MATH 3001 Study Guide - Final Guide: Linear Map, Integrating Factor
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2 y2 = ln|t2 + 1| + c, y(t) = pc + ln(t2 + 1), (b) we have an initial condition y(1) = 2, so we take the positive solution from above: Substituting back into the general solution gives y(t) =p4 ln 2 + ln(t2 + 1), (t2 + 1)(cid:21) Picard"s theorem: t/f + explanation: both conditions of picard"s theorem hold for the following ivp: dv dt. If we treat this de as v = f (t, v), we have f (t, v) = kv2/3. This function is continuous for all values of t and v, and so picard"s theorem guarantees that a solution exists for any initial condition. 3v1/3 , we see that this function is discontinuous at v = 0. However, given our initial condition v(3) = 4, it is possible to draw a rectangle around the ic such that fy is continous everywhere in the rectangle.
