What is inverse oflaplace (3s+4)/(s^2+2) ?

The answer to inverse oflaplace (3s+4)/(s^2+2) is 3cos(sqrt(2)t)+2sqrt(2)sin(sqrt(2)t)

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inverse oflaplace (3s+4)/(s^2+2)

inverse laplace

\frac{3 s + 4}{s ^{2} + 2}

3 cos (2 t) + 22 sin (2 t)

Solution steps

L^{- 1} {\frac{3 s + 4}{s ^{2} + 2}}

Expand

= L^{- 1} {\frac{3 s}{s ^{2} + 2} + \frac{4}{s ^{2} + 2}}

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)}

= 3 L^{- 1} {\frac{s}{s ^{2} + 2}} + 4 L^{- 1} {\frac{1}{s ^{2} + 2}}

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

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

= 3 cos (2 t) + 4 \cdot \frac{1}{2} sin (2 t)

Refine

3 cos (2 t) + 4 \frac{1}{2} sin (2 t) : 3 cos (2 t) + 22 sin (2 t)

= 3 cos (2 t) + 22 sin (2 t)

x \to 0 lim (x^{2} + 3 x - 6)

\frac{5}{2} y^{4} y^{^{'}} - \frac{y ^{5} + 4}{x ^{3}} = 0, y (1) = 0

(- 6, - 5), (4, 4)

\frac{d}{d x} (sin (x) csc (x))

y^{^{'}} = y (x y^{3} + 4)

What is inverse oflaplace (3s+4)/(s^2+2) ?
The answer to inverse oflaplace (3s+4)/(s^2+2) is 3cos(sqrt(2)t)+2sqrt(2)sin(sqrt(2)t)