Ordinary Diff Eq Exam Review Solutions 15

Ordinary Diff Eq Exam Review Solutions 15 - 1 Solutions ....

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Unformatted text preview: 1 Solutions . At and hence 17 (2010) we have . 41. The initial population is hours we have or compute the population after 42. We have . Since the population doubles in . Now we can . Hence hours: . So for . We get . Now we solve the equation years. 43. In the logistic growth equation . To determine we use A simple calculation give . Thus to get . . Now the population in 2010 is 44. In the logistics equation Since ing this equation for gives and 45. Let . Then . The equation equation together imply and . Thus we get . Solv- . Now . The equation implies implies that . These . Cross multiplying and simplifying leads to . Solving for gives the result. Now replace the formula for into . Simplifying gives . The formula for follows after taking the natural log of both sides. 46. We have , , and ing the result of the previous problem we get . Us- Section 1.4 1. This equation is already in standard form with . An antiderivative of is so the integrating factor is . If we multiply the differential equation by , we get the equation and the left hand side of this equation is a perfect derivative, namely, . Thus, . Now take antiderivatives of both sides and multiply by . This gives ...
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