What is laplacetransform f(t)=t^2+6t-3 ?

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

Popular >

laplacetransform f(t)=t^2+6t-3

laplace transform

f (t) = t^{2} + 6 t - 3

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

Solution steps

L {t^{2} + 6 t - 3}

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^{2}} + 6 L {t} - L {3}

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

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

L {3} : \frac{3}{s}

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

Refine

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

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

f (x) = 2 x^{4}, (- 2, 32)

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

\frac{d}{d x} (18 x e^{x^{2}})

t^{2} y^{^{''}} - 2 y = 3 t^{2} - 1, y (1) = \frac{3}{2}, y^{^{'}} (1) = 3

\frac{s}{( s + 4 ) ^{2} ( s ^{2} - 9 )}

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