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

The answer to inverse oflaplace (s+1)/(s^2(s^2+1)) is \H(t)+t-cos(t)-sin(t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (s+1)/(s^2(s^2+1))

inverse laplace

\frac{s + 1}{s ^{2} ( s ^{2} + 1 )}

H (t) + t - cos (t) - sin (t)

Solution steps

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

Take the partial fraction of

\frac{s + 1}{s ^{2} ( s ^{2} + 1 )} : \frac{1}{s} + \frac{1}{s ^{2}} + \frac{- s - 1}{s ^{2} + 1}

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

Expand

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

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

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

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

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

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

= H (t) + t - cos (t) - sin (t)

x \to 1 lim (\frac{∣ x - 1 ∣}{x - 1})

\frac{d y}{d x} = 136 x - x y

\frac{d y}{d x} = \frac{2 x}{1 + 2 y}

x^{^{''}} - 3 x^{^{'}} - 10 x = 20 t

\int_{1}^{\infty} x^{- \frac{1}{9}} d x

What is inverse oflaplace (s+1)/(s^2(s^2+1)) ?
The answer to inverse oflaplace (s+1)/(s^2(s^2+1)) is \H(t)+t-cos(t)-sin(t)