Chapter1Section2

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

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Unformatted text preview: 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 L = y , we find a particular solution y H x L = y + x 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 L = 2. 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 back- wards, 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 After working through a few problems, you should quickly realize that the constant...
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This note was uploaded on 01/16/2012 for the course MATH 306 taught by Professor Keithemmert during the Fall '11 term at Tarleton.

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

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