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)

\frac{d y}{d t} = - 3 y + 4 cos (2 t)

(7 x y^{2} - 2 y) + (7 x^{2} y - 2 x) y^{^{'}} = 0

\int 20 x d x

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

\int_{0}^{4} (e^{- 2 x} + 1) d x

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)