What is laplacetransform (t^2-3t+2)sin(3t) ?

The answer to laplacetransform (t^2-3t+2)sin(3t) is (6s^4-18s^3+126s^2-162s+432)/((s^2+9)^3)

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laplacetransform (t^2-3t+2)sin(3t)

laplace transform

(t^{2} - 3 t + 2) sin (3 t)

\frac{6 s ^{4} - 18 s ^{3} + 126 s ^{2} - 162 s + 432}{( s ^{2} + 9 ) ^{3}}

Solution steps

L {sin (3 t) (t^{2} - 3 t + 2)}

Expand

(t^{2} - 3 t + 2) sin (3 t) : t^{2} sin (3 t) - 3 t sin (3 t) + 2 sin (3 t)

= L {t^{2} sin (3 t) - 3 t sin (3 t) + 2 sin (3 t)}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= L {t^{2} sin (3 t)} - 3 L {t sin (3 t)} + 2 L {sin (3 t)}

L {t^{2} sin (3 t)} : - \frac{18 ( - s ^{2} + 3 )}{( s ^{2} + 9 ) ^{3}}

L {t sin (3 t)} : \frac{6 s}{( s ^{2} + 9 ) ^{2}}

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

= - \frac{18 ( - s ^{2} + 3 )}{( s ^{2} + 9 ) ^{3}} - 3 \cdot \frac{6 s}{( s ^{2} + 9 ) ^{2}} + 2 \cdot \frac{3}{s ^{2} + 9}

Simplify

- \frac{18 ( - s ^{2} + 3 )}{( s ^{2} + 9 ) ^{3}} - 3 \frac{6 s}{( s ^{2} + 9 ) ^{2}} + 2 \frac{3}{s ^{2} + 9} : \frac{6 s ^{4} - 18 s ^{3} + 126 s ^{2} - 162 s + 432}{( s ^{2} + 9 ) ^{3}}

= \frac{6 s ^{4} - 18 s ^{3} + 126 s ^{2} - 162 s + 432}{( s ^{2} + 9 ) ^{3}}

y = sin (9 x) + cos (2 x)

\frac{d}{d x} (3 arcsin (\frac{x}{2}))

\int 2 sin (3 x) d x

n = 0 \sum \infty e^{(- 5 n)}

y^{^{'''}} - y^{^{'}} = sin (x)

What is laplacetransform (t^2-3t+2)sin(3t) ?
The answer to laplacetransform (t^2-3t+2)sin(3t) is (6s^4-18s^3+126s^2-162s+432)/((s^2+9)^3)