What is inverse oflaplace ((3s+1))/(s^2(s^2+4)) ?

The answer to inverse oflaplace ((3s+1))/(s^2(s^2+4)) is 3/4 \H(t)+t/4-3/4 cos(2t)-1/8 sin(2t)

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inverse oflaplace ((3s+1))/(s^2(s^2+4))

inverse laplace

\frac{( 3 s + 1 )}{s ^{2} ( s ^{2} + 4 )}

\frac{3}{4} H (t) + \frac{t}{4} - \frac{3}{4} cos (2 t) - \frac{1}{8} sin (2 t)

Solution steps

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

Take the partial fraction of

\frac{3 s + 1}{s ^{2} ( s ^{2} + 4 )} : \frac{3}{4 s} + \frac{1}{4 s ^{2}} + \frac{- 3 s - 1}{4 ( s ^{2} + 4 )}

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

Expand

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

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)}

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

L^{- 1} {\frac{1}{s}} : H (t)

L^{- 1} {\frac{1}{s ^{2}}} : t

L^{- 1} {\frac{s}{s ^{2} + 4}} : cos (2 t)

L^{- 1} {\frac{1}{s ^{2} + 4}} : \frac{1}{2} sin (2 t)

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

Refine

\frac{3}{4} H (t) + \frac{1}{4} t - \frac{3}{4} cos (2 t) - \frac{1}{4} \frac{1}{2} sin (2 t) : \frac{3}{4} H (t) + \frac{t}{4} - \frac{3}{4} cos (2 t) - \frac{1}{8} sin (2 t)

= \frac{3}{4} H (t) + \frac{t}{4} - \frac{3}{4} cos (2 t) - \frac{1}{8} sin (2 t)

\int ln (x) + sin (π x) d x

\frac{2 s + 3}{( s - 7 ) ^{4}}

(3 y^{2} + 2 x) d x + (6 x y + 16 y^{3}) d y = 0

\int sin (x) - 5 cos (x) d x

y^{^{'}} - 4 y = 4 e^{6 t}

What is inverse oflaplace ((3s+1))/(s^2(s^2+4)) ?
The answer to inverse oflaplace ((3s+1))/(s^2(s^2+4)) is 3/4 \H(t)+t/4-3/4 cos(2t)-1/8 sin(2t)