What is inverse oflaplace (3s^2)/((s^2+1)^2) ?

The answer to inverse oflaplace (3s^2)/((s^2+1)^2) is (3tcos(t)+3sin(t))/2

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inverse oflaplace (3s^2)/((s^2+1)^2)

inverse laplace

\frac{3 s ^{2}}{( s ^{2} + 1 ) ^{2}}

\frac{3 t cos ( t ) + 3 sin ( t )}{2}

Solution steps

L^{- 1} {\frac{3 s ^{2}}{( s ^{2} + 1 ) ^{2}}}

Take the partial fraction of

\frac{3 s ^{2}}{( s ^{2} + 1 ) ^{2}} : \frac{3}{s ^{2} + 1} - \frac{3}{( s ^{2} + 1 ) ^{2}}

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

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

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

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

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

= 3 sin (t) - \frac{3}{2} (sin (t) - t cos (t))

Simplify

3 sin (t) - \frac{3}{2} (sin (t) - t cos (t)) : \frac{3 t cos ( t ) + 3 sin ( t )}{2}

= \frac{3 t cos ( t ) + 3 sin ( t )}{2}

\int y \cdot e^{x \cdot y} d x

\int \frac{1}{( x + 5 ) ^{2}} d x

\int 2 x \frac{1}{x} d x

\frac{\partial}{\partial x} (e^{x} ln (y))

What is inverse oflaplace (3s^2)/((s^2+1)^2) ?
The answer to inverse oflaplace (3s^2)/((s^2+1)^2) is (3tcos(t)+3sin(t))/2