What is laplacetransform e^{4t}cos(2t) ?

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

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laplacetransform e^{4t}cos(2t)

laplace transform

e^{4 t} cos (2 t)

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

Solution steps

L {e^{4 t} cos (2 t)}

Apply transform rule:

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

then

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

For

e^{4 t} cos (2 t) : f (t) = cos (2 t), a = 4

= L {cos (2 t)} (s - 4)

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

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

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

\frac{d y}{d x} = 5 x e^{y}

\int sin (15 x) d x

x \to \frac{3 π}{2} - lim (tan (x))

\frac{x ^{2} + 2 x - 7}{( x + 1 ) ^{2}}

\int (\frac{( ln ( x ) ) ^{2}}{x ^{2}}) d x

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