What is inverse oflaplace 1/((s-7)^3) ?

The answer to inverse oflaplace 1/((s-7)^3) is (e^{7t}t^2)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 1/((s-7)^3)

inverse laplace

\frac{1}{( s - 7 ) ^{3}}

\frac{e ^{7 t} t ^{2}}{2}

Solution steps

L^{- 1} {\frac{1}{( s - 7 ) ^{3}}}

Apply inverse transform rule:

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

then

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

For

\frac{1}{( s - 7 ) ^{3}} : a = 7, F (s) = \frac{1}{s ^{3}}

= e^{7 t} L^{- 1} {\frac{1}{s ^{3}}}

L^{- 1} {\frac{1}{s ^{3}}} : \frac{t ^{2}}{2}

= \frac{e ^{7 t} t ^{2}}{2}

n = 1 \sum \infty \frac{2 n}{n ^{2} + 1}

y = e^{- 3 x}, x \geq 0

\frac{3}{4 x ^{5}}

x \to \infty lim (1^{2 x})

x \to 0 + lim (\frac{7}{1 + e ^{\frac{1}{x}}})

What is inverse oflaplace 1/((s-7)^3) ?
The answer to inverse oflaplace 1/((s-7)^3) is (e^{7t}t^2)/2