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Unformatted text preview: 1 AMS 361: Applied Calculus IV by Prof. Y. Deng Quiz 2 Tuesday (10/18/2011) at 5:20PM6:40PM in 001 ESS (1) Closed Book with 1page (doublesided 8.5x11) selfprepared formulas. (2) Do any 2of the 3 problems.If all 3 are attempted, only the 2 best will be graded. (3) No points for guessing work and for solutions without appropriate intermediate steps; Partial credits are given only for steps that are relevant to the solutions. (4) No name, no grade and no question asked or answered. SB ID: Class ID: Name: Problems Score Remarks Q11 Q12 Q13 Total Score Q21 (7.5 Points): The first order linear DE can be expressed alternatively as ࠵ ࠵ ࠵ − ࠵ ( ࠵ ) ࠵¡ + ࠵¡ = ࠵ where functions ࠵ ࠵ ≠ ࠵ and ࠵ ࠵ ≠ ࠵ are given. Please do the following: (1) Check if this DE is Exact (1.5 Points);(2) If no, convert it to an Exact DE (1.5 Points); (3) Solve the DE using the Exact Equation method and your solution may be expressed in terms of the functions ࠵ ࠵ and ࠵ ࠵ (3.0 Points); (4) If ࠵ ࠵ = ࠵ ࠵ and ࠵ ࠵ = ࠵¡¢ ࠵ ࠵ , get the specific solution (1.5 Points). Solution: (1) Check if this DE is Exact (1.5 Points) ࠵ ࠵ , ࠵ = ࠵ ࠵ ࠵ − ࠵ ࠵ , ࠵ ࠵ , ࠵ = 1 ࠵ ( ࠵ ) = ࠵¡ ( ࠵ , ࠵ ) ࠵¡ ≠ ࠵¡ ࠵ , ࠵ ࠵¡ = So, this DE is not an Exact DE. (2) If no, convert it to an Exact DE (1.5 Points) For converting given DE into Exact DE, we need to calculate two ratios. 2 ࠵ ࠵ = !" ( ! , ! ) !" − !" ! , ! !" ࠵ ( ࠵ , ࠵ ) = ࠵ ( ࠵ ) Because f(x) is a function of x, we have the integration factor ࠵ ࠵ = exp ࠵ ࠵ ࠵¡ = exp ࠵ ࠵ ࠵¡ By multiplying the integration factor into given DE, we can obtain the Exact DE. By multiplying the integration factor into given DE, we can obtain the Exact DE....
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