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

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

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

inverse laplace

\frac{s + 1}{s ^{2} + 3 s + 2}

e^{- \frac{3 t}{2}} cosh (\frac{t}{2}) - e^{- \frac{3 t}{2}} sinh (\frac{t}{2})

Solution steps

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

Expand

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

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

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

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

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

= e^{- \frac{3 t}{2}} cosh (\frac{t}{2}) - \frac{1}{2} e^{- \frac{3 t}{2}} \cdot 2 sinh (\frac{t}{2})

Refine

e^{- \frac{3 t}{2}} cosh (\frac{t}{2}) - \frac{1}{2} e^{- \frac{3 t}{2}} 2 sinh (\frac{t}{2}) : e^{- \frac{3 t}{2}} cosh (\frac{t}{2}) - e^{- \frac{3 t}{2}} sinh (\frac{t}{2})

= e^{- \frac{3 t}{2}} cosh (\frac{t}{2}) - e^{- \frac{3 t}{2}} sinh (\frac{t}{2})

\frac{\partial}{\partial y} (e^{x} ln (x y))

\int e^{\frac{x}{10}} d x

\int 3 x^{2} + 3 x^{2} ln (y) d x

\int_{1}^{4} (6 x + x) d x

\frac{d}{d x} (\frac{cos ( x ) \cdot e ^{x}}{tan ( x )})

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