Inverse Laplace transform

If two continuous functions have the same transform, they are equal; that is why the inverse is well defined and can be read from a table. It is also linear: \(\mathcal L^{-1}\{\alpha F+\beta G\}=\alpha f+\beta g\), so it suffices to recognize each summand.

Rarely does \(F(s)\) come ready for the table. The usual tools are partial fractions, which split a quotient of polynomials into terms like \(\frac{A}{s-a}\), and completing the square, which turns \(s^2+2s+5\) into \((s+1)^2+4\) to recognize \(e^{-t}\sin 2t\). When adjusting numerators remember that \(\sin bt\) needs \(b\) on top: \(\frac{3}{s^2+4}=\frac32\cdot\frac{2}{s^2+4}\).

Guiding example: from \(Y(s)=\frac{s-3}{(s-1)(s-2)}=\frac{2}{s-1}-\frac{1}{s-2}\) we get \(y(t)=2e^{t}-e^{2t}\), the solution of the initial value problem that produced \(Y\). The poles of \(Y\) at \(s=1\) and \(s=2\) are exactly the exponents that appear.

A common mistake is getting the exponent sign wrong: \(\frac1{s+3}\) corresponds to \(e^{-3t}\), not \(e^{3t}\). The inverse is defined for \(t\ge0\), where the transform lives.