What is laplacetransform-4t^2+15t+8 ?

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

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laplacetransform-4t^2+15t+8

laplace transform

- 4 t^{2} + 15 t + 8

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

Solution steps

L {- 4 t^{2} + 15 t + 8}

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}} + 15 L {t} + L {8}

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

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

L {8} : \frac{8}{s}

= - 4 \cdot \frac{2}{s ^{3}} + 15 \cdot \frac{1}{s ^{2}} + \frac{8}{s}

Refine

- 4 \frac{2}{s ^{3}} + 15 \frac{1}{s ^{2}} + \frac{8}{s} : - \frac{8}{s ^{3}} + \frac{15}{s ^{2}} + \frac{8}{s}

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

\int (5 x^{2} + 1)^{2} (10 x) d x

\frac{d}{d x} (\frac{x}{( x + y ) ^{2}})

\frac{d}{d x} (x^{2} + x)

4 cos^{2} (\frac{π}{4} - x)

x \to 0 + lim (e^{c s c (x)})

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