What is inverse oflaplace {5/(s^2+49)} ?

The answer to inverse oflaplace {5/(s^2+49)} is 5/7 sin(7t)

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inverse oflaplace {5/(s^2+49)}

inverse laplace

{\frac{5}{s ^{2} + 49}}

\frac{5}{7} sin (7 t)

Solution steps

L^{- 1} {\frac{5}{s ^{2} + 49}}

= L^{- 1} {\frac{5}{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{5}{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{5}{7} sin (7 t)

(a + 3 y^{2} + b)^{\frac{1}{2}}

f (x) = \frac{π ^{2}}{x ^{2} + 4 x}

\int - \frac{1}{27} sin (3 x) d x

\int \frac{3 - 5 x}{9 - 5 x ^{2}} d x

\frac{d}{d x} (arccos (\frac{6}{x}))

What is inverse oflaplace {5/(s^2+49)} ?
The answer to inverse oflaplace {5/(s^2+49)} is 5/7 sin(7t)