What is inverse oflaplace 3/(s-5)-2/(s+7) ?

The answer to inverse oflaplace 3/(s-5)-2/(s+7) is 3e^{5t}-2e^{-7t}

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inverse oflaplace 3/(s-5)-2/(s+7)

inverse laplace

\frac{3}{s - 5} - \frac{2}{s + 7}

3 e^{5 t} - 2 e^{- 7 t}

Solution steps

L^{- 1} {\frac{3}{s - 5} - \frac{2}{s + 7}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{3}{s - 5}} - L^{- 1} {\frac{2}{s + 7}}

L^{- 1} {\frac{3}{s - 5}} : 3 e^{5 t}

L^{- 1} {\frac{2}{s + 7}} : 2 e^{- 7 t}

= 3 e^{5 t} - 2 e^{- 7 t}

\int b^{x} d x

\frac{d y}{d x} = \frac{( x ^{2} + 1 )}{e ^{y}}

\int (\frac{x ^{2}}{1 - x ^{2}}) d x

2 t cos (t)

- y d x + (x + x y) d y = 0

What is inverse oflaplace 3/(s-5)-2/(s+7) ?
The answer to inverse oflaplace 3/(s-5)-2/(s+7) is 3e^{5t}-2e^{-7t}