121616949-math.245 - 9.11 before Z 1 dy = y Differential...

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9.11 Differential equations 231 before: Z 1 y dy = Z k dt ln | y | = kt + C | y | = e kt e C y = ± e kt e C y = Ae kt . Again, if we allow A = 0 this includes the constant solution, and we can simply say that y = Ae kt is the general solution. With an initial value we can easily solve for A to get the solution of the initial value problem. In particular, if the initial value is given for time t = 0, y (0) = y 0 , then A = y 0 and the solution is y = y 0 e kt . Exercises 1. Which of the following equations are separable? a. ˙ y = sin( ty ) b. ˙ y = e t e y c. y ˙ y = t d. ˙ y = ( t 3 - t ) sin - 1 y e. ˙ y = t 2 ln y + 4 t 3 ln y 2. Solve ˙ y = 1 / (1 + t 2 ). 3. Solve the initial value problem ˙ y = t n with y (0) = 1. Do not exclude the case n = - 1. 4. Solve ˙ y = ln t . 5. Identify the constant solutions (if any) of ˙ y = t sin y . 6. Identify the constant solutions (if any) of ˙ y = te y . 7. Solve ˙ y = t/y . 8. Solve ˙ y = y 2 - 1. 9. Solve ˙ y = t/ ( y 3 - 5). You may leave your solution in implicit form: that is, you may stop once you have done the integration, without solving for
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Unformatted text preview: . 10. Find a non-constant solution of the initial value problem ˙ y = y 1 / 3 , y (0) = 0, using separation of variables. Note that the constant function y ( t ) = 0 also solves the initial value problem. Hence, an initial value problem need not have a unique solution. 11. Solve the equation for Newton’s law of cooling leaving M and k unknown. 12. After 10 minutes in Jean-Luc’s ready room, his tea has cooled to 40 ◦ Celsius. The room temperature is 25 ◦ Celsius. If k = 1, what was the initial temperature of the tea? 13. Solve the logistic equation ˙ y = ky ( M-y ). (This is a somewhat more reasonable population model in most cases than the simpler ˙ y = ky .) Sketch the graph of the solution to this equation when y (0) = 1 / 2....
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