Ordinary Differential Equations Multiple Choice Questions 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}))
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})).
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})
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 ))
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(2sqrt{2t})-sin(sqrt{2t})})
b) (e^{-t} {cos(2sqrt{2t})-sin(2sqrt{2t})})
c) (e^{-t} {cos(2sqrt{2t})-cos(sqrt{2t})})
d) (e^{-2t} {cos(2sqrt{2t})-sin(2sqrt{2t})})
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(2sqrt{2t})-sin(2sqrt{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
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.
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
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 ))
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))
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})
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
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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