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

The answer to inverse oflaplace 5/((s^2+5^2)^2) is 1/50 (sin(5t)-5tcos(5t))

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inverse oflaplace 5/((s^2+5^2)^2)

inverse laplace

\frac{5}{( s ^{2} + 5 ^{2} ) ^{2}}

\frac{1}{50} (sin (5 t) - 5 t cos (5 t))

Solution steps

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

= L^{- 1} {\frac{1}{50} \cdot \frac{2 \cdot 5 ^{3}}{( s ^{2} + 5 ^{2} ) ^{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}{50} L^{- 1} {\frac{2 \cdot 5 ^{3}}{( s ^{2} + 5 ^{2} ) ^{2}}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{2 a ^{3}}{( s ^{2} + a ^{2} ) ^{2}}} = sin (a t) - a t cos (a t)

L^{- 1} {\frac{2 \cdot 5 ^{3}}{( s ^{2} + 5 ^{2} ) ^{2}}} = sin (5 t) - 5 t cos (5 t)

= \frac{1}{50} (sin (5 t) - 5 t cos (5 t))

n = 3 \sum \infty \frac{3 ^{n}}{1 1 ^{n}}

\int \frac{y ^{2} - 49}{y} d y

\int sin (12 x) cos (5 x) d x

\int_{- \infty}^{0} \frac{1}{5 - 4 x} d x

\frac{d}{d x} (ln (13 x) + 13)

What is inverse oflaplace 5/((s^2+5^2)^2) ?
The answer to inverse oflaplace 5/((s^2+5^2)^2) is 1/50 (sin(5t)-5tcos(5t))