What is inverse oflaplace (16)/((s-4)^7) ?

The answer to inverse oflaplace (16)/((s-4)^7) is (e^{4t}t^6)/(45)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (16)/((s-4)^7)

inverse laplace

\frac{16}{( s - 4 ) ^{7}}

\frac{e ^{4 t} t ^{6}}{45}

Solution steps

L^{- 1} {\frac{16}{( s - 4 ) ^{7}}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{16}{( s - 4 ) ^{7}} : a = 4, F (s) = \frac{16}{s ^{7}}

= e^{4 t} L^{- 1} {\frac{16}{s ^{7}}}

L^{- 1} {\frac{16}{s ^{7}}} : \frac{t ^{6}}{45}

= \frac{e ^{4 t} t ^{6}}{45}

\int sin^{\frac{3}{2}} (x) cos^{3} (x) d x

\int_{e}^{3} \frac{1}{16 - x ^{2}} d x

\frac{d}{d x} (5 x - 2)

\int_{0}^{π} 1 + sin (x) d x

\frac{d y}{d t} = t \cdot y

What is inverse oflaplace (16)/((s-4)^7) ?
The answer to inverse oflaplace (16)/((s-4)^7) is (e^{4t}t^6)/(45)