What is inverse oflaplace (150)/(s(s^2+8s+25)) ?

The answer to inverse oflaplace (150)/(s(s^2+8s+25)) is 6\H(t)-6e^{-4t}cos(3t)-8e^{-4t}sin(3t)

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inverse oflaplace (150)/(s(s^2+8s+25))

inverse laplace

\frac{150}{s ( s ^{2} + 8 s + 25 )}

6 H (t) - 6 e^{- 4 t} cos (3 t) - 8 e^{- 4 t} sin (3 t)

Solution steps

L^{- 1} {\frac{150}{s ( s ^{2} + 8 s + 25 )}}

Take the partial fraction of

\frac{150}{s ( s ^{2} + 8 s + 25 )} : \frac{6}{s} + \frac{- 6 s - 48}{s ^{2} + 8 s + 25}

= L^{- 1} {\frac{6}{s} + \frac{- 6 s - 48}{s ^{2} + 8 s + 25}}

Expand

\frac{- 6 s - 48}{s ^{2} + 8 s + 25} : - 6 \cdot \frac{s + 4}{( s + 4 ) ^{2} + 9} - 24 \cdot \frac{1}{( s + 4 ) ^{2} + 9}

= L^{- 1} {\frac{6}{s} - 6 \cdot \frac{s + 4}{( s + 4 ) ^{2} + 9} - 24 \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{6}{s}} - 6 L^{- 1} {\frac{s + 4}{( s + 4 ) ^{2} + 9}} - 24 L^{- 1} {\frac{1}{( s + 4 ) ^{2} + 9}}

L^{- 1} {\frac{6}{s}} : 6 H (t)

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

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

= 6 H (t) - 6 e^{- 4 t} cos (3 t) - 24 e^{- 4 t} \frac{1}{3} sin (3 t)

Refine

6 H (t) - 6 e^{- 4 t} cos (3 t) - 24 e^{- 4 t} \frac{1}{3} sin (3 t) : 6 H (t) - 6 e^{- 4 t} cos (3 t) - 8 e^{- 4 t} sin (3 t)

= 6 H (t) - 6 e^{- 4 t} cos (3 t) - 8 e^{- 4 t} sin (3 t)

\frac{\partial}{\partial x} (\frac{1}{f ( x )})

\frac{d}{d x} (\frac{4 x}{7 - cot ( x )})

x \to 1 - lim (\frac{1}{1 - x})

\frac{d}{d x} (ln (x^{2} + 4))

\frac{\partial}{\partial x} (e^{- 2 x} sin (y))

What is inverse oflaplace (150)/(s(s^2+8s+25)) ?
The answer to inverse oflaplace (150)/(s(s^2+8s+25)) is 6\H(t)-6e^{-4t}cos(3t)-8e^{-4t}sin(3t)