What is inverse oflaplace (4s-18)/(s^2-6s+10) ?

The answer to inverse oflaplace (4s-18)/(s^2-6s+10) is 4e^{3t}cos(t)-6e^{3t}sin(t)

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inverse oflaplace (4s-18)/(s^2-6s+10)

inverse laplace

\frac{4 s - 18}{s ^{2} - 6 s + 10}

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

Solution steps

L^{- 1} {\frac{4 s - 18}{s ^{2} - 6 s + 10}}

Expand

\frac{4 s - 18}{s ^{2} - 6 s + 10} : 4 \cdot \frac{s - 3}{( s - 3 ) ^{2} + 1} - 6 \cdot \frac{1}{( s - 3 ) ^{2} + 1}

= L^{- 1} {4 \cdot \frac{s - 3}{( s - 3 ) ^{2} + 1} - 6 \cdot \frac{1}{( s - 3 ) ^{2} + 1}}

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 - 3}{( s - 3 ) ^{2} + 1}} - 6 L^{- 1} {\frac{1}{( s - 3 ) ^{2} + 1}}

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

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

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

y = x + 5

\frac{d y}{d x} + a y = 0

x \to - 2 lim (3 x^{4} + 2 x^{2} - x + 1)

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3}, y (0) = 1, y^{^{'}} (0) = 0

\frac{d y}{d x} = \frac{( y - x + 1 )}{y - x - 6}

What is inverse oflaplace (4s-18)/(s^2-6s+10) ?
The answer to inverse oflaplace (4s-18)/(s^2-6s+10) is 4e^{3t}cos(t)-6e^{3t}sin(t)