Ordinary Diff Eq Exam Review Solutions 20

Ordinary Diff Eq Exam Review Solutions 20 - 22 1 Solutions...

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Unformatted text preview: 22 1 Solutions 16. The given 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 get for the general solution. 17. The given equation is in standard form, , tive is , and the integrating factor is by the integrating factor to get of which is a perfect derivative . Thus antiderivatives of both sides gives where is a constant. Now multiply by , an antideriva. Now multiply , the left hand side and taking to get for the general solution. 18. 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 . Now assume . Then taking antiderivatives of both sides gives where is a constant. Now multiply by to get for the general solution. If antiderivatives leads to and hence the general solution in this case. then taking is 19. In standard form we get . In this case , 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 and we get for the general solution. 20. Divide by to put the equation in the standard form In this case grating factor is get , an antiderivative is , and the inte. Now multiply by the integrating factor to ...
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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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