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What is the inverse Laplace transform of [math] \frac{e^{(-\tau)}}{s (s^2+s-2)}[/math]?
Can you tell me what the inverse Laplace transform of [math] \frac{e^{(-\tau)}}{s (s^2+s-2)}[/math] is?
Sure, the inverse Laplace transform of [math] \frac{e^{(-\tau)}}{s (s^2+s-2)}[/math] is [math] \frac{1}{2}e^{(-\tau)}(\cosh(\tau)-\sinh(\tau))u(t-\tau)-\frac{1}{2}e^{(-\tau)}(u(t-1)-u(t+2))[/math]. Is there anything else you would like to know?