What is inverse oflaplace (s-5)/((s^2-6s+25)) ?

The answer to inverse oflaplace (s-5)/((s^2-6s+25)) is e^{3t}cos(4t)-1/2 e^{3t}sin(4t)

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inverse oflaplace (s-5)/((s^2-6s+25))

inverse laplace

\frac{s - 5}{( s ^{2} - 6 s + 25 )}

e^{3 t} cos (4 t) - \frac{1}{2} e^{3 t} sin (4 t)

Solution steps

L^{- 1} {\frac{s - 5}{( s ^{2} - 6 s + 25 )}}

Expand

\frac{s - 5}{s ^{2} - 6 s + 25} : \frac{s - 3}{( s - 3 ) ^{2} + 16} - 2 \cdot \frac{1}{( s - 3 ) ^{2} + 16}

= L^{- 1} {\frac{s - 3}{( s - 3 ) ^{2} + 16} - 2 \cdot \frac{1}{( s - 3 ) ^{2} + 16}}

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 - 3}{( s - 3 ) ^{2} + 16}} - 2 L^{- 1} {\frac{1}{( s - 3 ) ^{2} + 16}}

L^{- 1} {\frac{s - 3}{( s - 3 ) ^{2} + 16}} : e^{3 t} cos (4 t)

L^{- 1} {\frac{1}{( s - 3 ) ^{2} + 16}} : e^{3 t} \frac{1}{4} sin (4 t)

= e^{3 t} cos (4 t) - 2 e^{3 t} \frac{1}{4} sin (4 t)

Refine

e^{3 t} cos (4 t) - 2 e^{3 t} \frac{1}{4} sin (4 t) : e^{3 t} cos (4 t) - \frac{1}{2} e^{3 t} sin (4 t)

= e^{3 t} cos (4 t) - \frac{1}{2} e^{3 t} sin (4 t)

x^{4} - 4 x^{3} - 2, at x = 1

f (x) = 2 x e^{4 x}

n = 0 \sum \infty \frac{n ( n + 1 )}{2}

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

\int_{0}^{\frac{1}{2}} cos (π x) d x

What is inverse oflaplace (s-5)/((s^2-6s+25)) ?
The answer to inverse oflaplace (s-5)/((s^2-6s+25)) is e^{3t}cos(4t)-1/2 e^{3t}sin(4t)