Ordinary Diff Eq Exam Review Solutions 16

Ordinary Diff Eq Exam Review Solutions 16 - 18 1 Solutions...

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Unformatted text preview: 18 1 Solutions for the general solution of the equation. To find the constant to satisfy the initial condition , substitute into the general solution to get . Hence , and the solution of the initial value problem is 2. Divide by to put the equation in the standard form In this case , an antiderivative is , and the integrating factor is . (We do not need since we are working near where .) 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. Letting gives so and 3. This equation is already in standard form. In this case 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 and taking antiderivatives of both sides gives where is a constant. Now multiply by the general solution. Letting gives 4. Divide by , an . . Thus to get so for 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 ...
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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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