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)

f (x) = x^{2} - 8, at x = 4

x y^{^{'}} = 3 y - 6 x^{2}

\int cos (π x) cos (4 π x) d x

3 x, 3, \frac{3}{x}

\int \frac{x}{2 ln ( x )} d x

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)