What is laplacetransform 4t^2-5sin(3t) ?

The answer to laplacetransform 4t^2-5sin(3t) is 8/(s^3)-(15)/(s^2+9)

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laplacetransform 4t^2-5sin(3t)

laplace transform

4 t^{2} - 5 sin (3 t)

\frac{8}{s ^{3}} - \frac{15}{s ^{2} + 9}

Solution steps

L {4 t^{2} - 5 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)}

= 4 L {t^{2}} - 5 L {sin (3 t)}

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

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

= 4 \cdot \frac{2}{s ^{3}} - 5 \cdot \frac{3}{s ^{2} + 9}

Refine

4 \frac{2}{s ^{3}} - 5 \frac{3}{s ^{2} + 9} : \frac{8}{s ^{3}} - \frac{15}{s ^{2} + 9}

= \frac{8}{s ^{3}} - \frac{15}{s ^{2} + 9}

f (x) = - 3 x^{3} - 2 x^{2} - 1, at x = - 1

\frac{\partial}{\partial x} (\frac{x z}{x ^{2} + y ^{2}})

\int cos (3 x) (2 + sin (3 x))^{4} d x

\frac{d}{d a} (sin (a) + sin (2 a))

\frac{d}{d x} (\frac{23}{x ^{4}})

What is laplacetransform 4t^2-5sin(3t) ?
The answer to laplacetransform 4t^2-5sin(3t) is 8/(s^3)-(15)/(s^2+9)