What is inverse oflaplace (25s)/((s^2+9)(s+4)) ?

The answer to inverse oflaplace (25s)/((s^2+9)(s+4)) is 4cos(3t)+3sin(3t)-4e^{-4t}

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inverse oflaplace (25s)/((s^2+9)(s+4))

inverse laplace

\frac{25 s}{( s ^{2} + 9 ) ( s + 4 )}

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

Solution steps

L^{- 1} {\frac{25 s}{( s ^{2} + 9 ) ( s + 4 )}}

Take the partial fraction of

\frac{25 s}{( s ^{2} + 9 ) ( s + 4 )} : \frac{4 s + 9}{s ^{2} + 9} - \frac{4}{s + 4}

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

Expand

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

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

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

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

L^{- 1} {\frac{1}{s + 4}} : e^{- 4 t}

= 4 cos (3 t) + 9 \cdot \frac{1}{3} sin (3 t) - 4 e^{- 4 t}

Refine

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

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

1 + sin^{2} (x)

y^{^{'}} - \frac{2}{t} y = t^{2}

\frac{d}{d x} (- \frac{8 x}{( 4 x ^{2} + 1 ) ^{2}})

\frac{d}{d x} (arcsin (coth (x)))

ln (7 x)

What is inverse oflaplace (25s)/((s^2+9)(s+4)) ?
The answer to inverse oflaplace (25s)/((s^2+9)(s+4)) is 4cos(3t)+3sin(3t)-4e^{-4t}