Ordinary Diff Eq Exam Review Solutions 28

Ordinary Diff Eq Exam Review Solutions 28 - 30 1 Solutions...

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Unformatted text preview: 30 1 Solutions 2. Since the numerator and denominator are homogeneous of degree 1 the quotient is homogeneous. Let . Clearly, . Then and so is an equilibrium solution. Separating the variables and using partial fractions gives . Integrat- ing gives . Simplifying and exponentiating gives , . Now let then for . The equilibrium solution gives as another solution. 3. Since the numerator and denominator are homogeneous of degree 2 the quotient is homogeneous. Let . Then . So . There are two equilibrium solutions . Separating the variables and using partial fractions gives . Integrating and simplifying gives . Solving for gives for When equilibrium solution , for , and so , we get or , which is the same as the . The equilibrium solution gives . Thus we can write the solutions as , and . 4. Since the numerator and denominator are homogeneous of degree 2 the quotient is homogeneous. Let . Subtract . Then from both sides to get . Separate the variables to get Integrating gives , and simplify to get , . Now exponentiate, substitute . 5. Since the numerator and denominator are homogeneous of degree 2 the quotient is homogeneous. Let from both sides to get . Separating variables gives . Then . Subtract . The equilibrium solutions are and integrating gives . Exponentiating gives we get . Now so solutions become . These occur when included in the general formula. and by simplifying . The equilibrium , so are already ...
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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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