What is inverse oflaplace 1/(s+1(-2s+10)) ?

The answer to inverse oflaplace 1/(s+1(-2s+10)) is -e^{10t}

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inverse oflaplace 1/(s+1(-2s+10))

inverse laplace

\frac{1}{s + 1 ( - 2 s + 10 )}

- e^{10 t}

Solution steps

L^{- 1} {\frac{1}{s + 1 \cdot ( - 2 s + 10 )}}

Factor

s + 1 \cdot (- 2 s + 10) : - (s - 10)

= L^{- 1} {\frac{1}{- ( s - 10 )}}

= L^{- 1} {- 1 \cdot \frac{1}{s - 10}}

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)}

= - 1 \cdot L^{- 1} {\frac{1}{s - 10}}

Use Inverse Laplace Transform table:

L^{- 1} {\frac{1}{s - a}} = e^{a t}

L^{- 1} {\frac{1}{s - 10}} = e^{10 t}

= - 1 \cdot e^{10 t}

= - e^{10 t}

n = 1 \sum \infty \frac{( - 5 ) ^{n}}{3 ^{2 n + 1}}

\frac{d}{d x} (32 cos (x))

n = 1 \sum \infty \frac{x ^{n}}{1 6 ^{n}}

f (x) = - \frac{4 x}{x ^{2} + 1}, at x = 1

\frac{d y}{d x}, x^{\frac{2}{3}} + y^{\frac{2}{3}} = 20

What is inverse oflaplace 1/(s+1(-2s+10)) ?
The answer to inverse oflaplace 1/(s+1(-2s+10)) is -e^{10t}