Ordinary Diff Eq Exam Review Solutions 17

Ordinary Diff Eq Exam Review Solutions 17 - 1 Solutions 19...

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Unformatted text preview: 1 Solutions 19 derivative . (Note that this is just the original left hand side of the equation. Thus if we had recognized that the left hand side was already a perfect derivative, the preliminary steps could have been skipped for this problem, and we could have proceeded directly to the next step.) Thus the equation can be written as and taking antiderivatives of both sides gives where is a constant. Now divide by to get for the general solution. . Now let 5. The general solution from Problem 4 is . So and . 6. Divide by to get to put the equation in the standard form , an antiderivative is , and the In this case integrating factor is . Now multiply the standard form equation by the integrating factor to get , the left hand side of which is a perfect derivative . Thus To integrate we consider the cases and separately. Case A simple substitution gives : Hence, . and so Case Use integration by parts to get : . Then . 7. We first put the equation in standard form and get In this case factor is , an antiderivative is , and the integrating . Now multiply by the integrating factor to get the left hand side of which is a perfect derivative and taking antiderivatives of both sides gives where is a constant. Now divide by general solution. . Thus to get for the new: please check 8. In this case multiply to get and the integrating factor is , which simplifies to . Now ...
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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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