What is laplacetransform g(t)=sin(2t)sin(t) ?

The answer to laplacetransform g(t)=sin(2t)sin(t) is (4s)/((s^2+9)(s^2+1))

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laplacetransform g(t)=sin(2t)sin(t)

laplace transform

g (t) = sin (2 t) sin (t)

\frac{4 s}{( s ^{2} + 9 ) ( s ^{2} + 1 )}

Solution steps

L {sin (2 t) sin (t)}

Rewrite using trig identities

L {- \frac{1}{2} cos (3 t) + \frac{1}{2} cos (t)}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= - \frac{1}{2} L {cos (3 t)} + \frac{1}{2} L {cos (t)}

L {cos (3 t)} : \frac{s}{s ^{2} + 9}

L {cos (t)} : \frac{s}{s ^{2} + 1}

= - \frac{1}{2} \cdot \frac{s}{s ^{2} + 9} + \frac{1}{2} \cdot \frac{s}{s ^{2} + 1}

Simplify

- \frac{1}{2} \frac{s}{s ^{2} + 9} + \frac{1}{2} \frac{s}{s ^{2} + 1} : \frac{4 s}{( s ^{2} + 9 ) ( s ^{2} + 1 )}

= \frac{4 s}{( s ^{2} + 9 ) ( s ^{2} + 1 )}

\frac{d}{d x} (tan (ln (a x + b)))

\frac{d}{d x} (x^{3} - \frac{1}{x ^{2}})

x \to - \infty lim (x e^{x} - 1)

\frac{x ^{3}}{5 - x ^{2}}

\frac{d}{d x} (\frac{1}{3} x^{3} + 7 x^{2} + 50 x)

What is laplacetransform g(t)=sin(2t)sin(t) ?
The answer to laplacetransform g(t)=sin(2t)sin(t) is (4s)/((s^2+9)(s^2+1))