What is laplacetransform 3t^2+13t ?

The answer to laplacetransform 3t^2+13t is 6/(s^3)+(13)/(s^2)

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laplacetransform 3t^2+13t

laplace transform

3 t^{2} + 13 t

\frac{6}{s ^{3}} + \frac{13}{s ^{2}}

Solution steps

L {3 t^{2} + 13 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)}

= 3 L {t^{2}} + 13 L {t}

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

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

= 3 \cdot \frac{2}{s ^{3}} + 13 \cdot \frac{1}{s ^{2}}

Refine

3 \frac{2}{s ^{3}} + 13 \frac{1}{s ^{2}} : \frac{6}{s ^{3}} + \frac{13}{s ^{2}}

= \frac{6}{s ^{3}} + \frac{13}{s ^{2}}

x \to 0 lim (\frac{( x ^{2} + 5 x )}{x})

y = \frac{3}{x ^{2} - 5}

\int \frac{2 y}{x + 1} d x

x \to 0 - lim (- x + 5)

y^{^{''}} + 16 y = 16 sec (24 t)

What is laplacetransform 3t^2+13t ?
The answer to laplacetransform 3t^2+13t is 6/(s^3)+(13)/(s^2)