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
Explanation: In the given question
= L(sin3 t)
= \(L \left (\frac{3 sint}{4}\right )-L \left (\frac{1}{4} sin(3t)\right )\)
= \(\frac{3}{4(s^2+1)}-\frac{3}{4(s^2+9)}\).
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
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
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
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
Explanation: In the given question,
L(t+sin(2t))
=\(\frac{1}{s}+\frac{2}{(s^2+4)}\).
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
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
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 \(L\left (\frac{d}{dt}(\frac{sint}{t})\right)\).
a) s×cot-1 s-1
b) s×tan-1 s-1
c) s×cot(s)-1
d) s×tan(s)-1
View Answer
Explanation: In the given question,
L(sint)=\(\frac{1}{s^2+1}\)
By effect of division by t,
\(L(\frac{sint}{t})=\int_{s}^{\infty}\frac{1}{s^2+1} ds\)
=\(\frac{\pi}{2}-tan^{-1}s\)
=cot-1s
By effect of derivative of Laplace Transform,
\(L\left (\frac{d}{dt}\left (\frac{sint}{t}\right )\right )=s×cot^{-1}s-1\).
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
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,
\(L(\int_{0}^{t}sin(u) cos(2u)du)=\frac{1}{2s} \left [\frac{3}{s^2+9}-\frac{1}{s^2+1}\right ]\)
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
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
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
Explanation: In the given question,
The formula is,
L(cat)=\(\frac{1}{s-a log(c)}\)
L(32t)=\(\frac{1}{s-2 log(3)}\).
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