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

The answer to inverse oflaplace s/((s^2+5^2)^2) is (tsin(5t))/(10)

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

inverse laplace

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

\frac{t sin ( 5 t )}{10}

Solution steps

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

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{s}{( s ^{2} + 5 ^{2} ) ^{2}} : F (s) = \frac{1}{( s ^{2} + 5 ^{2} ) ^{2}}

= \frac{d}{d t} (L^{- 1} {\frac{1}{( s ^{2} + 5 ^{2} ) ^{2}}}) + L^{- 1} {\frac{1}{( s ^{2} + 5 ^{2} ) ^{2}}} (0)

L^{- 1} {\frac{1}{( s ^{2} + 5 ^{2} ) ^{2}}} : \frac{1}{250} (sin (5 t) - 5 t cos (5 t))

\frac{d}{d t} (\frac{1}{250} (sin (5 t) - 5 t cos (5 t))) = \frac{t sin ( 5 t )}{10}

Evaluate

\frac{1}{250} (sin (5 t) - 5 t cos (5 t))

at point

0 = 0

= \frac{t sin ( 5 t )}{10}

\int \frac{4 + x ^{2}}{x} d x

\frac{d}{d x} (\frac{x}{( x - 1 )})

x \to 0 lim (\frac{4 e ^{x} - 4}{x})

\frac{d}{d r} (\frac{1}{2 r} + \frac{1}{3 r ^{\frac{2}{3}}})

\frac{d}{d x} (f (h, x) (g (h, x) (h (x))))

What is inverse oflaplace s/((s^2+5^2)^2) ?
The answer to inverse oflaplace s/((s^2+5^2)^2) is (tsin(5t))/(10)