What is inverse oflaplace (16)/(s^2+3s+16) ?

The answer to inverse oflaplace (16)/(s^2+3s+16) is (32}{sqrt(55)}e^{-\frac{3t)/2}sin((sqrt(55)t)/2)

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inverse oflaplace (16)/(s^2+3s+16)

inverse laplace

\frac{16}{s ^{2} + 3 s + 16}

\frac{32}{55} e^{- \frac{3 t}{2}} sin (\frac{55 t}{2})

Solution steps

L^{- 1} {\frac{16}{s ^{2} + 3 s + 16}}

Expand

\frac{16}{s ^{2} + 3 s + 16} : 16 \cdot \frac{1}{( s + \frac{3}{2} ) ^{2} + \frac{55}{4}}

= L^{- 1} {16 \cdot \frac{1}{( s + \frac{3}{2} ) ^{2} + \frac{55}{4}}}

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)}

= 16 L^{- 1} {\frac{1}{( s + \frac{3}{2} ) ^{2} + \frac{55}{4}}}

L^{- 1} {\frac{1}{( s + \frac{3}{2} ) ^{2} + \frac{55}{4}}} : e^{- \frac{3 t}{2}} \frac{2}{55} sin (\frac{55 t}{2})

= 16 e^{- \frac{3 t}{2}} \frac{2}{55} sin (\frac{55 t}{2})

Refine

16 e^{- \frac{3 t}{2}} \frac{2}{55} sin (\frac{55 t}{2}) : \frac{32}{55} e^{- \frac{3 t}{2}} sin (\frac{55 t}{2})

= \frac{32}{55} e^{- \frac{3 t}{2}} sin (\frac{55 t}{2})

\int_{0}^{2} \frac{x}{x ^{2} + 1} d x

(\frac{y}{2} + 2 x) d x + (- 1) d y = 0

y = 3, y = x, x = 0

\frac{x d y}{d x} + x y = 1 - y

\frac{d y}{d x} = - 2 y

What is inverse oflaplace (16)/(s^2+3s+16) ?
The answer to inverse oflaplace (16)/(s^2+3s+16) is (32}{sqrt(55)}e^{-\frac{3t)/2}sin((sqrt(55)t)/2)