What is inverse oflaplace 1/(s^2+25) ?

The answer to inverse oflaplace 1/(s^2+25) is 1/5 sin(5t)

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inverse oflaplace 1/(s^2+25)

inverse laplace

\frac{1}{s ^{2} + 25}

\frac{1}{5} sin (5 t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} + 25}}

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

Use Inverse Laplace Transform table:

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

L^{- 1} {\frac{5}{s ^{2} + 5 ^{2}}} = sin (5 t)

= \frac{1}{5} sin (5 t)

f (x) = x^{3} - 5 x^{2}

\frac{\partial}{\partial x} (e^{y} + 4 x^{3} y)

\frac{- 1}{s}

\int - cos (3 x) d x

\frac{1}{x - 1}, 22

What is inverse oflaplace 1/(s^2+25) ?
The answer to inverse oflaplace 1/(s^2+25) is 1/5 sin(5t)