What is inverse oflaplace 1/(s^2+7) ?

The answer to inverse oflaplace 1/(s^2+7) is 1/(sqrt(7))sin(sqrt(7)t)

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inverse oflaplace 1/(s^2+7)

inverse laplace

\frac{1}{s ^{2} + 7}

\frac{1}{7} sin (7 t)

Solution steps

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

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

Use the constant multiplication property of Inverse Laplace Transform:

For function

f (t)

and constant

a : L^{- 1} {a \cdot f (t)} = a \cdot L^{- 1} {f (t)}

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

Use Inverse Laplace Transform table:

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

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

= \frac{1}{7} sin (7 t)

\int (n + 3)^{3} d n

\int ln^{2} (x - 9) d x

ln (x^{2}) - \frac{1}{x ^{2}} - 5

y^{3} = 126 - x^{3}, at x = 5

\frac{d}{d x} (2^{3} e^{2})

What is inverse oflaplace 1/(s^2+7) ?
The answer to inverse oflaplace 1/(s^2+7) is 1/(sqrt(7))sin(sqrt(7)t)