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

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

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

inverse laplace

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

e^{- \frac{t}{2}} cos (\frac{t}{2}) + e^{- \frac{t}{2}} sin (\frac{t}{2})

Solution steps

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

Expand

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

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

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

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

= e^{- \frac{t}{2}} cos (\frac{t}{2}) + \frac{1}{2} e^{- \frac{t}{2}} \cdot 2 sin (\frac{t}{2})

Refine

e^{- \frac{t}{2}} cos (\frac{t}{2}) + \frac{1}{2} e^{- \frac{t}{2}} 2 sin (\frac{t}{2}) : e^{- \frac{t}{2}} cos (\frac{t}{2}) + e^{- \frac{t}{2}} sin (\frac{t}{2})

= e^{- \frac{t}{2}} cos (\frac{t}{2}) + e^{- \frac{t}{2}} sin (\frac{t}{2})

\int \frac{6}{( x - 2 ) ( x + 4 )} d x

\frac{\partial}{\partial x} ((3 x + 4 y)^{5})

y = \frac{x ^{5}}{1 - x ^{4}}

x \to \infty lim (sin (0))

n = 2 \sum \infty \frac{ln ( n )}{7 n}

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