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We will nd the solution by rst nding the general

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Unformatted text preview: t problems, but usually y1 and y2 are more convenient since they do not involve complex numbers. Example. Consider now the initial value problem 16 y ￿￿ − 8 y ￿ + 145 y = 0, y (0) = −2, y ￿ (0) = 1. We will find the solution by first finding the general solution to the differential equation, then finding a particular solution corresponding to the initial conditions. First we find the general solution to the differential equation. We do so by assuming that the solution is in the form y (t) = ert for some constant r. The characteristic equation is 16 r2 − 8 r + 145 = 0. Upon completing the square, we can find...
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