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Unformatted text preview: 1 AMS 361: Applied Calculus IV by Prof Y. Deng Homework 6 Assignment Date: Wednesday (02/29/2012) Collection Date: Wednesday (03/21/2012) Grade: Each problem is worth 10 points. Problem 6.1 Find the general solution of the following equation ࠵ − 3 ! ࠵ − 2 ! ࠵ ! − ࠵ + 2 ࠵ ( ࠵ ) = where ࠵ ≡ ࠵ ࠵¡ Solution: From the DE, general solution is given by sum of general solutions of the three DE’s ࠵ − 3 ! ࠵ = ࠵ − 2 ! ࠵ = ࠵ ! − ࠵ + 2 ࠵ = For ࠵ − 3 ! ࠵ = , solution is ࠵ = ࠵ ! + ࠵ ! ࠵ ࠵ ! ! For ࠵ − 2 ! ࠵ = , solution is ࠵ = ࠵ ! + ࠵ ! ࠵ + ࠵ ! ࠵ ! ࠵ ! ! For ࠵ ! − ࠵ + 2 ࠵ = , solution is ࠵ = ࠵ ! ࠵ ! ! + ࠵ ! ࠵ ! ! , where ࠵ ! , ࠵ ! are zeros of the characteristic polynomial ࠵ ! − ࠵ + 2 = ⇒ ࠵ ! , ! = ! ! ± ! ! ! If we like to express the solution in realvalue (at least it appears), it can be rewritten as ࠵ = ࠵ ! ! ࠵ ! sin ! ! ࠵ + ࠵ ! cos ! ! ࠵ The general solution to the original DE is then ࠵ = ࠵ ! + ࠵ ! ࠵ ࠵ ! ! + ࠵ ! + ࠵ ! ࠵ + ࠵ ! ࠵ ! ࠵ ! ! + ࠵ ! ! ࠵ ! sin 7 2 ࠵ + ࠵ ! cos 7 2 ࠵ . Problem 6.2 Find the general solution of the following equation ࠵ ࠵ ࠵¡ − ࠵ ! ࠵ ࠵ = where ࠵ = constant and ࠵ = positive integer 2 Solution: The solution to ࠵ ! !" − ࠵ ࠵ ! = has multiplicity ࠵ , so the general solution will contain terms similar but linearly independent from the first order equation....
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This note was uploaded on 03/22/2012 for the course AMS 361 taught by Professor Staff during the Spring '08 term at SUNY Stony Brook.
 Spring '08
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