What is inverse oflaplace (400)/(s^2+12s+400) ?

The answer to inverse oflaplace (400)/(s^2+12s+400) is (200)/(sqrt(91))e^{-6t}sin(2sqrt(91)t)

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inverse oflaplace (400)/(s^2+12s+400)

inverse laplace

\frac{400}{s ^{2} + 12 s + 400}

\frac{200}{91} e^{- 6 t} sin (291 t)

Solution steps

L^{- 1} {\frac{400}{s ^{2} + 12 s + 400}}

Expand

\frac{400}{s ^{2} + 12 s + 400} : 400 \cdot \frac{1}{( s + 6 ) ^{2} + 364}

= L^{- 1} {400 \cdot \frac{1}{( s + 6 ) ^{2} + 364}}

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)}

= 400 L^{- 1} {\frac{1}{( s + 6 ) ^{2} + 364}}

L^{- 1} {\frac{1}{( s + 6 ) ^{2} + 364}} : e^{- 6 t} \frac{1}{2 91} sin (291 t)

= 400 e^{- 6 t} \frac{1}{2 91} sin (291 t)

Refine

400 e^{- 6 t} \frac{1}{2 91} sin (291 t) : \frac{200}{91} e^{- 6 t} sin (291 t)

= \frac{200}{91} e^{- 6 t} sin (291 t)

f (x) = x^{2} + 4, (- 3, 13)

\int \frac{cos ^{3} ( 3 t )}{sin ^{5} ( 3 t )} d t

y^{^{'}} + \frac{2 y}{t - 2} = \frac{sin ( t )}{t - 2}

\frac{\partial}{\partial x} (3 x y + 4 x z + 5 yz - 3)

\frac{d}{d x} (3 cot (x))

What is inverse oflaplace (400)/(s^2+12s+400) ?
The answer to inverse oflaplace (400)/(s^2+12s+400) is (200)/(sqrt(91))e^{-6t}sin(2sqrt(91)t)