What is laplacetransform {t^2-3t-2e^{-t}sin(3t)} ?

The answer to laplacetransform {t^2-3t-2e^{-t}sin(3t)} is 2/(s^3)-3/(s^2)-6/((s+1)^2+9)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

laplacetransform {t^2-3t-2e^{-t}sin(3t)}

laplace transform

{t^{2} - 3 t - 2 e^{- t} sin (3 t)}

\frac{2}{s ^{3}} - \frac{3}{s ^{2}} - \frac{6}{( s + 1 ) ^{2} + 9}

Solution steps

L {t^{2} - 3 t - 2 e^{- t} sin (3 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)}

= L {t^{2}} - 3 L {t} - 2 L {e^{- t} sin (3 t)}

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

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

L {e^{- t} sin (3 t)} : \frac{3}{( s + 1 ) ^{2} + 9}

= \frac{2}{s ^{3}} - 3 \cdot \frac{1}{s ^{2}} - 2 \cdot \frac{3}{( s + 1 ) ^{2} + 9}

Refine

\frac{2}{s ^{3}} - 3 \frac{1}{s ^{2}} - 2 \frac{3}{( s + 1 ) ^{2} + 9} : \frac{2}{s ^{3}} - \frac{3}{s ^{2}} - \frac{6}{( s + 1 ) ^{2} + 9}

= \frac{2}{s ^{3}} - \frac{3}{s ^{2}} - \frac{6}{( s + 1 ) ^{2} + 9}

7 x^{2} cos (x) cot (x)

x \to 0 lim (e^{x} + 3 x \frac{1}{x})

\frac{d}{d t} (4 e^{\frac{t}{2}})

arcsin (6 x + 1)

\frac{d}{d x} (\frac{3}{x - 1})

What is laplacetransform {t^2-3t-2e^{-t}sin(3t)} ?
The answer to laplacetransform {t^2-3t-2e^{-t}sin(3t)} is 2/(s^3)-3/(s^2)-6/((s+1)^2+9)