What is inverse oflaplace (25)/(x(x^2+6x+25)) ?

The answer to inverse oflaplace (25)/(x(x^2+6x+25)) is \H(t)-e^{-3t}cos(4t)-3/4 e^{-3t}sin(4t)

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inverse oflaplace (25)/(x(x^2+6x+25))

inverse laplace

\frac{25}{x ( x ^{2} + 6 x + 25 )}

H (t) - e^{- 3 t} cos (4 t) - \frac{3}{4} e^{- 3 t} sin (4 t)

Solution steps

L^{- 1} {\frac{25}{x ( x ^{2} + 6 x + 25 )}}

Take the partial fraction of

\frac{25}{x ( x ^{2} + 6 x + 25 )} : \frac{1}{x} + \frac{- x - 6}{x ^{2} + 6 x + 25}

= L^{- 1} {\frac{1}{x} + \frac{- x - 6}{x ^{2} + 6 x + 25}}

Expand

\frac{- x - 6}{x ^{2} + 6 x + 25} : - \frac{x + 3}{( x + 3 ) ^{2} + 16} - 3 \cdot \frac{1}{( x + 3 ) ^{2} + 16}

= L^{- 1} {\frac{1}{x} - \frac{x + 3}{( x + 3 ) ^{2} + 16} - 3 \cdot \frac{1}{( x + 3 ) ^{2} + 16}}

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{1}{x}} - L^{- 1} {\frac{x + 3}{( x + 3 ) ^{2} + 16}} - 3 L^{- 1} {\frac{1}{( x + 3 ) ^{2} + 16}}

L^{- 1} {\frac{1}{x}} : H (t)

L^{- 1} {\frac{x + 3}{( x + 3 ) ^{2} + 16}} : e^{- 3 t} cos (4 t)

L^{- 1} {\frac{1}{( x + 3 ) ^{2} + 16}} : e^{- 3 t} \frac{1}{4} sin (4 t)

= H (t) - e^{- 3 t} cos (4 t) - 3 e^{- 3 t} \frac{1}{4} sin (4 t)

Refine

H (t) - e^{- 3 t} cos (4 t) - 3 e^{- 3 t} \frac{1}{4} sin (4 t) : H (t) - e^{- 3 t} cos (4 t) - \frac{3}{4} e^{- 3 t} sin (4 t)

= H (t) - e^{- 3 t} cos (4 t) - \frac{3}{4} e^{- 3 t} sin (4 t)

\frac{d x}{d t} = x - 12

\frac{d}{d x} (sin (\frac{1}{x}) \cdot x^{3})

cos (sin (tan (5 x)))

x \to - 1 lim (x^{3} - 1)

\int \frac{cos ( \frac{π}{x ^{15}} )}{x ^{16}} d x

What is inverse oflaplace (25)/(x(x^2+6x+25)) ?
The answer to inverse oflaplace (25)/(x(x^2+6x+25)) is \H(t)-e^{-3t}cos(4t)-3/4 e^{-3t}sin(4t)