What is laplacetransform e^{2t}cos(3t-5) ?

The answer to laplacetransform e^{2t}cos(3t-5) is (cos(5)(s-2)+3sin(5))/((s-2)^2+9)

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laplacetransform e^{2t}cos(3t-5)

laplace transform

e^{2 t} cos (3 t - 5)

\frac{cos ( 5 ) ( s - 2 ) + 3 sin ( 5 )}{( s - 2 ) ^{2} + 9}

Solution steps

L {e^{2 t} cos (3 t - 5)}

Apply transform rule:

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

then

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

For

e^{2 t} cos (3 t - 5) : f (t) = cos (3 t - 5), a = 2

= L {cos (3 t - 5)} (s - 2)

L {cos (3 t - 5)} : \frac{cos ( 5 ) s + 3 sin ( 5 )}{s ^{2} + 9}

= \frac{cos ( 5 ) ( s - 2 ) + 3 sin ( 5 )}{( s - 2 ) ^{2} + 9}

f (x) = 2 x^{4} - 11 x^{2}, at x = 2

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

x^{2} y^{^{''}} - 6 x y^{^{'}} + 10 y = - 8 x^{3}

\int_{- 4}^{4} \frac{1}{2} (1 6^{2} - x^{4}) d x

\int cos (3 x + 1) d x

What is laplacetransform e^{2t}cos(3t-5) ?
The answer to laplacetransform e^{2t}cos(3t-5) is (cos(5)(s-2)+3sin(5))/((s-2)^2+9)