What is laplacetransform e^{-2t}(2cos(3t)-sin(3t)) ?

The answer to laplacetransform e^{-2t}(2cos(3t)-sin(3t)) is (2(s+2))/((s+2)^2+9)-3/((s+2)^2+9)

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laplacetransform e^{-2t}(2cos(3t)-sin(3t))

laplace transform

e^{- 2 t} (2 cos (3 t) - sin (3 t))

\frac{2 ( s + 2 )}{( s + 2 ) ^{2} + 9} - \frac{3}{( s + 2 ) ^{2} + 9}

Solution steps

L {e^{- 2 t} (2 cos (3 t) - sin (3 t))}

Apply transform rule:

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

then

L {e^{a t} f (t)} = F (s - a)

For

e^{- 2 t} (2 cos (3 t) - sin (3 t)) : f (t) = (2 cos (3 t) - sin (3 t)), a = - 2

= L {(2 cos (3 t) - sin (3 t))} (s + 2)

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

= \frac{2 ( s + 2 )}{( s + 2 ) ^{2} + 9} - \frac{3}{( s + 2 ) ^{2} + 9}

x \to \infty lim (\frac{5}{x})

\int_{1}^{4} \frac{x + 4}{2 x} d x

6 x (x^{2} + 1)^{2}

x^{^{'}}

1 - 5, 2 x^{2} + 11, 0

What is laplacetransform e^{-2t}(2cos(3t)-sin(3t)) ?
The answer to laplacetransform e^{-2t}(2cos(3t)-sin(3t)) is (2(s+2))/((s+2)^2+9)-3/((s+2)^2+9)