What is laplacetransform f(t)=4e^{-7t}cos(9t) ?

The answer to laplacetransform f(t)=4e^{-7t}cos(9t) is (4(s+7))/((s+7)^2+81)

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laplacetransform f(t)=4e^{-7t}cos(9t)

laplace transform

f (t) = 4 e^{- 7 t} cos (9 t)

\frac{4 ( s + 7 )}{( s + 7 ) ^{2} + 81}

Solution steps

L {4 e^{- 7 t} cos (9 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)}

= 4 L {e^{- 7 t} cos (9 t)}

L {e^{- 7 t} cos (9 t)} : \frac{s + 7}{( s + 7 ) ^{2} + 81}

= 4 \cdot \frac{s + 7}{( s + 7 ) ^{2} + 81}

Refine

4 \frac{s + 7}{( s + 7 ) ^{2} + 81} : \frac{4 ( s + 7 )}{( s + 7 ) ^{2} + 81}

= \frac{4 ( s + 7 )}{( s + 7 ) ^{2} + 81}

\frac{d}{d x} (e^{x} (cos (x)))

\frac{\partial}{\partial y} (4 x^{2} y^{3} + 4)

\frac{d}{d t} (t^{2} sin (wt))

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

\int \frac{1}{9 x ^{2} + 36 x + 35} d x

What is laplacetransform f(t)=4e^{-7t}cos(9t) ?
The answer to laplacetransform f(t)=4e^{-7t}cos(9t) is (4(s+7))/((s+7)^2+81)