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,
=\(L^{-1} \left (\frac{s-1+1}{(s-1)^7}\right)\)
=\(e^t L^{-1} \left (\frac{s+1}{s^7}\right)\)
=\(e^t L^{-1} \left (\frac{1}{s^7}+\frac{1}{s^6}\right)\)
=\(e^t \left (\frac{t^6}{6!}+\frac{t^5}{5!}\right)\)

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,
\(L^{-1} \left (\frac{1}{(s^2+4)(s^2+9)}\right)\)
=\(\frac{1}{5} L^{-1} \left (\frac{5}{(s^2+4)(s^2+9)}\right)\)
=\(\frac{1}{5} L^{-1} \left (\frac{(s^2+9)-(s^2+4)}{(s^2+4)(s^2+9)}\right)\)
=\(\frac{1}{5} L^{-1} \left (\frac{1}{(s^2+4)}\right )-\frac{1}{5} L^{-1} \left (\frac{1}{(s^2+9)}\right)\)
=\(\frac{1}{5} \left (\frac{sin⁡(2t)}{2}-\frac{sin⁡(3t)}{3}\right)\).

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,
=\(L^{-1} \left (\frac{2}{9(s-1)} + \frac{-2}{9(s+2)} + \frac{1}{3(s+2)^2}\right)\)
=\(\frac{2}{9} e^t-\frac{2}{9} e^{-2t}+\frac{1}{3} e^{-2t}×t\).

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
=\(e^{-2t} L^{-1} \left (\frac{3s+2}{s^2+21}\right )\)
=\(e^{-2t} L^{-1} \left (\frac{3s}{s^2+21}\right )+e^{-2t} L^{-1} \left (\frac{2}{s^2+21}\right )\)
=\(e^{-2t} (3cos⁡(\sqrt{21}t+\frac{2sin⁡(\sqrt{21}t)}{\sqrt{21}})\).
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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn