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Unformatted text preview: t problems, but usually y1 and y2 are more convenient since they do not involve complex
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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- Spring '09