What is inverse oflaplace (3(1-4x))/(58(x^2+9)) ?

The answer to inverse oflaplace (3(1-4x))/(58(x^2+9)) is (sin(3t)-12cos(3t))/(58)

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inverse oflaplace (3(1-4x))/(58(x^2+9))

inverse laplace

\frac{3 ( 1 - 4 x )}{58 ( x ^{2} + 9 )}

\frac{sin ( 3 t ) - 12 cos ( 3 t )}{58}

Solution steps

L^{- 1} {\frac{3 ( - 4 x + 1 )}{58 ( x ^{2} + 9 )}}

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{3}{58} L^{- 1} {\frac{1 - 4 x}{x ^{2} + 9}}

L^{- 1} {\frac{1 - 4 x}{x ^{2} + 9}} : \frac{1}{3} sin (3 t) - 4 cos (3 t)

= \frac{3}{58} (\frac{1}{3} sin (3 t) - 4 cos (3 t))

Refine

\frac{3}{58} (\frac{1}{3} sin (3 t) - 4 cos (3 t)) : \frac{sin ( 3 t ) - 12 cos ( 3 t )}{58}

= \frac{sin ( 3 t ) - 12 cos ( 3 t )}{58}

sin (ln (cos (t)))

\int x^{2} 6 x + 6 d x

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

(2 x^{3} - x^{2} y) + (2 x^{2} y - x^{3}) y^{^{'}} = 0

\frac{d}{d x} (arctan (\frac{x + y}{1 - x y}))

What is inverse oflaplace (3(1-4x))/(58(x^2+9)) ?
The answer to inverse oflaplace (3(1-4x))/(58(x^2+9)) is (sin(3t)-12cos(3t))/(58)