Chapter1Section2

# Chapter1Section2 - 1.2: Integrals as General and Particular...

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1.2: Integrals as General and Particular Solutions ± Basic Definitions Consider a first order differential equation, dy dx = f H x L where the right hand side does not depend upon y H x L . The general solution is y H x L = ± f H x L ± x + C where C is some constant. For the first order initial value problem, dy dx = f H x L , y H x 0 L = y 0 , we find a particular solution y H x L = y 0 + ± x 0 x f H u L ± u or we find the general solution and apply the initial conditions to find the unknown constant C . ± Example Suppose that du dt = 2 t 2 - 3 t + 5. A. Find the general solution. B. Find the particular solution given u H 0 L = 2.

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± Motion Suppose x H t L is the p o s i t i o n of some critter. Then v H t L = x ' H t L is the v e l o c i t y and v H t L is the s p e e d . Note that velocity is really speed with a direction - negative is backwards, and positive is forwards. Also, this is a differential equation: dx dt = v H t L which can be solved: x H t L = ± v H t L ± t + C . Finally, a H t L = v ' H t L = x '' H t L is the a c c e l e r a t i o n . Again, we have differential equations and x ' H t L = v H t L = ± a H t L ± t + C 1 x H t L = ± v H u L ± u + C 2 = ± ± H a H u L ± u + C 1 L ± t + C 2
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## This note was uploaded on 01/13/2012 for the course MATH 306 taught by Professor Keithemmert during the Spring '11 term at Tarleton.

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Chapter1Section2 - 1.2: Integrals as General and Particular...

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