ − tan−1 12
πœ‹
2
−
Crescent Academy…….………………….…..For Research in Education
π‘π‘œπ‘ π‘Žπ‘‘ −π‘π‘œπ‘ π‘π‘‘
13. Find L.T. of
𝑑
𝑒 −π‘Žπ‘‘ −π‘π‘œπ‘ π‘Žπ‘‘
14. Find L.T. of
1
𝑠 2 +𝑏 2
2
𝑠 2 +π‘Ž 2
𝑠 2 +π‘Ž 2
Ans. log
Ans. log
𝑑
sin 2 2𝑑
𝑠+π‘Ž
15. Find L.T. of
𝑑
[M16/AutoMechCivil/5M]
Solution:
𝐿
sin 2 2 𝑑
𝑑
=
=
=
=
=
∞
𝐿 sin2 2𝑑 𝑑𝑠
𝑠
∞
1−π‘π‘œπ‘  4𝑑
𝐿
𝑑𝑠
𝑠
2
1 ∞1
𝑠
−
𝑑𝑠
2
2 𝑠 𝑠
𝑠 +16
1
1
2
π‘™π‘œπ‘”π‘  − 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 −3𝑑
∞
∞
1−π‘π‘œπ‘  3𝑑
1
𝑠 2 +6𝑠+18
2
𝑠 2 +6𝑠+9
Ans. log
𝑑
𝑠𝑖𝑛𝑑 𝑠𝑖𝑛 5𝑑
17. Find L.T. of
𝑑
[M19/Extc/4M]
Solution:
𝐿
𝑠𝑖𝑛𝑑𝑠𝑖𝑛 5𝑑
𝑑
1
2𝑠𝑖𝑛𝑑𝑠𝑖𝑛 5𝑑
2
1
𝑑
cos −4𝑑 −π‘π‘œπ‘  6𝑑
2
1
𝑑
π‘π‘œπ‘  4𝑑−π‘π‘œπ‘  6𝑑
= 𝐿
= 𝐿
= 𝐿
=
=
=
=
=
2
1
2
1
2
1
𝑑
∞
𝐿 π‘π‘œπ‘ 4𝑑 − π‘π‘œπ‘ 6𝑑
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
∞ −𝑑
𝑒
0
π‘π‘œπ‘ π‘Žπ‘‘ −π‘π‘œπ‘ π‘π‘‘
𝑑
26
1
𝑏 2 +1
2
π‘Ž 2 +1
𝑑𝑑 = π‘™π‘œπ‘”
Crescent Academy…….………………….…..For Research in Education
π‘π‘œπ‘ π‘π‘‘ −π‘π‘œπ‘ π‘Žπ‘‘
19. Find L.T. of
𝑑
[N13/ElexExtcElectBiomInst/4M]
Solution:
𝐿
π‘π‘œπ‘ π‘π‘‘ −π‘π‘œπ‘ π‘Žπ‘‘
=
𝑑
=
=
=
=
∞
𝐿 π‘π‘œπ‘ π‘π‘‘ − π‘π‘œπ‘ π‘Žπ‘‘ 𝑑𝑠
𝑠
∞
𝑠
𝑠
−
𝑑𝑠
2
2
2
𝑠 𝑠 +𝑏
𝑠 +π‘Ž 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 +π‘Ž 2 𝑠
𝑠 2 +𝑏 2
0 − log
= log
+ π‘Ž2
𝑠 2 +π‘Ž 2
𝑠 2 +π‘Ž 2
𝑠 2 +𝑏 2
π‘π‘œπ‘  3𝑑−π‘π‘œπ‘  2𝑑
𝑒 −𝑑
𝑑
𝑑𝑑
[N15/ElexExctElectBiomInst][N17/IT/6M]
Solution:
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 =
π‘π‘œπ‘  3𝑑−π‘π‘œπ‘  2𝑑
𝑑
∞ −𝑠𝑑 π‘π‘œπ‘  3𝑑−π‘π‘œπ‘  2𝑑
𝑑𝑑
𝑒
0
𝑑
π‘π‘œπ‘  3𝑑−π‘π‘œπ‘  2𝑑
=𝐿
=
=
=
𝑑
∞
𝐿 π‘π‘œπ‘ 3𝑑 − π‘π‘œπ‘ 2𝑑
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,
∞ −𝑑
𝑒
0
π‘π‘œπ‘  3𝑑−π‘π‘œπ‘  2𝑑
𝑑
1
5
2
10
𝑑𝑑 = π‘™π‘œπ‘”
1
1
2
2
= log
∞ π‘π‘œπ‘  6𝑑−π‘π‘œπ‘  4𝑑
21. Evaluate using L.T.: 0
𝑑𝑑
𝑑
[M14/ElexExtcElectBiomInst/5M]
Solution:
By definition of L.T.,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 =
π‘π‘œπ‘  6𝑑−π‘π‘œπ‘  4𝑑
𝑑
∞ −𝑠𝑑 π‘π‘œπ‘  6𝑑−π‘π‘œπ‘  4𝑑
𝑒
𝑑𝑑
0
𝑑
=𝐿
=
S.E/Paper Solutions
π‘π‘œπ‘  6𝑑−π‘π‘œπ‘  4𝑑
∞
𝐿
𝑠
𝑑
π‘π‘œπ‘ 6𝑑 − π‘π‘œπ‘ 4𝑑 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,
∞ π‘π‘œπ‘  6𝑑−π‘π‘œπ‘  4𝑑
𝑑𝑑
0
𝑑
1
16
2
36
= π‘™π‘œπ‘”
= π‘™π‘œπ‘”
∞ π‘π‘œπ‘ π‘Žπ‘‘ −π‘π‘œπ‘ π‘π‘‘
22. Evaluate using L.T.: 0
[M14/AutoMechCivil/6M]
Solution:
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 =
𝑑
4
6
= log
2
3
𝑑𝑑
π‘π‘œπ‘ π‘Žπ‘‘ −π‘π‘œπ‘ π‘π‘‘
∞ −𝑠𝑑 π‘π‘œπ‘ π‘Žπ‘‘ −π‘π‘œπ‘ π‘π‘‘
𝑒
0
𝑑
𝑑
𝑑𝑑 = 𝐿
=
=
=
π‘π‘œπ‘ π‘Žπ‘‘ −π‘π‘œπ‘ π‘π‘‘
𝑑
∞
𝐿 π‘π‘œπ‘ π‘Žπ‘‘ − π‘π‘œπ‘ π‘π‘‘
𝑠
∞ 𝑠
𝑠
− 2 2 𝑑𝑠
𝑠 𝑠 2 +π‘Ž 2
𝑠 +𝑏
1
1
2
2
2
1
log 𝑠 + π‘Ž
= log
2
𝑑𝑠
2
− log 𝑠 + 𝑏
2
2
𝑠 2 +𝑏 2
∞
𝑠
𝑠 2 +π‘Ž 2
Put 𝑠 = 0,
∞ π‘π‘œπ‘ π‘Žπ‘‘ −π‘π‘œπ‘ π‘π‘‘
0
𝑑
1
𝑏2
2
π‘Ž2
𝑑𝑑 = π‘™π‘œπ‘”
= π‘™π‘œπ‘”
𝑏
π‘Ž
∞ π‘π‘œπ‘  4𝑑−π‘π‘œπ‘  3𝑑
𝑑𝑑
0
𝑑
∞ −3𝑑 π‘π‘œπ‘  7𝑑−π‘π‘œπ‘  11𝑑
𝑒
𝑑𝑑
0
𝑑
Ans. log
23. Evaluate using L.T.:
24. Evaluate
3
4
[M19/IT/6M]
Solution:
By definition of L.T.,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 =
π‘π‘œπ‘  7𝑑−π‘π‘œπ‘  11𝑑
𝑑
∞ −𝑠𝑑 π‘π‘œπ‘  7𝑑−π‘π‘œπ‘  11𝑑
𝑒
𝑑𝑑
0
𝑑
=𝐿
=
=
=
π‘π‘œπ‘  7𝑑−π‘π‘œπ‘  11𝑑
𝑑
∞
𝐿 π‘π‘œπ‘ 7𝑑 − π‘π‘œπ‘ 11𝑑
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,
∞ −3𝑑 π‘π‘œπ‘  7𝑑−π‘π‘œπ‘  11𝑑
𝑒
0
𝑑
1
130
2
58
𝑑𝑑 = log
1
65
2
29
= log
1−π‘π‘œπ‘ π‘‘
25. Find L.T. of
[N18/IT/4M]
Solution:
𝐿
1−π‘π‘œπ‘ π‘‘
𝑑
=
=
=
=
𝑑
∞
𝐿 1 − π‘π‘œπ‘ π‘‘ 𝑑𝑠
𝑠
∞1
𝑠
− 2 𝑑𝑠
𝑠 𝑠
𝑠 +1
1
π‘™π‘œπ‘”π‘  − log 𝑠 2 +
2
∞
1
𝑠2
2
1
log
𝑠 2 +1
𝑠 2 +1
= log
1
∞
𝑠
𝑠
𝑠2
2
sin 2 𝑑
∞
26. Evaluate using L.T.: 0 𝑒 −𝑑
𝑑𝑑
𝑑
[N13/AutoMechCivil/6M][N14/AutoMechCivil/6M]
[M15/ChemBiot/6M][N17/Comp/6M][M18/Elect/6M]
[N18/AutoMechCivil/6M]
Solution:
By definition of L.T.,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 =
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
sin 2 𝑑
sin 2 𝑑
𝑑
sin 2 𝑑
𝑑
sin 2 𝑑
𝑑
si n 2 𝑑
𝑑
𝑑𝑑 =
𝑑𝑑 =
𝑑𝑑 =
𝑑
∞ −𝑠𝑑 sin 2 𝑑
𝑒
𝑑𝑑
0
𝑑
∞ −𝑠𝑑 sin 2 𝑑
𝑒
𝑑𝑑
0
𝑑
=
sin 2 𝑑
∞
𝐿 sin2
𝑠
𝑑
∞
1−π‘π‘œπ‘  2𝑑
𝑑𝑠
𝐿
𝑠
2
𝑠
1 ∞1
− 2 𝑑𝑠
𝑠
2
𝑠
𝑠 +4
1
1
2
𝑑𝑑 = 𝐿
2
1
4
1
=
π‘™π‘œπ‘”π‘  − log 𝑠 + 4
2
log
= log
4
𝑠2
∞
𝑠 2 +4
𝑠 2 +4
𝑠
𝑠2
Put 𝑠 = 1,
∞ −𝑑 sin 2 𝑑
𝑒
𝑑𝑑
0
𝑑
1
= π‘™π‘œπ‘”5
4
27. Evaluate using L.T.:
[N18/Comp/4M]
S.E/Paper Solutions
𝑑 𝑑𝑠
∞ 𝑒 −𝑑 −𝑒 −2𝑑
0
𝑑
𝑑𝑑
29
∞
𝑠
Crescent Academy…….………………….…..For Research in Education
Solution:
By definition of L.T.,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 =
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
∞ −𝑠𝑑
𝑒
0
𝑒 −𝑑 −𝑒 −2𝑑
𝑑
𝑒 −𝑑 −𝑒 −2𝑑
𝑑
𝑒 −𝑑 −𝑒 −2𝑑
𝑑
𝑒 −𝑑 −𝑒 −2𝑑
𝑑
𝑒 −𝑑 −𝑒 −2𝑑
𝑑
𝑒 −𝑑 −𝑒 −2𝑑
𝑑
𝑒 −𝑑 −𝑒 −2𝑑
𝑑𝑑 = 𝐿
𝑑𝑑 =
=
𝑑
∞ 1
𝑠 𝑠+1
−
1
𝑠+2
∞
𝑠
𝐿 𝑒 −𝑑 − 𝑒 −2𝑑 𝑑𝑠
𝑑𝑠
𝑑𝑑 = log 𝑠 + 1 − log 𝑠 + 2
𝑑𝑑 = log
𝑑𝑑 = log
𝑠+1
∞
𝑠+2
𝑠+2
𝑠
∞
𝑠
𝑠+1
Put 𝑠 = 0,
∞ 𝑒 −𝑑 −𝑒 −2𝑑
0
𝑑
𝑑𝑑 = π‘™π‘œπ‘”2
∞ 𝑒 −𝑑 −π‘π‘œπ‘ π‘‘
28. Evaluate 0
𝑑 𝑒 4𝑑
[M19/Comp/6M]
Solution:
∞
𝑑𝑑
𝑒 −𝑑 −π‘π‘œπ‘ π‘‘
We have, 0 𝑒 −4𝑑 .
𝑑
By definition of L.T.,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 =
𝑑𝑑
𝑒 −𝑑 −π‘π‘œπ‘ π‘‘
𝑑
∞ −𝑠𝑑 𝑒 −𝑑 −π‘π‘œπ‘ π‘‘
𝑒
0
𝑑
𝑑𝑑 = 𝐿
=
=
=
𝑒 −𝑑 −π‘π‘œπ‘ π‘‘
𝑑
∞
−𝑑
𝐿
𝑒
− π‘π‘œπ‘ π‘‘ 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,
∞ −4𝑑 𝑒 −𝑑 −π‘π‘œπ‘ π‘‘
𝑒 .
0
𝑑
S.E/Paper Solutions
𝑑𝑑 = 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
𝑠𝑖𝑛𝑑
𝑑𝑑
𝑑
[N13/ElexExtcElectBiomInst/4M]
Solution:
𝑠𝑖𝑛𝑑
Let 𝑓 𝑑 =
𝑑
𝑠𝑖𝑛𝑑
∴ 𝑓 0 = lim𝑑→0
And,
𝐿𝑓 𝑑
=𝐿
∴𝐹 𝑠 =
=
=1
𝑑
𝑠𝑖𝑛𝑑
𝑑
∞
𝐿
𝑠𝑖𝑛𝑑
𝑠
∞ 1
𝑑𝑠
𝑠 𝑠 2 +1
tan−1 𝑠 ∞
𝑠
πœ‹
−1
=
= − tan
2
= cot −1 𝑠
𝑑𝑠
𝑠
Now,
𝐿
2.
𝑑
𝑠𝑖𝑛𝑑
𝑑𝑑
𝑑
Find L.T. of
𝑑
=𝐿
𝑓 𝑑
𝑑𝑑
= 𝐿 𝑓′ 𝑑
=𝑠𝐹 𝑠 −𝑓 0
= 𝑠 . cot −1 𝑠 – 𝑓 0
= 𝑠 cot −1 𝑠 − 1
𝑑
1−π‘π‘œπ‘  2𝑑
𝑑𝑑
𝑑
[N14/ElexExtcElectBiomInst/4M]
Solution:
1−π‘π‘œπ‘  2𝑑
Let 𝑓 𝑑 =
𝑑
∴ 𝑓 0 = lim𝑑→0
= lim𝑑→0
= 2𝑠𝑖𝑛0
∴𝑓 0 =0
And,
𝐿𝑓 𝑑
=𝐿
∴𝐹 𝑠 =
=
1−π‘π‘œπ‘  2𝑑 0
𝑑
2𝑠𝑖𝑛 2𝑑
0
1
1−π‘π‘œπ‘  2𝑑
∞
𝐿
𝑠
∞1
𝑠 𝑠
𝑑
1 − π‘π‘œπ‘ 2𝑑 𝑑𝑠
−
𝑠
𝑠 2 +4
1
𝑑𝑠
= π‘™π‘œπ‘”π‘  − 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−π‘π‘œπ‘  2𝑑
𝑑𝑑
𝑑
=𝐿
𝑑
𝑓 𝑑
=𝐿 𝑓 𝑑
=𝑠𝐹 𝑠 −𝑓 0
𝑑𝑑
′
𝑠
= log
𝑠 2 +4
2
= 𝑠 log
𝑑
𝑠2
𝑠 2 +4
–𝑓 0
𝑠
𝑠𝑖𝑛 3𝑑𝑠𝑖𝑛𝑑
Find L.T. of
4.
Find L.T. of
𝑑𝑑
𝑑
[M14/ElexExtcElectBiomInst/4M][N17/IT/5M]
Solution:
𝑑𝑑
𝑑
𝑑 𝑠𝑖𝑛 3𝑑
𝑠𝑖𝑛 3𝑑
Let 𝑓 𝑑 =
𝑑
∴ 𝑓 0 = lim𝑑→0
= lim𝑑→0
= 3π‘π‘œπ‘ 0
∴𝑓 0 =3
And,
𝐿𝑓 𝑑
=𝐿
𝑠𝑖𝑛 3𝑑 0
𝑑
0
3π‘π‘œπ‘  3𝑑
1
𝑠𝑖𝑛 3𝑑
𝑑
∞
𝐿 𝑠𝑖𝑛3𝑑
𝑠
∞ 3
𝑑𝑠
𝑠 𝑠 2 +9
∞
−1 𝑠
∴𝐹 𝑠 =
=
𝑑𝑠
= tan
=
3 𝑠
𝑠
− tan−1
2
3
−1 𝑠
πœ‹
= cot
3
Now,
𝐿
𝑑
𝑠𝑖𝑛 3𝑑
𝑑𝑑
𝑑
=𝐿
𝑑
𝑑𝑑
′
𝑓 𝑑
=𝐿 𝑓 𝑑
=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
π‘ π‘π‘œπ‘‘
3
−3
32
Crescent Academy…….………………….…..For Research in Education
5.
𝑑
𝑠𝑖𝑛 4𝑑
Find L.T. of
𝑑𝑑
𝑑
[N17/Elect/5M]
Solution:
𝑠𝑖𝑛 4𝑑
Let 𝑓 𝑑 =
𝑑
∴ 𝑓 0 = lim𝑑→0
= lim𝑑→0
= 4π‘π‘œπ‘ 0
∴𝑓 0 =4
And,
𝐿𝑓 𝑑
=𝐿
𝑠𝑖𝑛 4𝑑 0
𝑑
0
4π‘π‘œπ‘  4𝑑
1
𝑠𝑖𝑛 4𝑑
𝑑
∞
𝐿 𝑠𝑖𝑛4𝑑
𝑠
∞ 4
𝑑𝑠
𝑠 𝑠 2 +16
∞
−1 𝑠
∴𝐹 𝑠 =
=
𝑑𝑠
= tan
=
4 𝑠
𝑠
− tan−1
2
4
−1 𝑠
πœ‹
= cot
4
Now,
𝐿
𝑑
𝑠𝑖𝑛 4𝑑
𝑑𝑑
𝑑
𝑑
=𝐿
𝑑𝑑
′
𝑓 𝑑
=𝐿 𝑓 𝑑
=𝑠𝐹 𝑠 −𝑓 0
𝑠
= 𝑠 . cot −1 – 𝑓 0
=
6.
Find L.T. of
4
−1 𝑠
π‘ π‘π‘œπ‘‘
4
𝑑
sin 2 𝑑
𝑑𝑑
𝑑
−4
[M19/IT/5M]
Solution:
Let 𝑓 𝑑 =
sin 2 𝑑
=
𝑑
∴ 𝑓 0 = lim𝑑→0
= lim𝑑→0
= 𝑠𝑖𝑛0
∴𝑓 0 =0
And,
𝐿𝑓 𝑑
=𝐿
S.E/Paper Solutions
1−π‘π‘œπ‘  2𝑑
2𝑑
1−π‘π‘œπ‘  2𝑑 0
2𝑑
2𝑠𝑖𝑛 2𝑑
0
2
1−π‘π‘œπ‘  2𝑑
2𝑑
33
Crescent Academy…….………………….…..For Research in Education
∴𝐹 𝑠 =
=
=
=
∞
𝐿
2 𝑠
1 ∞1
2 𝑠 𝑠
1
1
2
1
4
1
1 − π‘π‘œπ‘ 2𝑑 𝑑𝑠
𝑠
−
𝑠 2 +4
𝑑𝑠
1
π‘™π‘œπ‘”π‘  − log 𝑠 2 + 4
2
log
= log
𝑠2
∞
𝑠 2 +4
𝑠 2 +4
𝑠
∞
𝑠
𝑠2
4
Now,
𝐿
𝑑
sin 2 𝑑
𝑑𝑑
𝑑
=𝐿
𝑑
𝑑𝑑
𝑓 𝑑
= 𝐿 𝑓′ 𝑑
=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
𝑑
π‘’π‘π‘œπ‘ π‘•π‘’π‘‘π‘’
0
𝑑 −2𝑑
𝑒
cos 2 𝑑 𝑑𝑑
0
𝑑
𝑒𝑒 −2𝑒 𝑠𝑖𝑛3𝑒𝑑𝑒
0
𝑑 𝑑 𝑑
𝑑 𝑠𝑖𝑛𝑑 𝑑𝑑 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
𝑑 𝑑 𝑑
0 0 0
𝑑𝑠𝑖𝑛𝑑 𝑑𝑑
3
=
=
1
𝑠3
1
𝑠3
=−
=−
=
5.
Ans.
Ans.
𝐿 𝑑𝑠𝑖𝑛𝑑
𝑑
. −1
𝑑𝑠
1
1 𝑑
𝐿 𝑠𝑖𝑛𝑑
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 𝑑. 𝑒 −3𝑑 . 𝑠𝑖𝑛𝑑
𝑠 +1
S.E/Paper Solutions
34
2
1
𝑠+2
+ .
2 𝑠
2𝑠 𝑠+2
3 2𝑠+4
1
𝑠
𝑑
𝑑
𝑠 𝑠 2 −1
1
Ans. .
Find L.T. of 0 𝑒. 𝑒 −3𝑒 . 𝑠𝑖𝑛 𝑒 𝑑𝑒
[N18/AutoMechCivil/5M]
Solution:
𝐿 0 𝑒. 𝑒 −3𝑒 . 𝑠𝑖𝑛𝑒 𝑑𝑒 =
We have,
1
𝐿 𝑠𝑖𝑛𝑑 = 2
𝑠 2 +1
𝑠 2 +4𝑠+13
2
𝑠 2 +4𝑠+8
Crescent Academy…….………………….…..For Research in Education
𝑑
𝐿 𝑑𝑠𝑖𝑛𝑑 = −1
= −1
1
𝑑𝑠 𝑠 2 +1
𝑠 2 +1 [0]−1[2𝑠]
= −1
∴ 𝐿{𝑑𝑠𝑖𝑛𝑑} =
𝐿 𝑠𝑖𝑛𝑑
𝑑𝑠
𝑑
𝑠 2 +1
2𝑠
𝑠 2 +1
2
2 𝑠+3
∴ 𝐿 𝑒 −3𝑑 𝑑 𝑠𝑖𝑛𝑑 =
6.
𝑑𝑒 =
2 𝑠+3
=
𝑠 2 +6𝑠+10 2
2 𝑠+3
𝑠
𝑠 2 +6𝑠+10 2
𝑑
Find L.T. of 0 𝑒. 𝑒 −3𝑒 . cos 2 2 𝑒 𝑑𝑒
[M16/ElexExctElectBiomInst/4M]
Solution:
𝐿 cos 2 2𝑑 = 𝐿
1+π‘π‘œπ‘  4𝑑
2
2
𝐿 𝑑 cos 2𝑑 = −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
𝑑
Find L.T. of 0 𝑒. 𝑒 −3𝑒 . 𝑠𝑖𝑛 4𝑒 𝑑𝑒
[N13/ElexExtcElectBiomInst/5M]
Solution:
𝑑
𝐿 0 𝑒. 𝑒 −3𝑒 . 𝑠𝑖𝑛4𝑒 𝑑𝑒 =
We have,
4
𝐿 𝑠𝑖𝑛4𝑑 = 2
𝐿 𝑑𝑠𝑖𝑛4𝑑 = −1
= −1
= −4
S.E/Paper Solutions
𝑑
𝑑𝑠
𝑑
𝑠
𝑠 2 +16
𝑠
𝑠 2 +16
1
𝑠+3
𝑑
𝑒. 𝑒 −3𝑒 . cos 2 2 𝑒 𝑑𝑒
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 2𝑑
−
2
𝐿 1 + π‘π‘œπ‘ 4𝑑 =
2
−
=
𝐿
𝑑
𝑑𝑠 2 𝑠
1
1
𝐿 𝑒 −3𝑑 𝑑 cos 2 2𝑑 =
1
=
𝑑𝑠
𝑑 1 1
= −1
7.
2 +1 2
𝑠+3
𝑑
𝑒. 𝑒 −3𝑒 . 𝑠𝑖𝑛𝑒
0
∴𝐿
2
1
𝑠
𝐿 𝑑. 𝑒 −3𝑑 . 𝑠𝑖𝑛4𝑑
𝐿 𝑠𝑖𝑛4𝑑
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{𝑑𝑠𝑖𝑛4𝑑} =
𝑠 2 +16
∴ 𝐿 𝑒 −3𝑑 𝑑 𝑠𝑖𝑛4𝑑 =
∴𝐿
8.
9.
10.
𝑠 2 +16
8𝑠
2
2
8 𝑠+3
𝑠+3
𝑑
−3𝑒
𝑒. 𝑒
. 𝑠𝑖𝑛4𝑒 𝑑𝑒
0
2 +16 2
=
=
8 𝑠+3
𝑠 2 +6𝑠+25 2
8 𝑠+3
𝑠 𝑠 2 +6𝑠+25
𝑑 −1 −𝑒
𝑒 𝑒 𝑠𝑖𝑛𝑒 𝑑𝑒
0
𝑑 1−𝑒 −π‘Žπ‘’
Find L.T. of 0
𝑑𝑒
𝑒
∞
𝑑 𝑠𝑖𝑛𝑒
Evaluate using L.T.: 0 𝑒 −𝑑 0
𝑒
2
1
Ans. cot −1 𝑠 + 1
Find L.T. of
𝑠
1
𝑠+π‘Ž
𝑠
𝑠
Ans. log
𝑑𝑒 𝑑𝑑
[N14/ChemBiot/6M][N15/ChemBiot/6M]
Solution:
By definition of L.T.,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
𝑑 𝑠𝑖𝑛𝑒
𝑑𝑒
0 𝑒
∞ −𝑠𝑑 𝑑 𝑠𝑖𝑛𝑒
𝑒
𝑑𝑒 𝑑𝑑
0 𝑒
0
Put 𝑓 𝑑 =
=𝐿
=
=
=
=
=
=
1
𝐿
𝑑 𝑠𝑖𝑛𝑒
0 𝑒
𝑠𝑖𝑛𝑑
𝑑𝑒
𝑑
∞
𝐿 𝑠𝑖𝑛𝑑 𝑑𝑠
𝑠 𝑠
1 ∞ 1
𝑑𝑠
𝑠 𝑠 𝑠 2 +1
1
tan−1 𝑠 ∞
𝑠
𝑠
1 πœ‹
− tan−1 𝑠
𝑠 2
1
cot −1 𝑠
𝑠
𝑠
1
Put 𝑠 = 1,
∞ −𝑑 𝑑 𝑠𝑖𝑛𝑒
πœ‹
1
𝑒
𝑑𝑒 𝑑𝑑 = cot −1 1 =
0
0
𝑒
11. Find L.T. of
12. Find L.T. of
13. Find L.T. of
1
4
𝑑 𝑠𝑖𝑛𝑒
𝑑𝑒
0 𝑒
𝑑 𝑒 𝑠𝑖𝑛 4𝑒
𝑒
𝑑𝑒
0
𝑒
𝑑
𝑒 −2𝑑 0 π‘‘π‘π‘œπ‘ 3𝑑𝑑𝑑
𝑑
π‘π‘œπ‘ π‘•π‘‘ 0 𝑒 π‘₯ 𝑠𝑖𝑛𝑕π‘₯𝑑π‘₯
1
Ans. cot −1 𝑠
𝑠
1
Ans. cot −1
𝑠
Ans.
14. Find L.T. of
[N15/ElexExtcElectBiomInst/4M]
Solution:
1
𝐿 𝑠𝑖𝑛𝑕𝑑 = 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 𝑑 𝑠𝑖𝑛𝑕𝑑 =
𝐿
𝐿
1
2 −1
=
1
𝑠−1
𝑑 π‘₯
1
1
𝑒 𝑠𝑖𝑛𝑕π‘₯𝑑π‘₯ = .
0
𝑠 𝑠 𝑠−2
𝑑 π‘₯
π‘π‘œπ‘ π‘•π‘‘ 0 𝑒 𝑠𝑖𝑛𝑕π‘₯𝑑π‘₯ = 𝐿
1
=
2
1
=
=
=
1
𝑠 𝑠−2
1
𝑠 2 𝑠−2
𝑑
𝑒 +𝑒 −𝑑
𝐿
= 𝐿
∞
=
𝑠 2 −2𝑠
2
1
𝑑 π‘₯
𝑒 𝑠𝑖𝑛𝑕π‘₯𝑑π‘₯
0
2
𝑑
𝑒 𝑑 + 𝑒 −𝑑 0 𝑒 π‘₯ 𝑠𝑖𝑛𝑕π‘₯𝑑π‘₯
𝑑
𝑑
𝑒 𝑑 0 𝑒 π‘₯ 𝑠𝑖𝑛𝑕π‘₯𝑑π‘₯ + 𝑒 −𝑑 0 𝑒 π‘₯ 𝑠𝑖𝑛𝑕π‘₯𝑑π‘₯
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
𝑑
15. Evaluate using L.T.: 0 𝑒 −𝑑 0 𝑒4 π‘ π‘–π‘›π‘•π‘’π‘π‘œπ‘ π‘•π‘’π‘‘π‘’ 𝑑𝑑
[N18/IT/4M][N18/Elect/6M]
Solution:
By definition of L.T.,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
𝑑 4
𝑒 π‘ π‘–π‘›π‘•π‘’π‘π‘œπ‘ π‘•π‘’π‘‘π‘’
0
∞ −𝑠𝑑 𝑑 4
𝑒
𝑒 π‘ π‘–π‘›π‘•π‘’π‘π‘œπ‘ π‘•π‘’π‘‘π‘’ 𝑑𝑑
0
0
Put 𝑓 𝑑 =
=𝐿
=
1
𝑠
1
𝑑 4
𝑒 π‘ π‘–π‘›π‘•π‘’π‘π‘œπ‘ π‘•π‘’π‘‘π‘’
0
4
𝐿 𝑑 𝑠𝑖𝑛𝑕𝑑 π‘π‘œπ‘ π‘•π‘‘
= 𝐿 𝑑4
=
=
=
𝑠
1
4𝑠
1
4𝑠
1
4𝑠
𝑒 𝑑 −𝑒 −𝑑
4
𝐿 𝑑 𝑒
2
2𝑑
𝑒 𝑑 +𝑒 −𝑑
2
−𝑒
−2𝑑
𝐿 𝑒 2𝑑 𝑑 4 − 𝑒 −2𝑑 𝑑 4
4!
𝑠−2
5
−
4!
𝑠+2
5
Put 𝑠 = 1,
∞ −𝑑
𝑒
0
𝑑
0
𝑒4 π‘ π‘–π‘›π‘•π‘’π‘π‘œπ‘ π‘•π‘’π‘‘π‘’ 𝑑𝑑 =
𝑑 𝑠𝑖𝑛 𝑒
16. Find L.T. of 𝑒 −𝑑 0
𝑑𝑒
𝑒
[M15/ElexExtcElectBiomInst/4M]
Solution:
We have,
1
𝐿 𝑠𝑖𝑛𝑑 = 2
𝐿
𝑠𝑖𝑛𝑑
𝑑
=
=
=
𝑠 +1
∞
𝐿 𝑠𝑖𝑛𝑑 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œ‹
=
𝐿
𝐿
2
𝑑 𝑠𝑖𝑛𝑒
𝑑𝑒
0 𝑒
𝑑 𝑠𝑖𝑛𝑒
𝑒 −𝑑 0
𝑒
− tan−1 𝑠 = cot −1 𝑠
=
1
𝑠𝑖𝑛𝑑
𝐿
𝑠
𝑑𝑒 =
1
1
= cot −1 𝑠
𝑑
𝑠+1
cot
𝑠
−1
(𝑠 + 1)
𝑑
17. Find L.T. of 𝑒 −3𝑑 0 𝑒𝑠𝑖𝑛3𝑒𝑑𝑒
[N14/ElexExtcElectBiomInst/4M]
Solution:
We have,
3
𝐿 𝑠𝑖𝑛3𝑑 = 2
𝑠 +9
𝐿 𝑑𝑠𝑖𝑛3𝑑 = −1
= −1
= −3
= −3
∴ 𝐿{𝑑𝑠𝑖𝑛3𝑑} =
∴𝐿
∴𝐿
𝑑
1
𝑑𝑠 𝑠 2 +9
𝑠 2 +9 [0]−1[2𝑠]
𝑠 2 +9
6𝑠
𝑠 2 +9
𝐿 𝑑𝑠𝑖𝑛4𝑑 = −1
𝑑
𝑑
0
= −1
= −4
= −4
∴ 𝐿{𝑑𝑠𝑖𝑛4𝑑} =
𝑑
𝑑𝑠
𝑑
2
2
𝑑
1
𝑒𝑠𝑖𝑛3𝑒𝑑𝑒 =
0
𝑠
−3𝑑 𝑑
𝑒
𝑒𝑠𝑖𝑛3𝑒𝑑𝑒
0
𝑠 +16
∴𝐿
3
𝑠 2 +9
𝑑𝑠
𝑑
18. Find the L.T. of 𝑒 −3𝑑
[N18/Comp/4M]
Solution:
We have,
4
𝐿 𝑠𝑖𝑛4𝑑 = 2
∴𝐿
𝐿 𝑠𝑖𝑛3𝑑
𝑑𝑠
𝑑
𝐿 𝑑𝑠𝑖𝑛3𝑑 =
=
6
𝑠+3
2 +9 2
1
𝑠
.
=
6𝑠
𝑠 2 +9 2
=
6
𝑠 2 +9 2
6
𝑠 2 +6𝑠+18 2
𝑠𝑖𝑛4𝑑 𝑑𝑑
𝐿 𝑠𝑖𝑛4𝑑
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
𝑑
1
1
8𝑠
8
𝑑𝑠𝑖𝑛4𝑑𝑑𝑑
=
𝐿
𝑑𝑠𝑖𝑛4𝑑
=
.
=
2
2
2
0
𝑠
𝑠
𝑠 +16
𝑠 +16
𝑑
8
8
𝑒 −3𝑑 0 𝑑 𝑠𝑖𝑛4𝑑 𝑑𝑑 =
= 2
𝑠+3 2 +16 2
𝑠 +6𝑠+25 2
S.E/Paper Solutions
38
2
Crescent Academy…….………………….…..For Research in Education
𝑑
𝑒
0
19. Find the L.T. of 𝑒 −𝑑
[M17/CompIT/5M]
Solution:
We have,
𝑠
𝐿 π‘π‘œπ‘ 2𝑑 = 2
𝑠 +4
𝐿 π‘‘π‘π‘œπ‘ 2𝑑 = −1
= −1
∴𝐿
𝑑
𝑒
0
∴ 𝐿 𝑒 −𝑑
𝐿 π‘π‘œπ‘ 2𝑑
𝑑𝑠
𝑑
𝑠
𝑑𝑠 𝑠 2 +4
𝑠 2 +4 1 −𝑠 2𝑠
= −1
∴ 𝐿{π‘‘π‘π‘œπ‘ 2𝑑} =
𝑑
𝑠 2 +4
𝑠2 − 4
𝑠 2 +4
2
2
π‘π‘œπ‘ 2𝑒𝑑𝑒 =
𝑑
𝑒
0
π‘π‘œπ‘ 2𝑒 𝑑𝑒
1
1
𝐿 π‘‘π‘π‘œπ‘ 2𝑑 =
𝑠
π‘π‘œπ‘ 2𝑒𝑑𝑒 =
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
𝑑
2
20. Find L.T. of 𝑒 −4𝑑 0 𝑒 𝑠𝑖𝑛3𝑒𝑑𝑒
[M15/AutoMechCivil/6M][N16/AutoMechCivil/4M][N17/Elex/5M]
[M18/Elex/6M]
Solution:
We have,
3
𝐿 𝑠𝑖𝑛3𝑑 = 2
𝑠 +9
𝑑
𝐿 𝑑𝑠𝑖𝑛3𝑑 = −1
= −1
𝑑𝑠 𝑠 2 +9
𝑠 2 +9 [0]−1[2𝑠]
= −3
∴𝐿
∴𝐿
3
𝑑𝑠 𝑠 2 +9
𝑑
1
= −3
∴ 𝐿{𝑑𝑠𝑖𝑛3𝑑} =
𝐿 𝑠𝑖𝑛3𝑑
𝑑𝑠
𝑑
𝑠 2 +9
6𝑠
𝑠 2 +9
2
𝑑
1
𝑒𝑠𝑖𝑛3𝑒𝑑𝑒
=
0
𝑠
𝑑
𝑒 −4𝑑 0 𝑒𝑠𝑖𝑛3𝑒𝑑𝑒
21. Find L.T. of 𝑒 −2𝑑
[N18/Extc/6M]
Solution:
We have,
𝑠
𝐿 π‘π‘œπ‘ 4𝑑 = 2
𝑑
𝑒
0
2
𝐿 𝑑𝑠𝑖𝑛3𝑑 =
=
6
𝑠+4
2 +9 2
𝑒 3𝑒 π‘π‘œπ‘ 4𝑒 𝑑𝑒
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 π‘‘π‘π‘œπ‘ 4𝑑 = −1
= −1
𝑑
𝑒
0
𝑠 2 +16
𝑠 2 − 16
𝑠−3 2 +16
1
𝑒 3𝑒 π‘π‘œπ‘ 4𝑒 𝑑𝑒 =
𝑑
𝑒
0
∴ 𝐿 𝑒 −2𝑑
2
𝑠 2 +16 2
𝑠−3 2 −16
∴ 𝐿 𝑑𝑒 3𝑑 π‘π‘œπ‘ 4𝑑 =
∴𝐿
𝑠
𝑑𝑠 𝑠 2 +16
𝑠 2 +16 1 −𝑠 2𝑠
= −1
∴ 𝐿{π‘‘π‘π‘œπ‘ 4𝑑} =
𝐿 π‘π‘œπ‘ 4𝑑
𝑑𝑠
𝑑
2
𝑠 2 −6𝑠−7
=
𝑠 2 −6𝑠+25
3𝑑
2
𝐿 𝑑𝑒 π‘π‘œπ‘ 4𝑑 =
𝑠
1
𝑒 3𝑒 π‘π‘œπ‘ 4𝑒 𝑑𝑒 =
=
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
𝑑
∞
22. Evaluate 0 𝑒 −4𝑑 0 𝑒𝑠𝑖𝑛𝑕𝑒 2 π‘π‘œπ‘ π‘•5𝑒 𝑒 3𝑒 𝑑𝑒 𝑑𝑑
[N17/Elect/8M]
Solution:
By definition of L.T.,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
𝑑 𝑠𝑖𝑛𝑒
0 𝑒
Put 𝑓 𝑑 =
∞ −𝑠𝑑 𝑑
𝑒
0
0
=
1
𝑠
1
= .
𝑑
0
𝑒𝑠𝑖𝑛𝑕2 𝑒 2 π‘π‘œπ‘ π‘•5𝑒. 𝑒 3𝑒 𝑑𝑒 𝑑𝑑 = 𝐿
𝑒𝑠𝑖𝑛𝑕2 𝑒 2 π‘π‘œπ‘ π‘•5𝑒. 𝑒 3𝑒 𝑑𝑒
𝐿 𝑑 2 sinh4 𝑑 π‘π‘œπ‘ π‘•5𝑑 𝑒 3𝑑
= 𝐿 𝑑2
𝑠
1
𝑑𝑒
1
𝑠 32
𝑒 𝑑 −𝑒 −𝑑
4
𝑒 5𝑑 +𝑒 −5𝑑
2
𝑒 3𝑑
2
𝐿 𝑑 2 𝑒 4𝑑 − 4𝑒 2𝑑 + 6 − 4𝑒 −2𝑑 + 𝑒 −4𝑑 𝑒 8𝑑 + 𝑒 −2𝑑
1
= 32𝑠 𝐿 𝑑 2 𝑒12𝑑 + 𝑒 2𝑑 − 4𝑒10𝑑 − 4 + 6𝑒 8𝑑 + 6𝑒 −2𝑑 − 4𝑒 6𝑑 − 4𝑒 −4𝑑 + 𝑒 4𝑑 + 𝑒 −6𝑑
=
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,
∞ −𝑠𝑑 𝑑
𝑒
0
0
𝑒𝑠𝑖𝑛𝑕2 𝑒 2 π‘π‘œπ‘ π‘•5𝑒. 𝑒 3𝑒 𝑑𝑒 𝑑𝑑 = ∞
23. Find Laplace Transform of 𝑓 𝑑 = 𝑑
[M19/AutoMechCivil/5M]
Solution:
We have,
4
𝐿 𝑠𝑖𝑛4𝑑 = 2
𝑠 +16
S.E/Paper Solutions
40
𝑑 −2𝑒
𝑒
𝑠𝑖𝑛4𝑒
0
𝑑𝑒
+
2
𝑠−4 3
+
2
𝑠+6 3
Crescent Academy…….………………….…..For Research in Education
4
𝐿 𝑒 −2𝑑 𝑠𝑖𝑛4𝑑 =
𝐿
𝐿
2 +16
=
4
𝑠 2 +4𝑠+20
𝑠+2
𝑑 −2𝑒
1
4
𝑒
𝑠𝑖𝑛4𝑒 𝑑𝑒 = . 2
0
𝑠 𝑠 +4𝑠+20
𝑑 −2𝑒
𝑑
4
𝑑 0𝑒
𝑠𝑖𝑛4𝑒 𝑑𝑒 = −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 𝑑 2 𝐻 𝑑 − 3
Find the Laplace transform of𝑠𝑖𝑛𝑑. 𝐻 𝑑 −
πœ‹
2
−𝐻 𝑑−
πœ‹
𝑠 +1
1
𝑠 2 +1
+ 𝑒 −πœ‹π‘ 
−𝑠
𝑠 2 +1
+
6
𝑠2
+
2
𝑠
1
𝑠 2 +1
Find the L.T. of 𝑒 −𝑑 π‘π‘œπ‘ π‘‘ 𝐻 𝑑 − πœ‹
[N14/ChemBiot/6M][M17/ElexExtcElectBiomInst/4M]
Solution:
𝐿 𝑒 −𝑑 π‘π‘œπ‘ π‘‘ 𝐻 𝑑 − πœ‹ = 𝑒 −πœ‹π‘  𝐿 𝑒 − 𝑑+πœ‹ cos 𝑑 + πœ‹
= 𝑒 −πœ‹π‘  𝐿 𝑒 −𝑑 . 𝑒 −πœ‹ π‘π‘œπ‘ π‘‘ π‘π‘œπ‘ πœ‹ − 𝑠𝑖𝑛𝑑 π‘ π‘–π‘›πœ‹
= 𝑒 −πœ‹π‘  . 𝑒 −πœ‹ 𝐿 𝑒 −𝑑 – π‘π‘œπ‘ π‘‘
= −𝑒 −πœ‹π‘  . 𝑒 −πœ‹ 𝐿 𝑒 −𝑑 π‘π‘œπ‘ π‘‘
𝑠+1
= −𝑒 −πœ‹π‘  . 𝑒 −πœ‹
2
= −𝑒
5.
+
9
𝑠
πœ‹
1
Ans. 𝑒 . 2 − 𝑒 −3 2 𝑠 .
𝑠 +1
𝑠
Find the L.T. of 𝑠𝑖𝑛𝑑 𝐻 𝑑 + π‘π‘œπ‘ π‘‘ − 𝑠𝑖𝑛𝑑 𝐻 𝑑 − πœ‹
[M19/IT/4M]
Solution:
𝐿 𝑠𝑖𝑛𝑑 𝐻 𝑑 + π‘π‘œπ‘ π‘‘ − 𝑠𝑖𝑛𝑑 𝐻 𝑑 − πœ‹
= 𝑒 −0𝑠 𝐿 sin 𝑑 + 0 + 𝑒 −πœ‹π‘  𝐿 cos 𝑑 + πœ‹ − sin 𝑑 + πœ‹
= 𝐿 𝑠𝑖𝑛𝑑 + 𝑒 −πœ‹π‘  𝐿{π‘π‘œπ‘ π‘‘ π‘π‘œπ‘ πœ‹ − 𝑠𝑖𝑛𝑑 π‘ π‘–π‘›πœ‹ − 𝑠𝑖𝑛𝑑 π‘π‘œπ‘ πœ‹ − π‘π‘œπ‘ π‘‘ π‘ π‘–π‘›πœ‹}
1
= 2 + 𝑒 −πœ‹π‘  𝐿 −π‘π‘œπ‘ π‘‘ + 𝑠𝑖𝑛𝑑
=
4.
𝑠3
3πœ‹
−2 𝑠
3.
2
Ans. 𝑒 −3𝑠
−πœ‹π‘ 
.𝑒
−πœ‹
.
𝑠+1 +1
𝑠+1
𝑠 2 +2𝑠+2
Find the L.T. of 𝑒 −𝑑 𝑠𝑖𝑛𝑑 𝐻 𝑑 − πœ‹
[M19/Comp/6M]
Solution:
𝐿 𝑒 −𝑑 𝑠𝑖𝑛𝑑 𝐻 𝑑 − πœ‹ = 𝑒 −πœ‹π‘  𝐿 𝑒 − 𝑑+πœ‹ sin 𝑑 + πœ‹
= 𝑒 −πœ‹π‘  𝐿 𝑒 −𝑑 . 𝑒 −πœ‹ 𝑠𝑖𝑛𝑑 π‘π‘œπ‘ πœ‹ + π‘π‘œπ‘ π‘‘ π‘ π‘–π‘›πœ‹
S.E/Paper Solutions
41
Crescent Academy…….………………….…..For Research in Education
= 𝑒 −πœ‹π‘  . 𝑒 −πœ‹ 𝐿 𝑒 −𝑑 – 𝑠𝑖𝑛𝑑
= −𝑒 −πœ‹π‘  . 𝑒 −πœ‹ 𝐿 𝑒 −𝑑 𝑠𝑖𝑛𝑑
1
= −𝑒 −πœ‹π‘  . 𝑒 −πœ‹
2
= −𝑒 −πœ‹π‘  . 𝑒 −πœ‹ .
6.
𝑠+1 +1
1
𝑠 2 +2𝑠+2
Find L.T. of 𝑑𝑒 −2𝑑 𝐻 𝑑 − 1
[N17/IT/4M]
Solution:
𝐿 𝑑𝑒 −2𝑑 𝐻 𝑑 − 1
= 𝑒 −𝑠 𝐿 𝑒 −2 𝑑+1 𝑑 + 1
= 𝑒 −𝑠 𝐿 𝑒 −2𝑑 . 𝑒 −2 𝑑 + 1
= 𝑒 −𝑠 . 𝑒 −2 𝐿 𝑒 −2𝑑 𝑑 + 𝑒 −2𝑑
1
1
= 𝑒 −𝑠−2
+
2
𝑠+2
𝑠+2
7.
Find the Laplace transform of: 𝑑. 𝐻 𝑑 − 4 + 𝑑 2 𝛿 𝑑 − 4
8.
Ans. 2 1 + 4𝑠 + 16𝑠 2
𝑠
Find the Laplace transform of: 𝑑 2 𝐻 𝑑 − 2 − π‘π‘œπ‘ π‘•π‘‘ 𝛿 𝑑 − 4
𝑒 −4𝑠
2
Ans. 𝑒 −2𝑠
9.
𝑠3
+
4
𝑠2
4
+
𝑠
− 𝑒 −4𝑠 π‘π‘œπ‘ π‘•4
Find the Laplace transform of: 1 + 2𝑑 − 3𝑑 2 + 4𝑑 3 𝐻 𝑑 − 2
[M15/ChemBiot/6M]
Solution:
𝐿 1 + 2t − 3t 2 + 4t 3 H t − 2
= 𝑒 −2𝑠 𝐿 1 + 2 𝑑 + 2 − 3 𝑑 + 2 2 + 4 𝑑 + 2 3
= 𝑒 −2𝑠 𝐿 1 + 2𝑑 + 4 − 3𝑑 2 − 12𝑑 − 12 + 4𝑑 3 + 24𝑑 2 + 48𝑑 + 32
= 𝑒 −2𝑠 𝐿 4𝑑 3 + 21𝑑 2 + 38𝑑 + 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 − 2𝑑 + 3𝑑 2 − 𝑑 3 𝐻 𝑑 − 3
Ans. 𝑒 −3𝑠 −
2
3
6
𝑠4
−
12
𝑠3
−
11
𝑠2
−
4
𝑠
11. Find the Laplace transform of: 1 + 3𝑑 − 𝑑 + 𝑑 𝐻(𝑑 − 4)
6
22
43
Ans. 𝑒 −4𝑠 4 + 3 + 2 +
𝑠
𝑠
𝑠
12. Find the Laplace transform of: 1 + 2𝑑 − 𝑑 2 + 𝑑 3 𝐻(𝑑 − 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 𝑑 + 4 − 𝑑 + 4 2 + 𝑑 + 4 3
= 𝑒 −4𝑠 𝐿 1 + 2𝑑 + 8 − 𝑑 2 − 8𝑑 − 16 + 𝑑 3 + 12𝑑 2 + 48𝑑 + 64
= 𝑒 −4𝑠 𝐿 𝑑 3 + 11𝑑 2 + 42𝑑 + 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: 𝑑 3 + 2𝑑 2 − 3𝑑 + 1 𝐻 𝑑 − 1
Ans. 𝑒 −𝑠
6
𝑠4
+
14. Find the Laplace transform of: 𝑑 2 𝐻 𝑑 − 2 + 𝑑 3 𝛿 𝑑 − 3
Ans. 𝑒 −2𝑠
15. Evaluate the following:
∞
0
𝑠3
+
4
𝑠2
+
𝑠3
4
𝑠
+
4
𝑠2
+
1
𝑠
+ 𝑒 −3𝑠 . 27
𝑒 −2𝑑 1 + 2𝑑 − 3𝑑 2 + 4𝑑 3 𝐻 𝑑 − 1 𝑑𝑑
Ans.
∞
0
2
10
−𝑑
2
3
31
4𝑒 2
16. Evaluate the following:
𝑒 1 + 2𝑑 − 𝑑 + 𝑑 𝐻 𝑑 − 1 𝑑𝑑
[N17/Comp/6M]
Solution:
By definition of L.T.,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = 1 + 2𝑑 − 𝑑 2 + 𝑑 3 𝐻 𝑑 − 1
∞ −𝑠𝑑
𝑒
1 + 2𝑑 − 𝑑 2 + 𝑑 3 𝐻 𝑑 − 1 𝑑𝑑 = 𝐿 1 + 2𝑑 − 𝑑 2 + 𝑑 3 𝐻 𝑑 − 1
0
= 𝑒 −𝑠 𝐿 (1 + 2 𝑑 + 1 − 𝑑 + 1 2 + 𝑑 + 1 3
= 𝑒 −𝑠 𝐿 1 + 2𝑑 + 2 − 𝑑 2 − 2𝑑 − 1 + 𝑑 3 + 3𝑑 2 + 3𝑑 + 1
= 𝑒 −𝑠 𝐿 𝑑 3 + 2𝑑 2 + 3𝑑 + 3
2.2!
3
3
3!
= 𝑒 −𝑠 4 + 3 + 2 +
𝑠
𝑠
𝑠
𝑠
Put 𝑠 = 1,
16
1 + 2𝑑 − 𝑑 2 + 𝑑 3 𝐻 𝑑 − 1 𝑑𝑑 = 𝑒 −1 6 + 4 + 3 + 3 =
∞ −𝑑
𝑒
0
17. Evaluate the following:
𝑒
∞
0
𝑒 −2𝑑 1 − 𝑑 + 𝑑 2 𝐻 𝑑 − 3 𝑑𝑑
Ans.
∞
0
5
𝑒6
18. Evaluate the following:
𝑒 −𝑑 1 + 3𝑑 + 𝑑 2 𝐻 𝑑 − 2 𝑑𝑑
[M18/Comp/6M]
Solution:
By definition of L.T.,
∞ −𝑠𝑑
𝑒 𝑓 𝑑 𝑑𝑑 = 𝐿 𝑓 𝑑
0
Put 𝑓 𝑑 = 1 + 3𝑑 + 𝑑 2 𝐻 𝑑 − 2
∞ −𝑠𝑑
𝑒
1 + 3𝑑 + 𝑑 2 𝐻 𝑑 − 2 𝑑𝑑 = 𝐿 1 + 3𝑑 + 𝑑 2 𝐻 𝑑 − 2
0
S.E/Paper Solutions
43
Crescent Academy…….………………….…..For Research in Education
= 𝑒 −2𝑠 𝐿 (1 + 3 𝑑 + 2 + 𝑑 + 2 2
= 𝑒 −2𝑠 𝐿 1 + 3𝑑 + 6 + 𝑑 2 + 4𝑑 + 4
= 𝑒 −2𝑠 𝐿 𝑑 2 + 7𝑑 + 11
2!
7
11
= 𝑒 −2𝑠 3 + 2 +
𝑠
𝑠
𝑠
Put 𝑠 = 1,
∞ −𝑑
𝑒 1 + 3𝑑 + 𝑑 2 𝐻 𝑑 − 2 𝑑𝑑 = 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 π‘“π‘œπ‘Ÿ 0 ≤ 𝑑 < π‘Ž
and 𝑓(𝑑) is periodic with period
−1 π‘“π‘œπ‘Ÿ π‘Ž < 𝑑 < 2π‘Ž
Find L.T. of 𝑓 𝑑 =
2a
[N18/IT/6M]
Solution:
We have, 𝑓(𝑑) is periodic with period 2a,
1
2π‘Ž −𝑠𝑑
𝐿𝑓 𝑑 =
𝑒 𝑓 𝑑 𝑑𝑑
−2π‘Žπ‘  0
𝐿𝑓 𝑑
π‘Ž −𝑠𝑑
2π‘Ž
𝑒 . 1𝑑𝑑 + π‘Ž
1−𝑒 −2π‘Žπ‘  0
π‘Ž
2π‘Ž
𝑒 −𝑠𝑑
𝑒 −𝑠𝑑
1
=
=
=
=
=
−
−𝑠 0
1−𝑒 −2π‘Žπ‘ 
1
𝑒 −π‘Žπ‘ 
𝑒0
−
=
−𝑠 π‘Ž
𝑒 −2π‘Žπ‘ 
−
1−𝑒 −2π‘Žπ‘ 
1
𝑠
1−𝑒 −π‘Žπ‘ 
1−𝑒 −2π‘Žπ‘ 
.
1−𝑒 −π‘Žπ‘ 
𝑠
1−𝑒 −π‘Žπ‘  1+𝑒 −π‘Žπ‘ 
1−𝑒 −π‘Žπ‘ 
1 1−𝑒 −π‘Žπ‘ 
𝑠
= .
1+𝑒 −π‘Žπ‘ 
𝑠
−𝑠
2
𝑠
1
1+𝑒 −π‘Žπ‘ 
2
1
= . tanh
𝑠
𝐸 π‘“π‘œπ‘Ÿ 0 ≤ 𝑑 <
Find L.T. of 𝑓 𝑑 =
𝑒 −𝑠𝑑 . −1𝑑𝑑
𝑒 −π‘Žπ‘ 
+
1−𝑒 −2π‘Žπ‘  −𝑠
−𝑠
−𝑠
1
1−2𝑒 −π‘Žπ‘  +𝑒 −2π‘Žπ‘ 
=
2.
1−𝑒
1
𝑝
−𝐸
π‘Žπ‘ 
2
𝐿𝑓 𝑑
=
=
=
=
=
𝑝
2
0
1−𝑒 −𝑝𝑠
2
π‘“π‘œπ‘Ÿ < 𝑑 < 𝑝
1
𝐸
1−𝑒 −𝑝𝑠
𝐸
1−𝑒 −𝑝𝑠
𝑒
−
𝑒 −𝑠𝑑
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 −𝑠𝑑 . −𝐸𝑑𝑑
𝑒 −𝑠𝑑
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 −πœƒ
and 𝑓 𝑑 + 𝑝 = 𝑓(𝑑)
2
𝑒 −𝑠𝑑 . 𝐸𝑑𝑑 +
1−𝑒 −πœƒ
𝑝
[N14/CompIT/6M]
Solution:
We have, 𝑓(𝑑) is periodic with period p,
1
𝑝 −𝑠𝑑
𝐿𝑓 𝑑 =
𝑒 𝑓 𝑑 𝑑𝑑
−𝑝𝑠 0
1−𝑒
1
since
−
𝑝𝑠
2
−𝑠
= tanh
πœƒ
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 −πœƒ
1+𝑒 −πœƒ
= tanh
𝑑 0<𝑑<1
Find L.T. of 𝑓 𝑑 =
and 𝑓 𝑑 + 2 = 𝑓 𝑑
0 1<𝑑<2
[M18/IT/5M][N18/AutoMechCivil/6M][N18/Comp/6M]
[M19/AutoMechCivil/6M]
Solution:
We have, 𝑓(𝑑) is periodic with period 2,
1
2 −𝑠𝑑
𝐿𝑓 𝑑 =
𝑒 𝑓 𝑑 𝑑𝑑
−2𝑠 0
𝐿𝑓 𝑑
=
=
=
=
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 −𝑠𝑑
𝑒 . 𝑑𝑑𝑑
0
1−𝑒 −2𝑠
1
1−𝑒 −2𝑠
1
𝑒 −𝑠𝑑
𝑑.
−𝑠
𝑒 −𝑠
−
2
2 −𝑠𝑑
𝑒 . 0𝑑𝑑
1
−𝑠𝑑 1
+
− 1 .
𝑒 −𝑠
πœƒ
𝑒
𝑠2
1
−0+
𝑠2
1−𝑒 −2𝑠 −𝑠
1
1−𝑠𝑒 −𝑠 −𝑒 −𝑠
0
𝑠2
𝑠2
1−𝑒 −2𝑠
𝑑
πœ‹−𝑑
Find L.T. of 𝑓 𝑑 =
0<𝑑<πœ‹
and 𝑓 𝑑 = 𝑓(𝑑 + 2πœ‹)
πœ‹ < 𝑑 < 2πœ‹
Solution:
We have, 𝑓(𝑑) is periodic with period 2πœ‹,
1
2πœ‹ −𝑠𝑑
𝐿𝑓 𝑑 =
𝑒 𝑓 𝑑 𝑑𝑑
−2πœ‹π‘  0
𝐿𝑓 𝑑
=
=
=
=
=
=
=
1−𝑒
1
πœ‹
0
1−𝑒 −2πœ‹π‘ 
1
1−𝑒 −2πœ‹π‘ 
1
𝑑.
πœ‹
𝑒 −𝑠𝑑
− 1 .
−𝑠
𝑒 −πœ‹π‘ 
−
𝑒 −πœ‹π‘ 
1−𝑒 −2πœ‹π‘ 
−𝑠
𝑠2
1
πœ‹(𝑒 −2πœ‹π‘  −𝑒 −πœ‹π‘  )
1−𝑒 −2πœ‹π‘ 
1
1−𝑒 −2πœ‹π‘ 
1−𝑒 −πœ‹π‘ 
𝑠
1−𝑒 −πœ‹π‘  2
𝑠2
−
2πœ‹
πœ‹
−𝑠𝑑 πœ‹
𝑒 −𝑠𝑑 . 𝑑𝑑𝑑 +
−
𝑒
𝑠2
−0+
+
+
πœ‹−𝑑 .
1
+ −πœ‹ .
2
𝑠
𝑒 −2πœ‹π‘  −2𝑒 −πœ‹π‘  +1
πœ‹π‘’ −πœ‹π‘ 
𝑠2
−πœ‹π‘ 
1−𝑒
𝑠
πœ‹π‘’ −πœ‹π‘ 
𝑠 2 1+𝑒 −πœ‹π‘ 
𝑠 1+𝑒 −π‘ πœ‹
−πœ‹π‘ 
−π‘ πœ‹
1−𝑒
−πœ‹π‘ π‘’
S.E/Paper Solutions
0
𝑒 −𝑠𝑑 . πœ‹ − 𝑑 𝑑𝑑
𝑠 2 1+𝑒 −πœ‹π‘ 
46
𝑒 −𝑠𝑑
−𝑠
𝑒 −2πœ‹π‘ 
−𝑠
− −1 .
+
𝑒 −2πœ‹π‘ 
𝑠2
𝑒 −𝑠𝑑
2πœ‹
𝑠2
πœ‹
−0−
𝑒 −πœ‹π‘ 
𝑠2
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5.
𝐾𝑑
Find L.T. of 𝑓 𝑑 = for 0 < 𝑑 < 𝑇 & 𝑓 𝑑 + 𝑇 = 𝑓 𝑑
𝑇
[N16/CompIT/6M]
Solution:
We have, 𝑓(𝑑) is periodic with period T,
𝑇 −𝑠𝑑
1
𝑒 𝑓 𝑑 𝑑𝑑
𝐿𝑓 𝑑 =
−𝑇𝑠 0
𝐿𝑓 𝑑
=
=
=
=
6.
𝑇 −𝑠𝑑 𝐾𝑑
𝑒 .
0
𝑇
1−𝑒 −𝑇𝑠
1
.
𝐾
1−𝑒 −𝑇𝑠 𝑇
1
𝐾
1−𝑒 −𝑇𝑠
.
𝑒 −𝑠𝑑
𝑑.
𝑇.
−𝑠
𝑒 −𝑇𝑠
𝑑𝑑
− 1 .
−
𝑒 −𝑇𝑠
𝐾
−𝑠
𝑠2
−𝑇𝑠
1−𝑠𝑇𝑒
−𝑒 −𝑇𝑠
𝑇 1−𝑒 −𝑇𝑠
𝑠2
𝑇
𝑒 −𝑠𝑑
𝑇
𝑠2
0
1
−0+
𝑠2
2𝑑
Find the Laplace Transform of 𝑓 𝑑 = , 0 ≤ 𝑑 ≤ 3, 𝑓 𝑑 + 3 = 𝑓 𝑑
3
[M17/CompIT/6M]
Solution:
We have, 𝑓(𝑑) is periodic with period 3,
3 −𝑠𝑑
1
𝑒 𝑓 𝑑 𝑑𝑑
𝐿𝑓 𝑑 =
−3𝑠 0
𝐿𝑓 𝑑
=
=
=
=
7.
1−𝑒
1
1−𝑒
1
3 −𝑠𝑑 2𝑑
𝑒 .
0
3
1−𝑒 −3𝑠
1
.
2
1−𝑒 −3𝑠 3
2
1
.
𝑑.
3.
𝑒 −𝑠𝑑
−𝑠
𝑒 −3𝑠
𝑑𝑑
− 1 .
−
𝑒 −3𝑠
2
−𝑠
𝑠2
1−3𝑠𝑒 −3𝑠 −𝑒 −3𝑠
3 1−𝑒 −3𝑠
𝑠2
1−𝑒 −3𝑠
3
𝑒 −𝑠𝑑
3
𝑠2
0
−0+
1
𝑠2
3𝑑
0<𝑑<2
where 𝑓 𝑑 has period 4. (i) Draw graph of
6
2<𝑑<4
𝑓 𝑑 (ii) Find 𝐿 𝑓 𝑑
Solution:
We have, 𝑓(𝑑) is periodic with period 4,
1
4 −𝑠𝑑
𝐿𝑓 𝑑 =
𝑒 𝑓 𝑑 𝑑𝑑
−4𝑠 0
If 𝑓 𝑑 =
𝐿𝑓 𝑑
=
=
=
=
=
1−𝑒
1
2 −𝑠𝑑
𝑒 . 3𝑑𝑑𝑑
0
1−𝑒 −4𝑠
1
1−𝑒 −4𝑠
1
3𝑑.
6
𝑒 −𝑠𝑑
−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 −𝑠𝑑
𝑒 . 6𝑑𝑑
2
2
4
𝑒 −𝑠𝑑
6𝑒 −𝑠𝑑
+
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
𝑑
8.
Find L.T. of 𝑓 𝑑 =
0<𝑑<π‘Ž
π‘Ž
2π‘Ž−𝑑
π‘Ž < 𝑑 < 2π‘Ž
π‘Ž
where 𝑓 𝑑 = 𝑓(𝑑 + 2π‘Ž)
Solution:
We have, 𝑓(𝑑) is periodic with period 2π‘Ž,
1
2π‘Ž −𝑠𝑑
𝐿𝑓 𝑑 =
𝑒 𝑓 𝑑 𝑑𝑑
−2π‘Žπ‘  0
𝐿𝑓 𝑑
=
=
=
=
=
1−𝑒
1
1−𝑒 −2π‘Žπ‘ 
1
π‘Ž 1−𝑒 −2π‘Žπ‘ 
1
π‘Ž 1−𝑒 −2π‘Žπ‘ 
1
π‘Ž
0
𝑑
π‘Ž
𝑑.
π‘Ž
𝑒 −𝑠𝑑
− 1 .
−𝑠
𝑒 −π‘Žπ‘ 
−
𝑒 −π‘Žπ‘ 
𝑒 −𝑠𝑑 .
𝑒 −𝑠𝑑
π‘Ž
𝑠2
0
−0+
−𝑠
𝑠2
−π‘Žπ‘ 
−2π‘Žπ‘ 
1−2𝑒
+𝑒
π‘Ž 1−𝑒 −2π‘Žπ‘ 
1
2π‘Ž
π‘Ž
𝑒 −𝑠𝑑 . 𝑑𝑑 +
2π‘Ž−𝑑
π‘Ž
𝑑𝑑
+ 2π‘Ž − 𝑑 .
1
𝑠2
+0+
𝑒 −𝑠𝑑
𝑒 −2π‘Žπ‘ 
𝑠2
−𝑠
− −1 .
− π‘Ž.
𝑒 −π‘Žπ‘ 
−
−𝑠
𝑒 −𝑠𝑑
2π‘Ž
𝑠2
π‘Ž
𝑒 −π‘Žπ‘ 
𝑠2
𝑠2
−π‘Žπ‘ 
2
1−𝑒
π‘Ž 1−𝑒 −2π‘Žπ‘ 
𝑠2
−π‘Žπ‘ 
2
1−𝑒
1
= .
=
=
9.
π‘Ž 𝑠 2 1−𝑒 −π‘Žπ‘  1+𝑒 −π‘Žπ‘ 
1 1−𝑒 −π‘Žπ‘ 
.
π‘Žπ‘  2 1+𝑒 −π‘Žπ‘ 
1
π‘Žπ‘ 
π‘Žπ‘  2
. tanh
2
πœ‹−𝑑 2
Find L.T. of 𝑓 𝑑 =
; 0 < 𝑑 < 2πœ‹ and 𝑓 𝑑 = 𝑓 𝑑 + 2πœ‹
2
Solution:
We have, 𝑓(𝑑) is periodic with period 2πœ‹,
1
2πœ‹ −𝑠𝑑
𝐿𝑓 𝑑 =
𝑒 𝑓 𝑑 𝑑𝑑
−2πœ‹π‘  0
𝐿𝑓 𝑑
=
=
=
=
1−𝑒
1
1−𝑒 −2πœ‹π‘ 
2πœ‹
0
1
πœ‹−𝑑
4 1−𝑒 −2πœ‹π‘ 
1
4 1−𝑒 −2πœ‹π‘ 
𝑒 −𝑠𝑑 .
πœ‹2
𝑒 −2πœ‹π‘ 
−𝑠
πœ‹−𝑑 2
−𝑠𝑑
2 𝑒
.
−𝑠
− 2πœ‹
1
πœ‹ 2 1−𝑒 −2πœ‹π‘ 
4 1−𝑒 −2πœ‹π‘ 
𝑠
π‘Ž 𝑠𝑖𝑛𝑝𝑑
10. Find L.T. of 𝑓 𝑑 =
0
2
𝑑𝑑
− 2 πœ‹ − 𝑑 −1 .
𝑒 −2πœ‹π‘ 
𝑠2
−
+ 2.
𝑠2
0<𝑑<
πœ‹
𝑝
<𝑑<
𝑝
48
−𝑠 3
2πœ‹ 1+𝑒 −2πœ‹π‘ 
Solution:
2πœ‹
We have, 𝑓(𝑑) is periodic with period ,
S.E/Paper Solutions
𝑒 −2πœ‹π‘ 
𝑒 −𝑠𝑑
𝑠2
− πœ‹2.
+
1
−𝑠
+ 2.
− 2πœ‹.
2 1−𝑒 −2πœ‹π‘ 
𝑒 −𝑠𝑑
2πœ‹
−𝑠 3 0
1
𝑠2
− 2.
𝑠3
πœ‹
𝑝
2πœ‹ and
𝑝
𝑓 𝑑 =𝑓 𝑑+
2πœ‹
𝑝
1
−𝑠 3
Crescent Academy…….………………….…..For Research in Education
𝐿𝑓 𝑑
=
1−𝑒
𝐿𝑓 𝑑
1−𝑒
2πœ‹
𝑠
𝑝
−
1−𝑒
−
πœ‹
𝑝
2πœ‹
𝑠
𝑝
𝑒 −𝑠𝑑
2πœ‹
𝑠
𝑝
−𝑠 2 +𝑝 2
𝑒
2πœ‹
− 𝑠
1−𝑒 𝑝
−
−
π‘ πœ‹
𝑝
𝑠 2 +𝑝 2
π‘Žπ‘
1−𝑒
𝑒 −𝑠𝑑 . π‘Žπ‘ π‘–π‘›π‘π‘‘π‘‘π‘‘ +
0
π‘Ž
=
𝑒 −𝑠𝑑 𝑓 𝑑 𝑑𝑑
0
π‘Ž
=
=
−
1
=
=
2πœ‹
𝑝
1
1+𝑒
2πœ‹
𝑠
𝑝
π‘Žπ‘
𝑒 −𝑠𝑑 . 0𝑑𝑑
πœ‹
𝑝
−𝑠. 𝑠𝑖𝑛𝑝𝑑 − 𝑝. π‘π‘œπ‘ π‘π‘‘
0
1
. 𝑝 −
𝑠 2 +𝑝 2
. (−𝑝)
π‘ πœ‹
𝑝
𝑠 2 +𝑝 2
1+𝑒
𝑠 2 +𝑝 2
−
2πœ‹
𝑝
πœ‹
𝑝
.
1−𝑒
−
−
π‘ πœ‹
𝑝
2πœ‹
𝑠
𝑝
𝑠𝑖𝑛2𝑑
0<𝑑<
πœ‹
2
and 𝑓 𝑑 = 𝑓 𝑑 + πœ‹
0
<𝑑<πœ‹
2
[N16/AutoMechCivil/6M]
Solution:
We have, 𝑓(𝑑) is periodic with period πœ‹,
πœ‹ −𝑠𝑑
1
𝑒 𝑓 𝑑 𝑑𝑑
𝐿𝑓 𝑑 =
−πœ‹π‘  0
11. Find L.T. of 𝑓 𝑑 =
𝐿𝑓 𝑑
=
=
=
=
=
πœ‹
1−𝑒
1
πœ‹
2
0
1−𝑒 −πœ‹π‘ 
1
𝑒 −𝑠𝑑
1−𝑒 −πœ‹π‘ 
−𝑠 2 +22
1
𝑒
−
π‘ πœ‹
2
1−𝑒 −πœ‹π‘ 
𝑠 2 +4
2
1+𝑒
1−𝑒 −πœ‹π‘ 
2
𝑠 2 +4
.
πœ‹
𝑒 −𝑠𝑑 . 𝑠𝑖𝑛2𝑑𝑑𝑑 +
−
πœ‹
2
−𝑠. 𝑠𝑖𝑛2𝑑 − 2. π‘π‘œπ‘ 2𝑑
. 2 −
−
1
. (−2)
𝑠 2 +4
π‘ πœ‹
2
π‘ πœ‹
2
1−𝑒 −πœ‹π‘ 
π‘ π‘–π‘›πœ”π‘‘
12. Find L.T. of 𝑓 𝑑 =
0
0<𝑑<
πœ‹
πœ”
<𝑑<
πœ‹
πœ”
2πœ‹
πœ”
Solution:
2πœ‹
We have, 𝑓(𝑑) is periodic with period ,
πœ”
S.E/Paper Solutions
πœ‹
2
0
𝑠 2 +4
1+𝑒
𝑒 −𝑠𝑑 . 0𝑑𝑑
49
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𝐿𝑓 𝑑
=
1−𝑒
𝐿𝑓 𝑑
1−𝑒
2πœ‹
𝑠
πœ”
−
1−𝑒
−
𝑒 −𝑠𝑑 𝑓 𝑑 𝑑𝑑
0
πœ‹
πœ”
2πœ‹
𝑠
πœ”
0
𝑒
−𝑠𝑑
𝑒 −𝑠𝑑
1
=
=
−
1
=
=
2πœ‹
𝑝
1
2πœ‹
𝑠
πœ”
−𝑠 2 +πœ” 2
1
𝑒
2πœ‹
− 𝑠
1−𝑒 πœ”
−
π‘ πœ‹
πœ”
πœ”
1+𝑒
2πœ‹
− 𝑠
1−𝑒 πœ”
−
π‘ πœ‹
πœ”
𝑠𝑖𝑛7𝑑
πœ‹
πœ”
0
1
𝑠 2 +πœ” 2
πœ”
𝑠 2 +πœ” 2
. (−πœ”)
1+𝑒
.
1−𝑒
0<𝑑<
πœ‹
2
𝑒 −𝑠𝑑 . 0𝑑𝑑
−𝑠. π‘ π‘–π‘›πœ”π‘‘ − πœ”. π‘π‘œπ‘ πœ”π‘‘
=
𝑠 2 +πœ” 2
13. Find L.T. of 𝑓 𝑑 =
. π‘ π‘–π‘›πœ”π‘‘π‘‘π‘‘ +
. πœ” −
𝑠 2 +πœ” 2
2πœ‹
πœ”
πœ‹
πœ”
−
π‘ πœ‹
πœ”
2πœ‹
𝑠
πœ”
πœ‹
2
and 𝑓 𝑑 = 𝑓 𝑑 + πœ‹
<𝑑<πœ‹
2
−
[N17/Elect/6M]
Solution:
We have, 𝑓(𝑑) is periodic with period πœ‹,
πœ‹ −𝑠𝑑
1
𝑒 𝑓 𝑑 𝑑𝑑
𝐿𝑓 𝑑 =
−πœ‹π‘  0
𝐿𝑓 𝑑
=
=
=
=
1−𝑒
1
πœ‹
2
0
1−𝑒 −πœ‹π‘ 
1
𝑒 −𝑠𝑑
1−𝑒 −πœ‹π‘ 
−𝑠 2 +72
1
𝑒
−
π‘ πœ‹
2
1−𝑒 −πœ‹π‘  𝑠 2 +49
1
1−𝑒 −πœ‹π‘ 
𝑠
πœ‹
𝑒 −𝑠𝑑 . 𝑠𝑖𝑛7𝑑𝑑𝑑 +
𝑠 2 +49
+
𝑒 −𝑠𝑑 . 2𝑑𝑑
πœ‹
2
−𝑠. 𝑠𝑖𝑛7𝑑 − 7. π‘π‘œπ‘ 7𝑑
1
𝑠 2 +49
7
𝑠 2 +49
−
. −7 + 2
2𝑒 −π‘ πœ‹
𝑠
+
𝑒 −π‘ πœ‹
−𝑠
=
1−𝑒 −2πœ‹π‘ 
1
= 1−𝑒 −2πœ‹π‘ 
=
=
1
πœ‹
0
𝑒 −𝑠𝑑 . 𝑠𝑖𝑛𝑑𝑑𝑑 +
. −𝑠. 𝑠𝑖𝑛𝑑 − 1. π‘π‘œπ‘ π‘‘
𝑠 2 +1
. 1 −
1
1−𝑒 −2πœ‹π‘  𝑠 2 +1
𝑠 2 +1
1
𝑒 −π‘ πœ‹ +1−𝑠.𝑒 −2πœ‹π‘  −𝑠
1−𝑒 −2πœ‹π‘ 
S.E/Paper Solutions
2πœ‹
πœ‹
−𝑠
πœ‹
2
𝑠 2 +1
50
0
−2
𝑒
−𝑠
πœ‹
2
−𝑠
𝑠𝑖𝑛𝑑
π‘π‘œπ‘ π‘‘
0<𝑑<πœ‹
πœ‹ < 𝑑 < 2πœ‹
𝑒 −𝑠𝑑 . π‘π‘œπ‘ π‘‘π‘‘π‘‘
πœ‹
𝑒 −𝑠𝑑
𝑒 −π‘ πœ‹
πœ‹
𝑠
Solution:
We have, 𝑓(𝑑) is periodic with period 2πœ‹,
1
2πœ‹ −𝑠𝑑
𝐿𝑓 𝑑 =
𝑒 𝑓 𝑑 𝑑𝑑
−2πœ‹π‘  0
𝐿𝑓 𝑑
𝑒 −𝑠𝑑
πœ‹
−𝑠
2𝑒 2
14. Find the Laplace Transforms of f(t), where 𝑓 𝑑 =
1−𝑒
1
+2
0
. 𝑠 −
πœ‹
−𝑠
𝑒 2
πœ‹
2
+
. −1 +
2πœ‹
𝑒 −𝑠𝑑
. −𝑠. π‘π‘œπ‘ π‘‘ + 1. 𝑠𝑖𝑛𝑑
𝑠 2 +1
𝑒 −2πœ‹π‘ 
𝑠 2 +1
. −𝑠 −
1
𝑠 2 +1
. (𝑠)
πœ‹
Crescent Academy…….………………….…..For Research in Education
15. Find the Laplace Transforms of f(t), where 𝑓 𝑑 = 𝑠𝑖𝑛𝑝𝑑 ; 𝑑 ≥ 0
[N15/CompIT/6M]
Solution:
We note that,
𝑓 𝑑+
πœ‹
= 𝑠𝑖𝑛𝑝 𝑑 +
𝑝
πœ‹
= sin(𝑝𝑑 + πœ‹) = 𝑠𝑖𝑛𝑝𝑑
𝑝
∴ 𝑓(𝑑) is periodic with period
𝐿𝑓 𝑑
𝐿𝑓 𝑑
=
=
1
πœ‹
− 𝑠
1−𝑒 𝑝
0
πœ‹
− 𝑠
𝑝
𝑒
πœ‹
− 𝑠
1−𝑒 𝑝
−
π‘ πœ‹
𝑝
𝑠 2 +𝑝 2
1+𝑒
1
1−𝑒
πœ‹
− 𝑠
𝑝
1
−
πœ‹
𝑝
0
. 𝑝 −
1
𝑠 2 +𝑝 2
. (−𝑝)
π‘ πœ‹
𝑝
𝑠 2 +𝑝 2
1+𝑒
𝑠 2 +𝑝 2
−𝑠. 𝑠𝑖𝑛𝑝𝑑 − 𝑝. π‘π‘œπ‘ π‘π‘‘
−𝑠 2 +𝑝 2
1
=
𝑒 −𝑠𝑑 . 𝑠𝑖𝑛𝑝𝑑𝑑𝑑
𝑒 −𝑠𝑑
1
1−𝑒
=
πœ‹
𝑝
πœ‹
− 𝑠
1−𝑒 𝑝
𝑝
𝑒 −𝑠𝑑 𝑓 𝑑 𝑑𝑑
0
1
=
=
πœ‹
𝑝
πœ‹
.
−
π‘ πœ‹
𝑝
=
π‘ πœ‹
−
1−𝑒 𝑝
1
𝑠 2 +𝑝 2
. coth
πœ‹π‘ 
2𝑝
16. Find the Laplace Transforms of f(t), where𝑓 𝑑 = 𝑠𝑖𝑛𝑑
[N18/Elect/5M]
Solution:
We note that,
𝑓 𝑑 + πœ‹ = 𝑠𝑖𝑛 𝑑 + πœ‹ = 𝑠𝑖𝑛𝑑
∴ 𝑓(𝑑) is periodic with period πœ‹
1
πœ‹ −𝑠𝑑
𝐿𝑓 𝑑 =
𝑒 𝑓 𝑑 𝑑𝑑
−πœ‹π‘  0
𝐿𝑓 𝑑
=
=
=
=
=
1−𝑒
1
1−𝑒 −πœ‹π‘ 
1
πœ‹ −𝑠𝑑
𝑒 . 𝑠𝑖𝑛𝑑𝑑𝑑
0
𝑒 −𝑠𝑑
1−𝑒 −πœ‹π‘  −𝑠 2 +1
1
𝑒 −π‘ πœ‹
. 1 −
1−𝑒 −πœ‹π‘  𝑠 2 +1
1+𝑒 −πœ‹π‘ 
1
1−𝑒 −πœ‹π‘  𝑠 2 +1
1
1+𝑒 −π‘ πœ‹
𝑠 2 +1
.
1−𝑒 −π‘ πœ‹
=
πœ‹
−𝑠. 𝑠𝑖𝑛𝑑 − 1. π‘π‘œπ‘ π‘‘
1
𝑠 2 +1
1
𝑠 2 +1
0
. (−1)
. coth
πœ‹π‘ 
2
17. Find the Laplace Transforms of f(t), where𝑓 𝑑 = π‘π‘œπ‘ π‘‘
Solution:
S.E/Paper Solutions
51
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We note that,
𝑓 𝑑 + πœ‹ = π‘π‘œπ‘  𝑑 + πœ‹ = π‘π‘œπ‘ π‘‘
∴ 𝑓(𝑑) is periodic with period πœ‹
1
πœ‹ −𝑠𝑑
𝐿𝑓 𝑑 =
𝑒 𝑓 𝑑 𝑑𝑑
−πœ‹π‘  0
𝐿𝑓 𝑑
=
=
=
=
=
=
1−𝑒
1
1−𝑒 −πœ‹π‘ 
1
πœ‹ −𝑠𝑑
𝑒 . π‘π‘œπ‘ π‘‘π‘‘π‘‘
0
𝑒 −𝑠𝑑
1−𝑒 −πœ‹π‘  −𝑠 2 +1
1
𝑒 −π‘ πœ‹
−𝑠. π‘π‘œπ‘ π‘‘ + 1. 𝑠𝑖𝑛𝑑
. 𝑠 −
1−𝑒 −πœ‹π‘  𝑠 2 +1
1+𝑒 −πœ‹π‘ 
𝑠
0
1
𝑠 2 +1
. (−𝑠)
1−𝑒 −πœ‹π‘  𝑠 2 +1
𝑠
1+𝑒 −π‘ πœ‹
𝑠 2 +1
𝑠
𝑠 2 +1
S.E/Paper Solutions
.
1−𝑒 −π‘ πœ‹
πœ‹π‘ 
. coth
πœ‹
2
52
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Type IX: Miscellaneous Solved Problems
1.
Find the Laplace Transforms of sin 𝑑
[N13/AutoMechCivil/5M][N13/Chem/5M][M18/Elex/5M]
Solution:
π‘₯3
We have, 𝑠𝑖𝑛π‘₯ = π‘₯ −
sin 𝑑 = 𝑑 −
𝑑2
𝐿 sin 𝑑 = 𝐿 𝑑
𝐿 sin 𝑑 =
𝐿 sin 𝑑 =
𝐿 sin 𝑑 =
𝐿 sin 𝑑 =
3
2
3
𝑠2
Γ
𝐿 sin 𝑑 =
−𝐿
πœ‹
πœ‹
3
5
𝑑2
𝑑2
5
2
5
6𝑠 2
31 1
. Γ
22 2
5
6𝑠 2
3
2
3
1−
2𝑠 2
πœ‹
1
+𝐿
+
+
+
4𝑠
1
− β‹― … ..
− β‹― … ..
7
1
2
7
120𝑠 2
53
.
22
120𝑠 2
− β‹― … ..
− β‹―…..
1
− β‹―…..
32𝑠 2
1 2
4𝑠
+
4𝑠
120
120𝑠 2
531
. . Γ
222
6𝑠
1
7
2
Γ
+
1−
1−
3
2𝑠 2
− β‹― ….
6
Γ
−
− β‹― ….
− β‹― … ..
120
−
1 1
Γ
2 2
3
𝑠2
1 1
Γ
2 2
3
𝑠2
𝐿 sin 𝑑 =
5
1
2
5!
5!
𝑑2
+
6
5
𝑑
+
3!
3
1
2
3
𝑑
sin 𝑑 = 𝑑 −
3!
π‘₯5
+
− β‹―…..
2!
−
3 . 𝑒 4𝑠
2𝑠 2
2.
Find the Laplace Transforms of
𝑠𝑖𝑛 𝑑
𝑑
Solution:
We have, 𝑠𝑖𝑛π‘₯ = π‘₯ −
𝑑
sin 𝑑 = 𝑑 −
1
sin 𝑑 = 𝑑 2 −
sin 𝑑
𝐿
𝐿
𝐿
3
𝑑2
6
𝑑
sin 𝑑
𝑑
sin 𝑑
𝑑
sin 𝑑
𝑑
3!
+
5
𝑑2
120
6
120
1
𝑠
1
6𝑠 2
𝑠
6𝑠 2
= −
S.E/Paper Solutions
1
π‘₯5
5
5!
− β‹― ….
− β‹― ….
5!
− β‹― … ..
− β‹― … ….
= 𝐿 1 −𝐿
1
+
𝑑
+
𝑑2
= −
3!
3
𝑑
= 1− +
π‘₯3
+
+
𝑑
+𝐿
6
2!
120𝑠 3
1
60𝑠 3
𝑑2
120
−β‹―…
− β‹― ….
53
−β‹―…
Crescent Academy…….………………….…..For Research in Education
3.
cos 𝑑
Find the Laplace Transforms of
𝑑
[M14/ChemBiot/5M][N16/ChemBiot/5M]
Solution:
π‘₯2
We have, π‘π‘œπ‘ π‘₯ = 1 −
cos 𝑑 = 1 −
+
2!
𝑑
𝑑2
cos 𝑑 = 1 − +
2
cos 𝑑
𝑑
𝐿
𝐿
𝐿
𝐿
𝐿
𝐿
𝐿
4.
1
=𝑑
cos 𝑑
𝑑
cos 𝑑
𝑑
cos 𝑑
𝑑
cos 𝑑
𝑑
cos 𝑑
𝑑
cos 𝑑
𝑑
cos 𝑑
𝑑
−2
−
24
1
𝑑2
2
2!
𝑑
2
𝑑
3
− β‹― … ….
1
=𝐿 𝑑
=
=
=
=
=
=
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œ‹
𝑠
−𝐿
3
2
3
2𝑠 2
1 1
Γ
2 2
3
2𝑠 2
1
2
Γ
−
−
− β‹― ….
− β‹― … ..
24
−2
− β‹― ….
4!
4
4!
𝑑2
+
π‘₯4
+
1−
1−
3
𝑑2
+𝐿
2
+
1−
1
𝑑2
Γ
5
2
5
1
2
5
24𝑠 2
31
.
22
24𝑠 2
+
2𝑠
1
1
+
4𝑠
1
− β‹― ….
− β‹―…..
− β‹―…..
32𝑠 2
1 2
4𝑠
+
4𝑠
−β‹―…
− β‹― … ..
24𝑠 2
31
. Γ
22
+
24
2!
− β‹―…..
1
𝑒 −4𝑠
Find the Laplace Transforms of π‘’π‘Ÿπ‘“ 𝑑
Solution:
2 π‘₯ −𝑒 2
By definition, π‘’π‘Ÿπ‘“π‘₯ =
𝑒 𝑑𝑒
0
πœ‹
π‘’π‘Ÿπ‘“ 𝑑 =
π‘’π‘Ÿπ‘“ 𝑑 =
π‘’π‘Ÿπ‘“ 𝑑 =
π‘’π‘Ÿπ‘“ 𝑑 =
π‘’π‘Ÿπ‘“ 𝑑 =
2
𝑑
πœ‹ 0
2
𝑑
πœ‹ 0
2
πœ‹
2
πœ‹
2
πœ‹
S.E/Paper Solutions
𝑒
−𝑒 2
1 − 𝑒2 +
𝑒−
𝑒3
3
𝑑 −
+
𝑑
𝑑−
1
2
𝑑𝑒
3
𝑒5
10
5
𝑑2
3
+
10
𝑒6
3!
− β‹―…
+
3
2!
−
+ β‹― . . 𝑑𝑒
𝑑
3
𝑑2
𝑒4
𝑑
10
5
0
−β‹―……
−β‹―…
54
Crescent Academy…….………………….…..For Research in Education
𝐿 π‘’π‘Ÿπ‘“ 𝑑 =
𝐿 π‘’π‘Ÿπ‘“ 𝑑 =
𝐿 π‘’π‘Ÿπ‘“ 𝑑 =
𝐿 π‘’π‘Ÿπ‘“ 𝑑 =
𝐿 π‘’π‘Ÿπ‘“ 𝑑 =
𝐿 π‘’π‘Ÿπ‘“ 𝑑 =
𝐿 π‘’π‘Ÿπ‘“ 𝑑 =
2
𝐿 𝑑
πœ‹
3
2
3
𝑠2
Γ
2
πœ‹
2
πœ‹
2
πœ‹
2
πœ‹
1
.
.
3
𝐿 π‘’π‘Ÿπ‘“ 𝑑 =
𝐿 π‘’π‘Ÿπ‘“ 𝑑 =
𝐿 π‘’π‘Ÿπ‘“ 𝑑 =
5.
1 1
Γ
2 2
3
𝑠2
1 1
Γ
2 2
3
𝑠2
5
𝑑2
1−
1
+𝐿
3
+
1−
3
2𝑠 2
3
𝑑2
5
2
5
3𝑠 2
31 1
. Γ
22 2
5
3𝑠 2
3
2
Γ
−
πœ‹
3𝑠
7
2
Γ
+
+
+
− β‹―…
7
1
2
7
10𝑠 2
53
.
22
10𝑠 2
3
− β‹―….
−β‹―…
2 4 𝑠2
1
1 −2
− β‹―……
− β‹―…..
1− . + . .
2 𝑠
−β‹―…
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 𝑑
Solution:
π‘’π‘Ÿπ‘“π‘₯ + π‘’π‘Ÿπ‘“π‘ π‘₯ = 1
erf𝑐 𝑑 = 1 − erf 𝑑
𝐿 erf𝑐 𝑑 = 𝐿 1 − 𝐿 erf 𝑑
1
1
𝑠
𝑠 𝑠+1
𝐿 erf𝑐 𝑑 = −
6.
−𝐿
−
𝑠2
𝐿 π‘’π‘Ÿπ‘“ 𝑑 =
1
2
Find the Laplace Transforms of 𝐽0 𝑑 π‘€π‘•π‘’π‘Ÿπ‘’π½0 𝑑 =
Solution:
𝐽0 𝑑 =
𝐽0 𝑑 =
π‘Ÿ 𝑑 2π‘Ÿ
∞ −1
0 π‘Ÿ! 2 2
−1 0 𝑑 0
−1
0!
2
𝐽0 𝑑 = 1 −
S.E/Paper Solutions
𝑑2
4
2
+
+
𝑑4
64
1!
1
2
𝑑 2
2
+
− β‹― … ..
55
−1
2
𝑑 4
2!
2
2
+ β‹―…
π‘Ÿ
∞ −1
0 π‘Ÿ! 2
𝑑 2π‘Ÿ
2
Crescent Academy…….………………….…..For Research in Education
𝐿 𝐽0 𝑑
=𝐿 1 −𝐿
𝐿 𝐽0 𝑑
= −
𝐿 𝐽0 𝑑
=
𝐿 𝐽0 𝑑
=
𝐿 𝐽0 𝑑
=
𝐿 𝐽0 𝑑
=
𝐿 𝐽0 𝑑
=
𝐿 𝐽0 𝑑
=
𝐿 𝐽0 𝑑
=
𝐿 𝐽0 𝑑
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
− β‹―….
8𝑠 4
1 3 1
2 4 𝑠4
− β‹―…..
𝑠2
1
−2
1
2
𝑠 𝑠 2 +1
1
𝑠
𝑠
=
S.E/Paper Solutions
1
− β‹― ….
− β‹―….
+ . .
2 𝑠2
1
1 −2
𝑑4
− β‹― ….
64𝑠 5
2
24
1− .
1+
𝑑2
𝑠 2 +1
𝑠 2 +1
56
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