Ordinary Diff Eq Exam Review Solutions 110

Ordinary Diff Eq Exam Review Solutions 110 - 112 1...

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Unformatted text preview: 112 1 Solutions . The general solution of on is . Thus and solve for to get , so that for . Continuity will then give , which will provide the initial condition for the next interval . The general solution to on is and the constant is obtained from the initial condition , which gives , so that for . Putting these three pieces together, we find that the solution is if if if 26. The general solution of on the interval is found by using the integrating factor . The general solution is and the initial condition gives . By continuity we must have , which will provide the initial condition for the next interval . The general solution of on is . Setting found from the interval equal to found from the interval and solving for gives . Thus the complete solution is if if 27. The general solution of on any interval is found by using the integrating factor . The general solution on the interval is and since the initial condition is , the solution on is . Continuity then given , which will be the initial condition for the interval . The general solution of on the interval is and the initial condition gives so that . Thus for . Continuity of at will then give , which will provide the initial condition for the next interval . The general solution of on is . Thus and solve for to get , so that for . Continuity will then give , which will provide the initial condition for the next interval . The general solution to on is and the constant is obtained from the initial condition , which gives , ...
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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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