What is laplacetransform t^3+2t^2-t+4 ?

The answer to laplacetransform t^3+2t^2-t+4 is 6/(s^4)+4/(s^3)-1/(s^2)+4/s

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laplacetransform t^3+2t^2-t+4

laplace transform

t^{3} + 2 t^{2} - t + 4

\frac{6}{s ^{4}} + \frac{4}{s ^{3}} - \frac{1}{s ^{2}} + \frac{4}{s}

Solution steps

L {t^{3} + 2 t^{2} - t + 4}

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^{3}} + 2 L {t^{2}} - L {t} + L {4}

L {t^{3}} : \frac{6}{s ^{4}}

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

L {t} : \frac{1}{s ^{2}}

L {4} : \frac{4}{s}

= \frac{6}{s ^{4}} + 2 \cdot \frac{2}{s ^{3}} - \frac{1}{s ^{2}} + \frac{4}{s}

Refine

\frac{6}{s ^{4}} + 2 \frac{2}{s ^{3}} - \frac{1}{s ^{2}} + \frac{4}{s} : \frac{6}{s ^{4}} + \frac{4}{s ^{3}} - \frac{1}{s ^{2}} + \frac{4}{s}

= \frac{6}{s ^{4}} + \frac{4}{s ^{3}} - \frac{1}{s ^{2}} + \frac{4}{s}

3 x^{2} y^{2} d x + (2 y^{3} + 4 y^{3}) d y = 0

\int_{0}^{3} e^{x^{2}} d x

x \to 0 lim (\frac{6}{x} - \frac{6}{x ^{2} + x})

\frac{\partial}{\partial x} (x^{7} y^{5} - x^{6} y^{4})

\frac{3}{5 s - 2}

What is laplacetransform t^3+2t^2-t+4 ?
The answer to laplacetransform t^3+2t^2-t+4 is 6/(s^4)+4/(s^3)-1/(s^2)+4/s