What is inverse oflaplace 1/s-(s+8)/(s^2+8s+25) ?

The answer to inverse oflaplace 1/s-(s+8)/(s^2+8s+25) is \H(t)-e^{-4t}(cos(3t)+4/3 sin(3t))

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inverse oflaplace 1/s-(s+8)/(s^2+8s+25)

inverse laplace

\frac{1}{s} - \frac{s + 8}{s ^{2} + 8 s + 25}

H (t) - e^{- 4 t} (cos (3 t) + \frac{4}{3} sin (3 t))

Solution steps

L^{- 1} {\frac{1}{s} - \frac{s + 8}{s ^{2} + 8 s + 25}}

Expand

\frac{s + 8}{s ^{2} + 8 s + 25} : \frac{s + 4}{( s + 4 ) ^{2} + 9} + 4 \cdot \frac{1}{( s + 4 ) ^{2} + 9}

= L^{- 1} {\frac{1}{s} - (\frac{s + 4}{( s + 4 ) ^{2} + 9} + 4 \cdot \frac{1}{( s + 4 ) ^{2} + 9})}

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{s + 4}{( s + 4 ) ^{2} + 9} + 4 \cdot \frac{1}{( s + 4 ) ^{2} + 9}}

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

L^{- 1} {\frac{s + 4}{( s + 4 ) ^{2} + 9} + 4 \cdot \frac{1}{( s + 4 ) ^{2} + 9}} : e^{- 4 t} (cos (3 t) + \frac{4}{3} sin (3 t))

= H (t) - e^{- 4 t} (cos (3 t) + \frac{4}{3} sin (3 t))

\int ∣ x - 3 ∣ d x

f (x) = (\frac{x + 2}{x - 2})^{5}

f (x) = x^{2} 1 - x^{2}

\frac{\partial}{\partial x} (3 x e^{x^{2} + y^{2}})

\int \frac{cos ( x )}{sin ^{2} ( x ) - 49} d x

What is inverse oflaplace 1/s-(s+8)/(s^2+8s+25) ?
The answer to inverse oflaplace 1/s-(s+8)/(s^2+8s+25) is \H(t)-e^{-4t}(cos(3t)+4/3 sin(3t))