Ordinary Diff Eq Exam Review Solutions 22

Ordinary Diff Eq Exam Review Solutions 22 - 24 1 Solutions...

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Unformatted text preview: 24 1 Solutions 24. In this case the given diﬀerential equation is in standard form. We have , 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. However, the right hand side has no closed form antiderivative. Using Corollary 8 to can write 25. Divide by to put the equation in the standard form 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 for where is a constant. Now multiply by to get the general solution. Letting gives so and gal lbs min gal gal output rate min 26. input rate: input rate output rate: Since input rate lbs . min lbs gal lbs . min output rate we get the initial value problem Simplifying and putting in standard form gives . The coeﬃ- cient function is , , and the integrating factor is . Thus . Integrating and simplifying gives The initial condition implies so . The concentration of the brine solution is now obtained by dividing by the volume which is gallons: . ...
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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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