MAT244H1 Study Guide - Midterm Guide: Jordan Bell, Phase Portrait
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MAT244H1 Full Course Notes
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August 9, 2013: f = y(cid:48)2 x3 . 2y(cid:48) x3 . (cid:18) 2y(cid:48) (cid:19) x3 so. Using y(0) = 1 and y(1) = 1 gives us c1 = 1 and c2 = 2. Hence y = c1x4 + c2. y = 2x4 + 1. 2y + = 2y(cid:48)(cid:48) so 2y(cid:48)(cid:48) + 2y = . If you are lucky you can solve this by looking at it (if you remember that cos and sin solve y(cid:48)(cid:48) + y = 0). But you can solve this by nding the homogeneous solutions and then using variation of parameters also. y = c1 cos x + c2 sin x + . Since y(0) = 0 and y( ) = 1, we get c1 + . So y = cos x us = 1 and so c1 = 1. Now we use the nal condition to gure out c2. 2 (you aren"t expected to be see this just in your head, it.
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