What is inverse oflaplace (2s+3)/(s^2+7) ?

The answer to inverse oflaplace (2s+3)/(s^2+7) is 2cos(sqrt(7)t)+3/(sqrt(7))sin(sqrt(7)t)

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inverse oflaplace (2s+3)/(s^2+7)

inverse laplace

\frac{2 s + 3}{s ^{2} + 7}

2 cos (7 t) + \frac{3}{7} sin (7 t)

Solution steps

L^{- 1} {\frac{2 s + 3}{s ^{2} + 7}}

Expand

= L^{- 1} {\frac{2 s}{s ^{2} + 7} + \frac{3}{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)}

= 2 L^{- 1} {\frac{s}{s ^{2} + 7}} + 3 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)

= 2 cos (7 t) + 3 \cdot \frac{1}{7} sin (7 t)

Refine

2 cos (7 t) + 3 \frac{1}{7} sin (7 t) : 2 cos (7 t) + \frac{3}{7} sin (7 t)

= 2 cos (7 t) + \frac{3}{7} sin (7 t)

y = \frac{x}{x + 1}

\int \frac{1}{x ( 8 x + 9 )} d x

t^{3} \frac{d y}{d t} + 3 t^{2} y = 2 cos (t), y (π) = 0

2 y^{^{'}} + y = 3 x^{2}

\int \frac{( 2 x - 3 ) ^{3}}{x} d x

What is inverse oflaplace (2s+3)/(s^2+7) ?
The answer to inverse oflaplace (2s+3)/(s^2+7) is 2cos(sqrt(7)t)+3/(sqrt(7))sin(sqrt(7)t)