What is laplacetransform sin(t-5) ?

The answer to laplacetransform sin(t-5) is (-sin(5)s+cos(5))/(s^2+1)

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laplacetransform sin(t-5)

laplace transform

sin (t - 5)

\frac{- sin ( 5 ) s + cos ( 5 )}{s ^{2} + 1}

Solution steps

L {sin (t - 5)}

Use Laplace Transform table:

L {sin (a t + b)} = \frac{s \cdot sin ( b ) + a \cdot cos ( b )}{s ^{2} + a ^{2}}

L {sin (t - 5)} = \frac{s sin ( - 5 ) + 1 \cdot cos ( - 5 )}{s ^{2} + 1 ^{2}}

= \frac{s sin ( - 5 ) + 1 \cdot cos ( - 5 )}{s ^{2} + 1 ^{2}}

= \frac{- sin ( 5 ) s + cos ( 5 )}{s ^{2} + 1}

y = (3 x^{2} + 3 x - 1)^{4}

(x^{2} - 3 y^{2}) d x + (2 x y) d y = 0

\frac{d}{d x} (\frac{x}{x})

f (x) = 4 x^{2} - 4 x + 1, (- 1, 8)

\frac{\partial}{\partial x} (y - \frac{27}{x ^{2}})

What is laplacetransform sin(t-5) ?
The answer to laplacetransform sin(t-5) is (-sin(5)s+cos(5))/(s^2+1)