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

This set of Ordinary Differential Equations Multiple Choice Questions & Answers (MCQs) focuses 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) $$\frac{1}{4} cos⁡(\frac{3t}{2})+\frac{1}{2} sin⁡(\frac{3t}{2})$$

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) $$e^{-2t}×\frac{t^2}{6}$$

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) $$e^{-t} \left (\frac{t^6}{6!}+\frac{t^5}{5!}\right )$$

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) $$e^{-t} \{cos⁡(2\sqrt{2t})-sin⁡(\sqrt{2t})\}$$
b) $$e^{-t} \{cos⁡(2\sqrt{2t})-sin⁡(2\sqrt{2t})\}$$
c) $$e^{-t} \{cos⁡(2\sqrt{2t})-cos(\sqrt{2t})\}$$
d) $$e^{-2t} \{cos⁡(2\sqrt{2t})-sin⁡(2\sqrt{2t})\}$$

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
=$$e^{-t} L^{-1} \left (\frac{s}{s^2+8}\right )-e^{-t} L^{-1} \left (\frac{1}{s^2+8}\right )$$
=$$e^{-t} \{cos⁡(2\sqrt{2t})-sin⁡(2\sqrt{2t})\}$$.

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

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.
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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

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) $$\frac{1}{5} \left (\frac{sin⁡(2t)}{2}-\frac{sin⁡(3t)}{3}\right )$$

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) $$\frac{1}{2} cos⁡(t)+cos⁡(\sqrt2t)+\frac{1}{2} cos⁡(\sqrt3t)$$

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) $$\frac{2}{9} e^t-\frac{2}{9} e^{-2t}+\frac{1}{3} e^{-2t}$$

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

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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