# 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) $$\frac{3}{4(s^2-1)}-\frac{3}{4(s^2+9)}$$

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) $$\frac{1}{s-2}+\frac{2}{(s-2)^3}$$

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) $$e^{-as}×\frac{24}{s^4}$$

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) $$\frac{8s+16}{(s^2+s-12)^2}$$

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) $$\frac{2}{s}+\frac{2}{(s^2+4)}$$

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) $$\frac{-1}{2}$$

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) $$\frac{1}{3} log⁡ \left (\frac{s+a}{s-a}\right )$$

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

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) $$\frac{1}{s} \left [\frac{3}{s^2+9}-\frac{1}{s^2+1}\right ]$$

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) $$\frac{s}{2((s-2)^2+4)}$$

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) $$\frac{4}{s\sqrt{s}}$$

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) $$\frac{1}{s+3 log⁡(2)}$$

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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