What is laplacetransform e^2tcos(2t) ?

The answer to laplacetransform e^2tcos(2t) is -(e^2(-s^2+4))/((s^2+4)^2)

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laplacetransform e^2tcos(2t)

laplace transform

e^{2} t cos (2 t)

- \frac{e ^{2} ( - s ^{2} + 4 )}{( s ^{2} + 4 ) ^{2}}

Solution steps

L {e^{2} t cos (2 t)}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= e^{2} L {t cos (2 t)}

L {t cos (2 t)} : - \frac{- s ^{2} + 4}{( s ^{2} + 4 ) ^{2}}

= e^{2} (- \frac{- s ^{2} + 4}{( s ^{2} + 4 ) ^{2}})

Refine

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

= - \frac{e ^{2} ( - s ^{2} + 4 )}{( s ^{2} + 4 ) ^{2}}

\int \frac{6}{7} x^{- \frac{1}{7}} d x

x \to 3 - lim ((x - 3)^{- 1})

\frac{\partial}{\partial y} (ln (10 x^{2} - 7 y^{2}))

f (x) = \frac{4 - 3 x - x ^{2}}{x ^{2} - 1}

What is laplacetransform e^2tcos(2t) ?
The answer to laplacetransform e^2tcos(2t) is -(e^2(-s^2+4))/((s^2+4)^2)