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

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

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

inverse laplace

\frac{25}{s ^{2} + 6 s + 25}

\frac{25}{4} e^{- 3 t} sin (4 t)

Solution steps

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

Expand

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

= L^{- 1} {25 \cdot \frac{1}{( s + 3 ) ^{2} + 16}}

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)}

= 25 L^{- 1} {\frac{1}{( s + 3 ) ^{2} + 16}}

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

= 25 e^{- 3 t} \frac{1}{4} sin (4 t)

Refine

25 e^{- 3 t} \frac{1}{4} sin (4 t) : \frac{25}{4} e^{- 3 t} sin (4 t)

= \frac{25}{4} e^{- 3 t} sin (4 t)

\int arctan (8 t) d t

(3 x y + 3 y - 4) d x + (x + 1)^{2} d y = 0

n = 0 \sum \infty n e^{- 8 n}

\frac{\partial}{\partial x} (e^{2 x - 3 y})

\frac{d}{d x} ((\frac{x ^{2} + 3}{x ^{2} - 3})^{4})

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