What is inverse oflaplace 1/((s-3)^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 1/((s-3)^2)

inverse laplace

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

e^{3 t} t

Solution steps

L^{- 1} {\frac{1}{( s - 3 ) ^{2}}}

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 - 3 ) ^{2}} : a = 3, F (s) = \frac{1}{s ^{2}}

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

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

= e^{3 t} t

\int \frac{2 cos ( x )}{5 + sin ^{2} ( x )} d x

\int 70 x e^{7 x^{2}} d x

x \to 0 lim (\frac{2 - x}{x})

sec (3 x)

\frac{x ^{2} + 1}{x + 4}

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