Ordinary Diff Eq Exam Review Solutions 11

Ordinary Diff Eq Exam Review Solutions 11 - 1 Solutions the...

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Unformatted text preview: 1 Solutions the equation or we can solve explicitly to get where the negative square root is used since . 11. In standard form we get . Clearly, solutions. Separating the variables gives 13 , are equilibrium Integrating both sides of this equation (using the substitution for the integral on the left) gives , Multiplying by , taking the exponential of both sides, and removing the absolute values gives where is a nonzero constant. However, when the equation becomes and hence . By considering an arbitrary constant (which we will call ), the implicit equation includes the two equilibrium solutions for . 12. The variables are already separated, so integrate both sides to get , a real constant. This can be simpliﬁed to . (where we replace by ) We leave the answer in implicit form. 13. The variables are already separated, so integrate both sides to get , a real constant. Simplifying gives leave the answer in implicit form . We 14. There is an equilibrium solution . Separating variables give and integrating gives . Thus , a real constant. This is equivalent to writing , a real constant, since twice an arbitrary constant is still an arbitrary constant. 15. In standard form we get so is a solution. Separating variables gives . The function is continuous on the interval and so has an antiderivative. Integration gives . Multiplying by and exponentiating gives where is a positive constant. Removing the absolute value signs gives , with . If we allow we get the equilibrium solution . Thus the solution can be written , any real constant. 16. An equilibrium solution is . Separating variables gives , a real constant. Simplifying and integrating gives gives , and the equilibrium solution . ...
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