What is inverse oflaplace (6s-5)/(s^2+7) ?

The answer to inverse oflaplace (6s-5)/(s^2+7) is 6cos(sqrt(7)t)-5/(sqrt(7))sin(sqrt(7)t)

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inverse oflaplace (6s-5)/(s^2+7)

inverse laplace

\frac{6 s - 5}{s ^{2} + 7}

6 cos (7 t) - \frac{5}{7} sin (7 t)

Solution steps

L^{- 1} {\frac{6 s - 5}{s ^{2} + 7}}

Expand

= L^{- 1} {\frac{6 s}{s ^{2} + 7} - \frac{5}{s ^{2} + 7}}

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)}

= 6 L^{- 1} {\frac{s}{s ^{2} + 7}} - 5 L^{- 1} {\frac{1}{s ^{2} + 7}}

L^{- 1} {\frac{s}{s ^{2} + 7}} : cos (7 t)

L^{- 1} {\frac{1}{s ^{2} + 7}} : \frac{1}{7} sin (7 t)

= 6 cos (7 t) - 5 \cdot \frac{1}{7} sin (7 t)

Refine

6 cos (7 t) - 5 \frac{1}{7} sin (7 t) : 6 cos (7 t) - \frac{5}{7} sin (7 t)

= 6 cos (7 t) - \frac{5}{7} sin (7 t)

y^{^{'}} - \frac{2 y}{x} = x^{3} y^{- 3}

\frac{d}{d x} (- e^{- x} x)

f a c t or t^{3} + t

y^{^{'}} = 2 sec^{2} (x) e^{2 y}

ma c l a u r in x - sin (x)

What is inverse oflaplace (6s-5)/(s^2+7) ?
The answer to inverse oflaplace (6s-5)/(s^2+7) is 6cos(sqrt(7)t)-5/(sqrt(7))sin(sqrt(7)t)