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)

t^{2} - 10 t + 31

\int 1 + 4 a^{2} x^{2} d x

\int ln (x - 2) d x

y = \frac{1}{x ^{3}}, (1, 1)

f (x) = \frac{5 x}{x - 3}, (4, 20)

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)