What is laplacetransform t^2sin^2(t) ?

The answer to laplacetransform t^2sin^2(t) is 1/(s^3)+(s(-s^2+12))/((s^2+4)^3)

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laplacetransform t^2sin^2(t)

laplace transform

t^{2} sin^{2} (t)

\frac{1}{s ^{3}} + \frac{s ( - s ^{2} + 12 )}{( s ^{2} + 4 ) ^{3}}

Solution steps

L {t^{2} sin^{2} (t)}

Rewrite using trig identities

L {\frac{1}{2} t^{2} - \frac{1}{2} t^{2} cos (2 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)}

= \frac{1}{2} L {t^{2}} - \frac{1}{2} L {t^{2} cos (2 t)}

L {t^{2}} : \frac{2}{s ^{3}}

L {t^{2} cos (2 t)} : - \frac{2 s ( - s ^{2} + 12 )}{( s ^{2} + 4 ) ^{3}}

= \frac{1}{2} \cdot \frac{2}{s ^{3}} - \frac{1}{2} (- \frac{2 s ( - s ^{2} + 12 )}{( s ^{2} + 4 ) ^{3}})

Refine

\frac{1}{2} \frac{2}{s ^{3}} - \frac{1}{2} (- \frac{2 s ( - s ^{2} + 12 )}{( s ^{2} + 4 ) ^{3}}) : \frac{1}{s ^{3}} + \frac{s ( - s ^{2} + 12 )}{( s ^{2} + 4 ) ^{3}}

= \frac{1}{s ^{3}} + \frac{s ( - s ^{2} + 12 )}{( s ^{2} + 4 ) ^{3}}

\frac{d y}{d x} = \frac{x}{25 y}, y (0) = - 5

2.5 (1 + e^{- 0.1 t})

\frac{d ^{2}}{d x ^{2}} (sin (3 x))

\int 3 sin^{3} (x co) s^{7} x d x

\frac{( 5 t + 300 )}{t ^{2}} + 25

What is laplacetransform t^2sin^2(t) ?
The answer to laplacetransform t^2sin^2(t) is 1/(s^3)+(s(-s^2+12))/((s^2+4)^3)