What is laplacetransform t*sin(5t+0.2pi) ?

The answer to laplacetransform t*sin(5t+0.2pi) is -(-0.58778…s^2-8.09016…s+14.69463…)/((s^2+25)^2)

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laplacetransform t*sin(5t+0.2pi)

laplace transform

t \cdot sin (5 t + 0.2 π)

- \frac{- 0.58778 \dots s ^{2} - 8.09016 \dots s + 14.69463 \dots}{( s ^{2} + 25 ) ^{2}}

Solution steps

L {t sin (5 t + 0.2 π)}

Use Laplace Transform table:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

For

t sin (5 t + 0.2 π) : f (t) = sin (5 t + 0.2 π), k = 1

= (- 1)^{1} \frac{d}{d s} (L {sin (5 t + 0.2 π)})

L {sin (5 t + 0.2 π)} : \frac{sin ( 0.62831 \dots ) s + 5 cos ( 0.62831 \dots )}{s ^{2} + 25}

\frac{d}{d s} (\frac{sin ( 0.62831 \dots ) s + 5 cos ( 0.62831 \dots )}{s ^{2} + 25}) = \frac{- 0.58778 \dots s ^{2} - 8.09016 \dots s + 14.69463 \dots}{( s ^{2} + 25 ) ^{2}}

= (- 1)^{1} \frac{- 0.58778 \dots s ^{2} - 8.09016 \dots s + 14.69463 \dots}{( s ^{2} + 25 ) ^{2}}

= - \frac{- 0.58778 \dots s ^{2} - 8.09016 \dots s + 14.69463 \dots}{( s ^{2} + 25 ) ^{2}}

f (x) = 7 + 4 x^{2} - 2 x^{3}, at x = 2

f (x) = x x + 3

\frac{d}{d x} (4 - 3 x - 5 x^{2})

\int_{0}^{1} t^{2} d t

What is laplacetransform t*sin(5t+0.2pi) ?
The answer to laplacetransform t*sin(5t+0.2pi) is -(-0.58778…s^2-8.09016…s+14.69463…)/((s^2+25)^2)