MAT244H1 Study Guide - Midterm Guide: Jordan Bell, Integrating Factor, Integral Equation

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25 Oct 2018
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MAT244H1 Full Course Notes
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June 2, 2013: to have (y )(cid:48) = y(cid:48) 2 t y, we need (cid:48)y = 2 t y and so. Integrating, ln = 2 ln t = ln(t 2). Since the left hand side of the equation is equal to (y )(cid:48), we have (yt 2)(cid:48) = tet. Then we integrate (integrate tet by parts) and get yt 2 = tet et + c. 1 = e1 e1 + c, so c = 1. Try an integrating factor that is only a function of t, so y = 0. 1 t ln = ln t = ln(t 1). Multiplying by this the equation becomes exact, and our new m and n are m = t4y5 and n = t5y4 + 1. Then integrating m with respect to t gives , Thus t5y4 + g(cid:48)(y) = t5y4 + 1. g(cid:48)(y) = 1, and g(y) = y. Using the initial condition y(0) = 1, t5y5.

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