What is inverse oflaplace (400)/((s(s^2+200))) ?

The answer to inverse oflaplace (400)/((s(s^2+200))) is 2\H(t)-2cos(10sqrt(2)t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (400)/((s(s^2+200)))

inverse laplace

\frac{400}{( s ( s ^{2} + 200 ) )}

2 H (t) - 2 cos (102 t)

Solution steps

L^{- 1} {\frac{400}{( s ( s ^{2} + 200 ) )}}

Take the partial fraction of

\frac{400}{s ( s ^{2} + 200 )} : \frac{2}{s} - \frac{2 s}{s ^{2} + 200}

= L^{- 1} {\frac{2}{s} - \frac{2 s}{s ^{2} + 200}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{2}{s}} - L^{- 1} {\frac{2 s}{s ^{2} + 200}}

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

L^{- 1} {\frac{2 s}{s ^{2} + 200}} : 2 cos (102 t)

= 2 H (t) - 2 cos (102 t)

x^{2} e^{3 x}

\frac{d}{d x} ((arctan (2 x))^{2})

\int \frac{( x - 1 )}{x ^{2}} d x

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

y = 2 - e^{x}, y = 10 e^{- x} - 5

What is inverse oflaplace (400)/((s(s^2+200))) ?
The answer to inverse oflaplace (400)/((s(s^2+200))) is 2\H(t)-2cos(10sqrt(2)t)