What is inverse oflaplace 1/(2x^2+x+1) ?

The answer to inverse oflaplace 1/(2x^2+x+1) is (2e^{-t/4}sin((sqrt(7)t)/4))/(sqrt(7))

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inverse oflaplace 1/(2x^2+x+1)

inverse laplace

\frac{1}{2 x ^{2} + x + 1}

\frac{2 e ^{- \frac{t}{4}} sin ( \frac{7 t}{4} )}{7}

Solution steps

L^{- 1} {\frac{1}{2 x ^{2} + x + 1}}

Expand

\frac{1}{2 x ^{2} + x + 1} : \frac{1}{2} \cdot \frac{1}{( x + \frac{1}{4} ) ^{2} + \frac{7}{16}}

= L^{- 1} {\frac{1}{2} \cdot \frac{1}{( x + \frac{1}{4} ) ^{2} + \frac{7}{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)}

= \frac{1}{2} L^{- 1} {\frac{1}{( x + \frac{1}{4} ) ^{2} + \frac{7}{16}}}

L^{- 1} {\frac{1}{( x + \frac{1}{4} ) ^{2} + \frac{7}{16}}} : e^{- \frac{t}{4}} \frac{4}{7} sin (\frac{7 t}{4})

= \frac{1}{2} e^{- \frac{t}{4}} \frac{4}{7} sin (\frac{7 t}{4})

Refine

\frac{1}{2} e^{- \frac{t}{4}} \frac{4}{7} sin (\frac{7 t}{4}) : \frac{2 e ^{- \frac{t}{4}} sin ( \frac{7 t}{4} )}{7}

= \frac{2 e ^{- \frac{t}{4}} sin ( \frac{7 t}{4} )}{7}

\frac{d N}{d t} = 5 - N, N (2) = 2

\int (\frac{e ^{x} - e ^{- x}}{2}) d x

\int \frac{\frac{1}{e ^{x} + e ^{2 x}}}{- e ^{- 3 x}} d x

x \to 2 - lim (\frac{x - 1}{∣ x - 1 ∣})

\frac{d}{d x} (3 x^{3} y^{2})

What is inverse oflaplace 1/(2x^2+x+1) ?
The answer to inverse oflaplace 1/(2x^2+x+1) is (2e^{-t/4}sin((sqrt(7)t)/4))/(sqrt(7))