Ordinary Diff Eq Exam Review Solutions 111

Ordinary Diff Eq Exam Review Solutions 111 - 1 Solutions so...

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Unformatted text preview: 1 Solutions so that for Putting these three pieces together, we find that the solution is 113 . if if if if 28. The general solution of on any interval is found by using the integrating factor . Using this integrating factor the general solution of on the interval is found to be . The initial condition actually occurs at the end of this interval, but by continuity we can substitute in this formula to get so . Hence on the interval the solution is . The general solution of on the interval is and this gives so and on . Putting these two pieces together, we find that the solution is if if 29. The characteristic polynomial of the equation is so the homogeneous equation has the solution for constants and . On the interval the equation has a particular solution so the general solution has the form . The initial conditions give and . Solving gives , so on . By continuity it follows that and and these constitute the initial values for the equation on the interval . The general solution on this interval is and at we get and . Solving for and gives and so that . Putting the two pieces together gives if 30. The characteristic polynomial of the equation is so the homogeneous equation has the solution for constants and , which is valid on the interval ...
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This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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