What is inverse oflaplace-(75)/(s^2+6s+25) ?

The answer to inverse oflaplace-(75)/(s^2+6s+25) is -75/4 e^{-3t}sin(4t)

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inverse oflaplace-(75)/(s^2+6s+25)

inverse laplace

- \frac{75}{s ^{2} + 6 s + 25}

- \frac{75}{4} e^{- 3 t} sin (4 t)

Solution steps

L^{- 1} {- \frac{75}{s ^{2} + 6 s + 25}}

Expand

\frac{75}{s ^{2} + 6 s + 25} : 75 \cdot \frac{1}{( s + 3 ) ^{2} + 16}

= L^{- 1} {- 75 \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)}

= - 75 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)

= - 75 e^{- 3 t} \frac{1}{4} sin (4 t)

Refine

- 75 e^{- 3 t} \frac{1}{4} sin (4 t) : - \frac{75}{4} e^{- 3 t} sin (4 t)

= - \frac{75}{4} e^{- 3 t} sin (4 t)

\frac{d y}{d x} = \frac{( x ^{2} + x y + y ^{2} )}{x ^{2}}

h \to 0 lim (\frac{h ^{2}}{3 ∣ h ∣})

\int \frac{1}{sin ( x ) - 1} d x

\frac{d}{d x} (ln (4 x + 9))

x \to - 6 lim (\frac{x ^{2} - 5}{6 - x})

What is inverse oflaplace-(75)/(s^2+6s+25) ?
The answer to inverse oflaplace-(75)/(s^2+6s+25) is -75/4 e^{-3t}sin(4t)