Laplace transform of sin(2t)+e^{t}sin(t)

Question

Laplace transform of

\sin (2 t) + e^{t} \sin (t)

Answer 1

Step 1

Given that: Let f (t) = \sin (2 t) + e^{t} \sin (t)

Step 2

So, Laplace transform of f (t) is given by:

L {f (t)} = L {\sin (2 t) + e^{t} \sin (t)}

= L {\sin (2 t)} + L {e^{t} \sin (t)}

= \frac{2}{2^{2} + s^{2}} + \frac{1}{1^{2} + (s - 1)^{2}} [∵ L {\sin k t} = \frac{k}{k^{2} + s^{2}}]

= \frac{2}{2^{2} + 4} + \frac{1}{s^{2} - 2 s + 2}

So, L {f (t)} = \frac{2}{2^{2} + 4} + \frac{1}{s^{2} - 2 s + 2}

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