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)

y (1 + x^{2}) y^{^{'}} - x (9 + y^{2}) = 0

\int cos (x) - cos^{4} (x) d x

\int 9 x^{7} e^{- x^{4}} d x

y^{^{''}} - y^{^{'}} + 4 y = 2 sin (2 x)

\int_{0}^{7} 49 + x^{2} d x

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