What is laplacetransform 6e^{-5t}*cos(t) ?

The answer to laplacetransform 6e^{-5t}*cos(t) is (6(s+5))/((s+5)^2+1)

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laplacetransform 6e^{-5t}*cos(t)

laplace transform

6 e^{- 5 t} \cdot cos (t)

\frac{6 ( s + 5 )}{( s + 5 ) ^{2} + 1}

Solution steps

L {6 e^{- 5 t} 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)}

= 6 L {e^{- 5 t} cos (t)}

L {e^{- 5 t} cos (t)} : \frac{s + 5}{( s + 5 ) ^{2} + 1}

= 6 \cdot \frac{s + 5}{( s + 5 ) ^{2} + 1}

Refine

6 \frac{s + 5}{( s + 5 ) ^{2} + 1} : \frac{6 ( s + 5 )}{( s + 5 ) ^{2} + 1}

= \frac{6 ( s + 5 )}{( s + 5 ) ^{2} + 1}

n = 1 \sum \infty 5 (- \frac{1}{5})^{n}

f (x) = 2 x^{2} + x, at x = - 2

x \to 0 + lim (ln (x) \cdot ln (1 + x))

4 x^{\frac{5}{4}} + 8 x^{\frac{3}{2}} + 9 x

ln (x - 1)

What is laplacetransform 6e^{-5t}*cos(t) ?
The answer to laplacetransform 6e^{-5t}*cos(t) is (6(s+5))/((s+5)^2+1)