What is laplacetransform t(5t^3+8t^2+10) ?

The answer to laplacetransform t(5t^3+8t^2+10) is -(-(120)/(s^5)-(48)/(s^4)-(10)/(s^2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

laplacetransform t(5t^3+8t^2+10)

laplace transform

t (5 t^{3} + 8 t^{2} + 10)

- (- \frac{120}{s ^{5}} - \frac{48}{s ^{4}} - \frac{10}{s ^{2}})

Solution steps

L {t (5 t^{3} + 8 t^{2} + 10)}

Use Laplace Transform table:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

For

t (5 t^{3} + 8 t^{2} + 10) : f (t) = (5 t^{3} + 8 t^{2} + 10), k = 1

= (- 1)^{1} \frac{d}{d s} (L {(5 t^{3} + 8 t^{2} + 10)})

L {(5 t^{3} + 8 t^{2} + 10)} : \frac{30}{s ^{4}} + \frac{16}{s ^{3}} + \frac{10}{s}

\frac{d}{d s} (\frac{30}{s ^{4}} + \frac{16}{s ^{3}} + \frac{10}{s}) = - \frac{120}{s ^{5}} - \frac{48}{s ^{4}} - \frac{10}{s ^{2}}

= (- 1)^{1} (- \frac{120}{s ^{5}} - \frac{48}{s ^{4}} - \frac{10}{s ^{2}})

= - (- \frac{120}{s ^{5}} - \frac{48}{s ^{4}} - \frac{10}{s ^{2}})

\frac{d}{d x} (z (x) x)

\int_{0}^{0.1} x arctan (5 x) d x

\int 8 x^{\frac{3}{5}} + 5 x^{- \frac{4}{5}} d x

\frac{d y}{d x} = \frac{( y + 1 )}{x}

\int e^{x} 9 - e^{2 x} d x

What is laplacetransform t(5t^3+8t^2+10) ?
The answer to laplacetransform t(5t^3+8t^2+10) is -(-(120)/(s^5)-(48)/(s^4)-(10)/(s^2))