Ordinary Differential Equations Questions and Answers – General Properties of Inverse Laplace Transform

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This set of Ordinary Differential Equations Multiple Choice Questions & Answers (MCQs) focuses on “General Properties of Inverse Laplace Transform”.

1. Find the \(L^{-1} (\frac{s+3}{4s^2+9})\).
a) \(\frac{1}{4} cos⁡(\frac{3t}{2})+\frac{1}{2} cos⁡(\frac{3t}{2})\)
b) \(\frac{1}{4} cos⁡(\frac{3t}{4})+\frac{1}{2} sin⁡(\frac{3t}{2})\)
c) \(\frac{1}{2} cos⁡(\frac{3t}{2})+\frac{1}{2} sin⁡(\frac{3t}{2})\)
d) \(\frac{1}{4} cos⁡(\frac{3t}{2})+\frac{1}{2} sin⁡(\frac{3t}{2})\)
View Answer

Answer: d
Explanation: In the given question
=\(\frac{1}{4} L^{-1}\left (\frac{s+3}{s^2+\frac{9}{4}}\right )\)
=\(\frac{1}{4} \Big\{L^{-1}\left (\frac{s}{s^2+\frac{9}{4}}\right)+L^{-1}\left (\frac{3}{s^2+\frac{9}{4}}\right)\Big\}\)
=\(\frac{1}{4} \Big\{cos⁡(\frac{3t}{2})+2 sin⁡(\frac{3t}{2})\Big\}\)
=\(\frac{1}{4} cos⁡(\frac{3t}{2})+\frac{1}{2} sin⁡(\frac{3t}{2})\).
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2. Find the \(L^{-1} (\frac{1}{(s+2)^4})\).
a) \(e^{-2t}×3\)
b) ⁡\(e^{-2t}×\frac{t^3}{3}\)
c) \(e^{-2t}×\frac{t^3}{6}\)
d) \(e^{-2t}×\frac{t^2}{6}\)
View Answer

Answer: c
Explanation: In the given question,
\(L^{-1} (\frac{1}{(s+2)^4})=e^{-2t} L^{-1} \frac{1}{s^4}\) —————– By the first shifting property
=\(e^{-2t}×\frac{t^3}{3!}\)
=\(e^{-2t}×\frac{t^3}{6}\).

3. Find the \(L^{-1} (\frac{s}{(s-1)^7})\).
a) \(e^{-t} \left (\frac{t^6}{5!}+\frac{t^5}{6!}\right )\)
b) \(e^t \left (\frac{t^6}{5!}+\frac{t^5}{6!}\right )\)
c) \(e^t \left (\frac{t^6}{6!}+\frac{t^5}{5!}\right )\)
d) \(e^{-t} \left (\frac{t^6}{6!}+\frac{t^5}{5!}\right )\)
View Answer

Answer: c
Explanation: In the given question,
=\(L^{-1} \left (\frac{s-1+1}{(s-1)^7}\right)\)
=\(e^t L^{-1} \left (\frac{s+1}{s^7}\right)\)
=\(e^t L^{-1} \left (\frac{1}{s^7}+\frac{1}{s^6}\right)\)
=\(e^t \left (\frac{t^6}{6!}+\frac{t^5}{5!}\right)\)
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4. Find the \(L^{-1} (\frac{s}{2s+9+s^2})\).
a) \(e^{-t} \{cos⁡(2\sqrt{2t})-sin⁡(\sqrt{2t})\}\)
b) \(e^{-t} \{cos⁡(2\sqrt{2t})-sin⁡(2\sqrt{2t})\}\)
c) \(e^{-t} \{cos⁡(2\sqrt{2t})-cos(\sqrt{2t})\}\)
d) \(e^{-2t} \{cos⁡(2\sqrt{2t})-sin⁡(2\sqrt{2t})\}\)
View Answer

Answer: b
Explanation: In the given question,
\(L^{-1} \left (\frac{s}{2s+9+s^2}\right )=L^{-1} \left (\frac{s}{(s+1)^2}+8)\right )\)
=\(L^{-1} \left (\frac{(s+1)-1}{(s+1)^2+8}\right )\)
=\(e^{-t} L^{-1} \left (\frac{(s-1)}{s^2+8}\right )\) ———————–By First Shifting Property
=\(e^{-t} L^{-1} \left (\frac{s}{s^2+8}\right )-e^{-t} L^{-1} \left (\frac{1}{s^2+8}\right )\)
=\(e^{-t} \{cos⁡(2\sqrt{2t})-sin⁡(2\sqrt{2t})\}\).

5. Find the \(L^{-1} \left (\frac{(s+1)}{(s+2)(s+3)}\right )\).
a) 2e-3t-e-2t
b) 3e-3t-e-2t
c) 2e-3t-3e-2t
d) 2e-2t-e-t
View Answer

Answer: a
Explanation: In the given question,
\(L^{-1} \left (\frac{(s+1)}{(s+2)(s+3)}\right )=L^{-1} \left (\frac{2(s+2)-(s+3)}{(s+2)(s+3)}\right )\)
=\(L^{-1} \left (\frac{2}{(s+3)}\right )+L^{-1} \left (\frac{1}{(s+2)}\right )\)
=2e-3t-e-2t.
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6. Find the \(L^{-1} \left (\frac{(3s+9)}{(s+1)(s-1)(s-2)}\right )\).
a) e-t+6et+5e2t
b) e-t-et+5e2t
c) e-3t-6et+5e2t
d) e-t-6et+5e2t
View Answer

Answer: d
Explanation: In the given question,
\(L^{-1} \left (\frac{(3s+9)}{(s+1)(s-1)(s-2)}\right )\)
=\(L^{-1} \left (\frac{1}{(s+1)}\right )-6L^{-1} \left (\frac{-6}{(s-1)}\right )+5L^{-1} \left (\frac{-6}{(s-2)}\right ) \)————-Using properties of Partial Fractions
=e-t-6et+5e2t.

7. Find the \(L^{-1} (\frac{1}{(s^2+4)(s^2+9)})\).
a) \(\frac{1}{5} \left (\frac{sin⁡(2t)}{2}-\frac{sin⁡(t)}{3}\right )\)
b) \(\frac{1}{5} \left (\frac{sin⁡(2t)}{2}+\frac{sin⁡(3t)}{3}\right )\)
c) \(\frac{1}{5} \left (\frac{sin⁡(t)}{2}-\frac{sin⁡(3t)}{3}\right )\)
d) \(\frac{1}{5} \left (\frac{sin⁡(2t)}{2}-\frac{sin⁡(3t)}{3}\right )\)
View Answer

Answer: d
Explanation: In the given question,
\(L^{-1} \left (\frac{1}{(s^2+4)(s^2+9)}\right)\)
=\(\frac{1}{5} L^{-1} \left (\frac{5}{(s^2+4)(s^2+9)}\right)\)
=\(\frac{1}{5} L^{-1} \left (\frac{(s^2+9)-(s^2+4)}{(s^2+4)(s^2+9)}\right)\)
=\(\frac{1}{5} L^{-1} \left (\frac{1}{(s^2+4)}\right )-\frac{1}{5} L^{-1} \left (\frac{1}{(s^2+9)}\right)\)
=\(\frac{1}{5} \left (\frac{sin⁡(2t)}{2}-\frac{sin⁡(3t)}{3}\right)\).
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8. Find the \(L^{-1} \left (\frac{s}{(s^2+1)(s^2+2)(s^2+3)}\right )\).
a) \(\frac{1}{2} cos⁡(t)-cos⁡(\sqrt3t)-\frac{1}{2} cos⁡(\sqrt3t)\)
b) \(\frac{1}{2} cos⁡(t)+cos⁡(\sqrt2t)-\frac{1}{2} cos⁡(\sqrt3t)\)
c) \(\frac{1}{2} cos⁡(t)-cos⁡(\sqrt2t)-\frac{1}{2} cos⁡(\sqrt3t)\)
d) \(\frac{1}{2} cos⁡(t)+cos⁡(\sqrt2t)+\frac{1}{2} cos⁡(\sqrt3t)\)
View Answer

Answer: c
Explanation: In the given question,
\(L^{-1} \left (\frac{s}{(s^2+1)(s^2+2)(s^2+3)}\right )\)
=\(L^{-1} \left (\frac{\frac{1}{2}}{(s^2+1)}+\frac{(-1)}{(s^2+2)}+\frac{\frac{(-1)}{2}}{(s^2+3)}\right )\) ——————-By method of Partial fractions
=\(\frac{1}{2} cos⁡(t)-cos⁡(\sqrt2t)-\frac{1}{2} cos⁡(\sqrt3t)\).

9. Find the \(L^{-1} \left (\frac{s+1}{(s-1)(s+2)^2}\right )\).
a) \(\frac{2}{7} e^t-\frac{2}{9} e^{-2t}+\frac{1}{3} e^{-2t}×t\)
b) \(\frac{2}{9} e^t-\frac{2}{9} e^{-2t}+\frac{1}{3} e^{-2t}×t\)
c) \(\frac{2}{9} e^t-\frac{2}{9} e^{-3t}+\frac{1}{3} e^{-2t}×t\)
d) \(\frac{2}{9} e^t-\frac{2}{9} e^{-2t}+\frac{1}{3} e^{-2t}\)
View Answer

Answer: b
Explanation: In the given question,
\(L^{-1} \left (\frac{s+1}{(s-1)(s+2)^2}\right )\)
Using properties of partial fractions-
s+1=A(s+2)2+B(s-1)(s+2)+C(s-1)
At s=1, A=\(\frac{2}{9}\)
At s=2, C=\(\frac{1}{3}\)
At s=0, B=\(\frac{-2}{9}\)
Re substituting all these values in the original fraction,
=\(L^{-1} \left (\frac{2}{9(s-1)} + \frac{-2}{9(s+2)} + \frac{1}{3(s+2)^2}\right)\)
=\(\frac{2}{9} e^t-\frac{2}{9} e^{-2t}+\frac{1}{3} e^{-2t}×t\).
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10. The \(L^{-1} \left (\frac{3s+8}{s^2+4s+25}\right )\) is \(e^{-st} (3cos⁡(\sqrt{21}t+\frac{2sin⁡(\sqrt{21}t)}{\sqrt{21}})\). What is the value of s?
a) 0
b) 1
c) 2
d) 3
View Answer

Answer: c
Explanation: In the given question,
\(L^{-1} \left (\frac{3s+8}{s^2+4s+25}\right )=L^{-1} \left (\frac{3(s+2)+2}{(s+2)^2+21}\right )\)
By the first shifting property
=\(e^{-2t} L^{-1} \left (\frac{3s+2}{s^2+21}\right )\)
=\(e^{-2t} L^{-1} \left (\frac{3s}{s^2+21}\right )+e^{-2t} L^{-1} \left (\frac{2}{s^2+21}\right )\)
=\(e^{-2t} (3cos⁡(\sqrt{21}t+\frac{2sin⁡(\sqrt{21}t)}{\sqrt{21}})\).

Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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