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Consider the differential equation y" - y' - 2y = 0. What is the solution to this equation?

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Final answer:

The general solution to the differential equation y'' - y' - 2y = 0 is y = c1e^(2t) + c2e^(-t), where c1 and c2 are constants.

Step-by-step explanation:

The solution to the given differential equation y'' - y' - 2y = 0 can be found by assuming a solution of the form y = e^(rt), where r is a constant. Plugging this solution back into the differential equation, we get r^2e^(rt) - re^(rt) - 2e^(rt) = 0. Simplifying this equation, we obtain the characteristic equation r^2 - r - 2 = 0. Factoring this equation, we find (r-2)(r+1) = 0, which gives us two possible values for r: r = 2 and r = -1. Therefore, the general solution to the differential equation is y = c1e^(2t) + c2e^(-t), where c1 and c2 are constants.

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