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

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This set of Ordinary Differential Equations online quiz focuses on “Table of General Properties of Laplace Transform”.

1. Find the L(sin3 t).
a) \(\frac{3}{4(s^2+1)}-\frac{1}{4(s^2+9)}\)
b) \(\frac{3}{4(s^2+1)}-\frac{3}{4(s^2+9)}\)
c) \(\frac{3}{4(s^2+1)}-\frac{9}{4(s^2+9)}\)
d) \(\frac{3}{4(s^2-1)}-\frac{3}{4(s^2+9)}\)
View Answer

Answer: b
Explanation: In the given question
= L(sin3 t)
= \(L \left (\frac{3 sint}{4}\right )-L \left (\frac{1}{4} sin⁡(3t)\right )\)
= \(\frac{3}{4(s^2+1)}-\frac{3}{4(s^2+9)}\).
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2. Find the \(L(e^{2t} (1+t)^2)\).
a) \(\frac{1}{s-2}+\frac{2}{(s-2)^3} + \frac{2}{(s-2)^2}\)
b) \(\frac{⁡3}{s-2}+\frac{2}{(s-2)^3} + \frac{2}{(s-2)^2}\)
c) \(\frac{1}{s-2}+\frac{2}{(s+2)^3} + \frac{2}{(s-2)^2}\)
d) \(\frac{1}{s-2}+\frac{2}{(s-2)^3} \)
View Answer

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

3. Find the Laplace Transform of g(t) which has value (t-1)3 for t>1 and 0 for t<1.
a) \(e^{-2as}×\frac{6}{s^4}\)
b) \(e^{-as}×\frac{24}{s^5}\)
c) \(e^{-as}×\frac{6}{s^4}\)
d) \(e^{-as}×\frac{24}{s^4}\)
View Answer

Answer: c
Explanation: In the given question,
We use the second shifting property.
Let f(t)=t3
\(L(f(t))=\frac{6}{s^4}\)
By the second shifting,
\(L(g(t))=e^{-as}×\frac{6}{s^4}\)
.

4. Find the L(t e-2t sinh⁡(4t)).
a) \(\frac{8s+16}{(s^2+2s-12)^2}\)
b) \(\frac{2s+16}{(s^2+2s-12)^2}\)
c) \(\frac{8s+16}{(s^2+21s-12)^2}\)
d) \(\frac{8s+16}{(s^2+s-12)^2}\)
View Answer

Answer: a
Explanation: In the given question,
L(t e-2t sinh⁡(4t))
L(sinh⁡(4t))=\(\frac{4}{s^2-16}\)
By effect of multiplication of t
L(t×sinh⁡(4t))=\((-1) \frac{d}{ds} \frac{4}{s^2-16}\)
L(t×sinh⁡(4t))=\(\frac{8s}{(s^2-16)^2}\)
By First shifting property
L(t e-2t sinh⁡(4t))=\(\frac{8(s+2)}{((s+2)^2-16)^2} = \frac{8s+16}{(s^2+2s-12)^2}\).

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

Answer: a
Explanation: In the given question,
L(t+sin⁡(2t))
=\(\frac{1}{s}+\frac{2}{(s^2+4)}\).
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6. The L(te-3t cos⁡(2t)cos⁡(3t)) is given by \(k\left [\frac{25-(s+3)^2}{((s+3)^2+25)^2} + \frac{(1-(s+3)^2)}{((s+3)^2+1)^2}\right ]\). Find the value of k.
a) 0
b) 1
c) \(\frac{1}{2}\)
d) \(\frac{-1}{2}\)
View Answer

Answer: d
Explanation: In the given question,
L(e-3t cos⁡(2t)cos⁡(3t))
L(cos⁡(2t) cos⁡(3t))
=\(\frac{1}{2} L(cos⁡(5t)+cos⁡(t))\)
=\(\frac{1}{2} \left (\frac{s}{(s^2+25)}\right )+\frac{1}{2} \left (\frac{s}{(s^2+1)}\right )\)
By effect of multiplication by t
L(t×cos⁡(2t) cos⁡(3t))=\((-1)×\frac{1}{2}×\frac{d}{ds} \frac{1}{2} \left (\frac{s}{(s^2+25)}\right )+\frac{1}{2} \left (\frac{s}{(s^2+1)}\right )\)
=\(\frac{-1}{2} \left [\frac{25-s^2}{(s^2+25)^2} + \frac{(1-s^2)}{(s^2+1)^2}\right]\)
By the effect of first shifting,
L(te-3t cos⁡(2t)cos⁡(3t))=\(\frac{-1}{2} \left [\frac{25-(s+3)^2}{((s+3)^2+25)^2} + \frac{1-(s+3)^2}{((s+3)^2+1)^2}\right ]\)
k=\(\frac{-1}{2}\).

7. Find the \(L\left (\frac{sinh⁡(at)}{t}\right )\).
a) \(\frac{1}{2} log⁡ \left (\frac{s×a}{s-a}\right )\)
b) \(\frac{1}{2} log⁡ \left (\frac{s-a}{s+a}\right )\)
c) \(\frac{1}{2} log⁡ \left (\frac{s+a}{s-a}\right )\)
d) \(\frac{1}{3} log⁡ \left (\frac{s+a}{s-a}\right )\)
View Answer

Answer: c
Explanation: In the given question,
L(sinh⁡(at))=\(\frac{a}{s^2-a^2}\)
By effect of division by t,
\(L\left (\frac{sinh⁡(at)}{t}\right )=\int_{s}^{\infty}\frac{a}{s^2-a^2} ds\)
=\(a×\frac{1}{2a}×log⁡ \left (\frac{s-a}{s+a}\right ) \) in limits s to ∞
=\(\frac{1}{2} log⁡(0)-\frac{1}{2} log⁡ \left (\frac{s-a}{s+a}\right )\)
=\(\frac{1}{2} log⁡ \left (\frac{s+a}{s-a}\right )\).

8. Find the \(L\left (\frac{d}{dt}(\frac{sin⁡t}{t})\right)\).
a) s×cot-1 s-1
b) s×tan-1 s-1
c) s×cot⁡(s)-1
d) s×tan⁡(s)-1
View Answer

Answer: a
Explanation: In the given question,
L(sint)=\(\frac{1}{s^2+1}\)
By effect of division by t,
\(L(\frac{sin⁡t}{t})=\int_{s}^{\infty}\frac{1}{s^2+1} ds\)
=\(\frac{\pi}{2}-tan^{-1}s\)
=cot-1s
By effect of derivative of Laplace Transform,
\(L\left (\frac{d}{dt}\left (\frac{sin⁡t}{t}\right )\right )=s×cot^{-1}s-1\).

9. Find the \(L(\int_{0}^{t}sin⁡(u) cos⁡(2u)du)\).
a) \(\frac{1}{2s} \left [\frac{3}{s^2+9}-\frac{1}{s^2+1}\right ]\)
b) \(\frac{1}{2s} \left [\frac{9}{s^2+9}-\frac{1}{s^2+1}\right ]\)
c) \(\frac{1}{2s} \left [\frac{3}{s^2+9}+\frac{1}{s^2+1}\right ]\)
d) \(\frac{1}{s} \left [\frac{3}{s^2+9}-\frac{1}{s^2+1}\right ]\)
View Answer

Answer: a
Explanation: In the given question,
L(sin⁡(t) cos⁡(2t))=\(\frac{1}{2} L(sin⁡(3t)-sin⁡(t))\)
=\(\frac{1}{2} \left [\frac{3}{s^2+9}-\frac{1}{s^2+1}\right ]\)
By Laplace Transformation of an integral,
\(L(\int_{0}^{t}sin⁡(u) cos⁡(2u)du)=\frac{1}{2s} \left [\frac{3}{s^2+9}-\frac{1}{s^2+1}\right ]\)
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10. Which of the following is not a term present in the Laplace Transform of e2t sin4 t.
a) \(\frac{3}{8s}\)
b) \(\frac{3}{8(s-2)}\)
c) \(\frac{s}{8((s-2)^2+16)}\)
d) \(\frac{s}{2((s-2)^2+4)}\)
View Answer

Answer: a
Explanation: In the given question,
sin4 t=\(\frac{1+cos^2(2t)-2cos⁡(2t)}{4}\)
L(sin4 t)=\(L\left (\frac{1+cos^2 (2t)-2 cos⁡(2t)}{4}\right )\)
=\(\frac{3}{8s}-\frac{s}{2(s^2+4)}+\frac{s}{8(s^2+16)}\)
By First Shifting Property
L(e2t sin4 t)=\(\frac{3}{8(s-2)}-\frac{s}{2((s-2)^2+4)}+\frac{s}{8((s-2)^2+16)}\)

11. If \((erf⁡(\sqrt{t}))=\frac{1}{s\sqrt{s}}\), then what is \(L(erf⁡(2\sqrt{t}))\)?
a) \(\frac{2}{\sqrt{s}}\)
b) \(\frac{1}{s\sqrt{s}}\)
c) \(\frac{2}{s\sqrt{s}}\)
d) \(\frac{4}{s\sqrt{s}}\)
View Answer

Answer: c
Explanation: In the given question,
By using scaling property of Laplace Transform,
\(L(erf⁡(2\sqrt{t}))=\frac{1}{4}×\frac{1}{\frac{s}{4}×\sqrt{\frac{s}{4}}}\)
=\(\frac{2}{s\sqrt{s}}\).

12. Find the value of L(32t).
a) \(\frac{1}{s-2 log⁡(3)}\)
b) \(\frac{1}{s+2 log⁡(3)}\)
c) \(\frac{1}{s-3 log⁡(2)}\)
d) \(\frac{1}{s+3 log⁡(2)}\)
View Answer

Answer: a
Explanation: In the given question,
The formula is,
L(cat)=\(\frac{1}{s-a log(c)}\)
L(32t)=\(\frac{1}{s-2 log⁡(3)}\).

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 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