inverselaplace (2s^2+5s+7)/((s+1)^3)

الحلّ

inverse laplace

\frac{2 s ^{2} + 5 s + 7}{( s + 1 ) ^{3}}

2 e^{- t} + e^{- t} t + e^{- t} \cdot 2 t^{2}

خطوات الحلّ

L^{- 1} {\frac{2 s ^{2} + 5 s + 7}{( s + 1 ) ^{3}}}

\frac{2 s ^{2} + 5 s + 7}{( s + 1 ) ^{3}}

خذ الكسر الجزئي لـ:

\frac{2}{s + 1} + \frac{1}{( s + 1 ) ^{2}} + \frac{4}{( s + 1 ) ^{3}}

= L^{- 1} {\frac{2}{s + 1} + \frac{1}{( s + 1 ) ^{2}} + \frac{4}{( s + 1 ) ^{3}}}

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 + 1}} + L^{- 1} {\frac{1}{( s + 1 ) ^{2}}} + L^{- 1} {\frac{4}{( s + 1 ) ^{3}}}

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

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

L^{- 1} {\frac{4}{( s + 1 ) ^{3}}} : e^{- t} \cdot 2 t^{2}

= 2 e^{- t} + e^{- t} t + e^{- t} \cdot 2 t^{2}

\frac{d}{d x} (\frac{4 5 x ^{3} + x}{( x - 1 )})

\frac{0.1 \cdot m}{s + 0.1 \cdot m}

\frac{x}{x + 4}

\frac{1}{s + 1} - s - 1

y = x^{5}, (- 1, - 1)