What is laplacetransform {5-e^{2t}+6t^2} ?

The answer to laplacetransform {5-e^{2t}+6t^2} is 5/s-1/(s-2)+(12)/(s^3)

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laplacetransform {5-e^{2t}+6t^2}

laplace transform

{5 - e^{2 t} + 6 t^{2}}

\frac{5}{s} - \frac{1}{s - 2} + \frac{12}{s ^{3}}

Solution steps

L {5 - e^{2 t} + 6 t^{2}}

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 {5} - L {e^{2 t}} + 6 L {t^{2}}

L {5} : \frac{5}{s}

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

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

= \frac{5}{s} - \frac{1}{s - 2} + 6 \cdot \frac{2}{s ^{3}}

Refine

\frac{5}{s} - \frac{1}{s - 2} + 6 \frac{2}{s ^{3}} : \frac{5}{s} - \frac{1}{s - 2} + \frac{12}{s ^{3}}

= \frac{5}{s} - \frac{1}{s - 2} + \frac{12}{s ^{3}}

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

\frac{d y}{d x} = x e^{y}, y (0) = 0

\int \frac{x ^{5} - x ^{3} + 1}{x ^{3} - x ^{2} - x + 1} d x

\int_{0}^{1} y^{4} d y

y = \frac{8}{4 + x ^{2}}, y = 1

What is laplacetransform {5-e^{2t}+6t^2} ?
The answer to laplacetransform {5-e^{2t}+6t^2} is 5/s-1/(s-2)+(12)/(s^3)