What is laplacetransform e^-cos(t) ?

The answer to laplacetransform e^-cos(t) is (e^-s)/(s^2+1)

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laplacetransform e^-cos(t)

laplace transform

e^{-} cos (t)

\frac{e ^{-} s}{s ^{2} + 1}

Solution steps

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

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

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

Refine

e^{-} \frac{s}{s ^{2} + 1} : \frac{e ^{-} s}{s ^{2} + 1}

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

e^{x 9 x + 10}

\int e^{c o s (x)} sin (x) d x

\frac{\partial}{\partial x} (- 2 sin (x))

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = 2, y^{^{'}} (0) = 2

(y + 4) y^{^{'}} = 2 x y + 4 y

What is laplacetransform e^-cos(t) ?
The answer to laplacetransform e^-cos(t) is (e^-s)/(s^2+1)