Let y(t)=int_0^tf(t)dt If the Laplace transform of y(t) is given Y(s)=frac{19}{(s^2+25)} , find f(t) a) f(t)=19 sin(5t) b) none c) f(t)=6 sin(2t) d) f(t)=20 cos(6t) e) f(t)=19 cos(5t)

Question

Let

y (t) = \int_{0}^{t} f (t) d t

If the Laplace transform of y(t) is given

Y (s) = \frac{19}{(s^{2} + 25)}

, find f(t)
a)

f (t) = 19 \sin (5 t)

b) none
c)

f (t) = 6 \sin (2 t)

d)

f (t) = 20 \cos (6 t)

e)

f (t) = 19 \cos (5 t)

Answer 1

Step 1
The Laplace transform of y(t) is given

Y (s) = \frac{19}{(s^{2} + 25)}

The Laplace transform is written as,

L {f (t)} = F (s)

L [\int_{0}^{t} f (t) d t] = \frac{1}{s} F (s)

It is given that ,

y (t) = \int_{0}^{t} f (t) d t

Take Laplace on both sides,

L {y (t)} = L {\int_{0}^{t} f (t) d t}

Step 2
Substitute Laplace of y(t) ,

\frac{1}{s} L {f (t)} = \frac{19}{(s^{2} + 25)}

L f (t) = \frac{19 s}{(s^{2} + 25)}

Take inverse on both sides,

f (t) = L^{- 1} [\frac{19}{(s^{2} + 25)}]

= 19 L^{- 1} [\frac{s}{(s^{2} + 25)}]

The inverse Laplace transform of

\cos (a t)

is

L^{- 1} [\frac{s}{(s^{2} + a^{2})}]

Step 3
Therefore ,

f (t) = L^{- 1} [\frac{19 s}{(s^{2} + 25)}]

= 19 L^{- 1} [\frac{s}{(s^{2} + 25)}]

= 19 \cos 5 t

The value of

f (t) = 19 \cos (5 t)

and the correct option is E.

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