What is inverse oflaplace (s+1)/(s^2-5s+6) ?

The answer to inverse oflaplace (s+1)/(s^2-5s+6) is -3e^{2t}+4e^{3t}

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inverse oflaplace (s+1)/(s^2-5s+6)

inverse laplace

\frac{s + 1}{s ^{2} - 5 s + 6}

- 3 e^{2 t} + 4 e^{3 t}

Solution steps

L^{- 1} {\frac{s + 1}{s ^{2} - 5 s + 6}}

Take the partial fraction of

\frac{s + 1}{s ^{2} - 5 s + 6} : - \frac{3}{s - 2} + \frac{4}{s - 3}

= L^{- 1} {- \frac{3}{s - 2} + \frac{4}{s - 3}}

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 - 2}} + L^{- 1} {\frac{4}{s - 3}}

L^{- 1} {\frac{3}{s - 2}} : 3 e^{2 t}

L^{- 1} {\frac{4}{s - 3}} : 4 e^{3 t}

= - 3 e^{2 t} + 4 e^{3 t}

y = x^{3} - 6 x^{2} + 9 x, y = x

\frac{\partial}{\partial x} (- x (x^{2} + y^{2} + z^{2})^{- \frac{3}{2}})

\int 100 - x d x

\int sec^{3} (d) x d x

x \to - \frac{5 π}{2} lim (sec (x))

What is inverse oflaplace (s+1)/(s^2-5s+6) ?
The answer to inverse oflaplace (s+1)/(s^2-5s+6) is -3e^{2t}+4e^{3t}