Ordinary Diff Eq Exam Review Solutions 21

Ordinary Diff Eq Exam Review Solutions 21 - 1 Solutions 23...

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Unformatted text preview: 1 Solutions 23 the left hand side of which is a perfect derivative . Thus and taking antiderivatives of both sides gives where is a constant. Now multiply by for the general solution. and we get 21. The given differential equation is in standard form, , an antiderivative is , and the integrating factor is . Now multiply by the integrating factor to get , the left hand side of which is a perfect derivative . Thus and taking antiderivatives of both sides gives where is a constant. Now multiply by to and we get for the general solution. 22. The given differential equation is in standard form, , an antiderivative is , and the integrating factor is . Now multiply by the integrating factor to get , the left hand side of which is a perfect derivative . Thus . Taking antiderivatives of both sides and using integration by parts gives where is a constant. Now multiply by to get for the general solution. Letting gives so and 23. Divide by to put the equation in the standard form In this case , an antiderivative is , and the 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 and taking antiderivatives of both sides gives where is a constant. Now multiply by and we get for the general solution. Letting gives so and ...
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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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