Ordinary Diff Eq Exam Review Solutions 10

Ordinary Diff Eq Exam Review Solutions 10 - 12 1 Solutions...

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Unformatted text preview: 12 1 Solutions 12. The equation equilibrium solutions. has no constant solution. Thus, there are no 13. The equation has two constant solutions: and 14. We substitute for all forces linear solution is into and . . Equality and the only to get . Thus and 15. We substitute into to get . Equality for all means that must be a constant function, which can occur only if the coefficient of is 0. This forces leaving us with the equation . This implies , where is an integer. Hence , is a family of linear solutions. Section 1.3 1. separable; and 2. In standard form we get . . This is separable; and . We cannot write 3. First write in standard form: product of a function of and a function of . It is not separable. as a 4. In standard form we get . We cannot write product of a function of and a function of . It is not separable. as a 5. Write in standard form to get: . It is separable; 6. We can factor to get and . Here we can write and . . It is separable; . 7. In standard form we get . We cannot write product of a function of and a function of . It is not separable 8. It is not separable as of and a function of . 9. In standard form we get: as a cannot be written as a product of a function It is separable; and 10. The variables are already separated, so integrate both sides of the equation to get , which we can rewrite as where is a constant. Since , substitute and to get that . Thus the solution is given implicitly by ...
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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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