Ordinary Diff Eq Exam Review Solutions 19

# Ordinary Diff Eq Exam Review Solutions 19 - 1 Solutions 13...

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Unformatted text preview: 1 Solutions 13. The given equation is in standard form, , and the integrating factor is by the integrating factor to get 21 , an antiderivative is . Now multiply 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. To satisfy the initial condition, , so . Thus, the solution of the initial value problem is 14. 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 is a constant. Now multiply by to get the general solution. where for 15. The given linear 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 and taking antiderivatives of both sides gives is a constant. Now multiply by to and we get for the general solution. Letting gives and . Thus where so ...
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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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