Inverse Laplace Transform
Formula: $E = mc^2$
Problem
Find the inverse Laplace transform of:
\[F(s) = \frac{3s - 4}{s^2 - 3s + 2}\]
Solution
Step 1: Factor the denominator
\[F(s) = \frac{3s - 4}{s^2 - 3s + 2} = \frac{3s - 4}{(s - 1)(s - 2)}\]
Step 2: Partial fraction decomposition
\[F(s) = \frac{A}{s - 1} + \frac{B}{s - 2}\]
Solving for A and B:
\[F(s) = \frac{1}{s - 1} + \frac{2}{s - 2}\]
Step 3: Apply inverse Laplace transform
\[f(t) = \mathcal{L}^{-1}\{F(s)\}\]
\[f(t) = \mathcal{L}^{-1}\left\{\frac{1}{s - 1} + \frac{2}{s - 2}\right\}\]
\[f(t) = \mathcal{L}^{-1}\left\{\frac{1}{s - 1}\right\} + 2\mathcal{L}^{-1}\left\{\frac{1}{s - 2}\right\}\]
Step 4: Use table of inverse transforms
Function
Laplace Transform
\(e^{at}\)
\(\frac{1}{s-a}\)
\[f(t) = e^t + 2e^{2t}\]
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