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)})
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)}).
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} )
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})
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})
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)})
Answer: a
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 (kleft [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})
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 (Lleft (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 ))
Answer: c
Explanation: In the given question,
L(sinh(at))=(frac{a}{s^2-a^2})
By effect of division by t,
(Lleft (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 (Lleft (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
Answer: a
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,
(Lleft (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 ])
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 ])
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)})
Answer: a
Explanation: In the given question,
sin4 t=(frac{1+cos^2(2t)-2cos(2t)}{4})
L(sin4 t)=(Lleft (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}{ssqrt{s}}), then what is (L(erf(2sqrt{t})))?
a) (frac{2}{sqrt{s}})
b) (frac{1}{ssqrt{s}})
c) (frac{2}{ssqrt{s}})
d) (frac{4}{ssqrt{s}})
Answer: c
Explanation: In the given question,
By using scaling property of Laplace Transform,
(L(erf(2sqrt{t}))=frac{1}{4}×frac{1}{frac{s}{4}×sqrt{frac{s}{4}}})
=(frac{2}{ssqrt{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)})
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)}).
Global Education & Learning Series – Ordinary Differential Equations.
To practice all areas of Ordinary Differential Equations for online Quizzes,