What is inverse oflaplace (3s^2)/((s^2+4)^2) ?

The answer to inverse oflaplace (3s^2)/((s^2+4)^2) is (3sin(2t)+6tcos(2t))/4

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inverse oflaplace (3s^2)/((s^2+4)^2)

inverse laplace

\frac{3 s ^{2}}{( s ^{2} + 4 ) ^{2}}

\frac{3 sin ( 2 t ) + 6 t cos ( 2 t )}{4}

Solution steps

L^{- 1} {\frac{3 s ^{2}}{( s ^{2} + 4 ) ^{2}}}

Take the partial fraction of

\frac{3 s ^{2}}{( s ^{2} + 4 ) ^{2}} : \frac{3}{s ^{2} + 4} - \frac{12}{( s ^{2} + 4 ) ^{2}}

= L^{- 1} {\frac{3}{s ^{2} + 4} - \frac{12}{( s ^{2} + 4 ) ^{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} + 4}} - L^{- 1} {\frac{12}{( s ^{2} + 4 ) ^{2}}}

L^{- 1} {\frac{3}{s ^{2} + 4}} : \frac{3}{2} sin (2 t)

L^{- 1} {\frac{12}{( s ^{2} + 4 ) ^{2}}} : \frac{3}{4} (sin (2 t) - 2 t cos (2 t))

= \frac{3}{2} sin (2 t) - \frac{3}{4} (sin (2 t) - 2 t cos (2 t))

Simplify

\frac{3}{2} sin (2 t) - \frac{3}{4} (sin (2 t) - 2 t cos (2 t)) : \frac{3 sin ( 2 t ) + 6 t cos ( 2 t )}{4}

= \frac{3 sin ( 2 t ) + 6 t cos ( 2 t )}{4}

\int \frac{1}{- x ^{2} + 4 x + 5} d x

\frac{\partial}{\partial x} (r + 2 r^{4} x + 4 x)

(4 + y^{2}) d x + (9 + x^{2}) d y = 0

\int \frac{x ^{4} + 2 x + 6}{x ^{3} + x ^{2} - 2 x} d x

\int \frac{11}{11 + e ^{x}} d x

What is inverse oflaplace (3s^2)/((s^2+4)^2) ?
The answer to inverse oflaplace (3s^2)/((s^2+4)^2) is (3sin(2t)+6tcos(2t))/4