# H1_130 - x = 2-1 3-2 ! x + 1-1 ! e t . First, note that x =...

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Math 415 Homework 1 2.1.31 y 0 + y = te - t + 1 -5 -4 -3 -2 -1 0 1 2 3 4 5 -2.5 2.5 Figure 1: The direction ﬁeld for the diﬀerential equation with the graph of y = te - t + 1, i.e. where the slope is zero. We can see that as t → ∞ solutions will tend towards the line y = 1. The integrating factor in this case will be μ ( t ) = e t , and with this included the diﬀerential equation becomes ( e t y ) 0 = t + e t . Integrating gives us e t y = t 2 / 2+ e t + C . Hence y ( t ) = 1 2 t 2 e - t +1+ Ce - t and direct inspection shows us that lim t →∞ y ( t ) = 1. 1.1.26 y 0 = - 2 + t - y -5 -4 -3 -2 -1 0 1 2 3 4 5 -2.5 2.5 Figure 2: The direction ﬁeld for the diﬀerential equation with the graph of y = - 2 + t . As t → ∞ , y ( t ) → ∞ . 1

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7.2.23 We want to verify that x = 1 0 ! e t + 2 1 1 ! te t is a solution to the diﬀerential equation
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Unformatted text preview: x = 2-1 3-2 ! x + 1-1 ! e t . First, note that x = 1 ! e t + 2 1 1 ! ( e t + te t ) = 3 2 ! e t + 2 1 1 ! te t . Now, on the other side of the diﬀerential equation we get: 2-1 3-2 ! x + 1-1 ! e t = 2-1 3-2 ! 1 ! e t + 2 1 1 ! te t ! + 1-1 ! e t = 2 · 1 + (-1) · 3 · 1 + (-2) · ! e t + 2 2 · 1 + (-1) · 1 3 · 1 + (-2) · 1 ! te t + 1-1 ! e t = 2 3 ! e t + 2 1 1 ! te t + 1-1 ! e t = 3 2 ! e t + 2 1 1 ! te t = x . So the given vector equation is a solution to the diﬀerential equation. 2...
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## This note was uploaded on 07/26/2011 for the course MATH 415 taught by Professor Costin during the Spring '07 term at Ohio State.

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H1_130 - x = 2-1 3-2 ! x + 1-1 ! e t . First, note that x =...

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