This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Solutions to Assignment 2, Math 20 January 27, 2006 Section 2.1: 21b) The original equation is y 1 2 y = 2 cos ( t ), y (0) = a . This is a first order, linear equation, and is not seperable, so we use integrating factors, with μ = e t 2 . This yields the solution y = 2 e t 2 Z e t 2 cos ( t ) dt + Ce t 2 . To find the integral, we use integration by parts twice, with u = e t 2 both times. Or we simply look up the integral in a table, and find the final solution to be y = 4 5 cos ( t ) + 8 5 sin ( t ) + Ce t 2 Substituting in the intitial value y (0) = a and noting that sin (0) = 0, cos (0) = 1, we find that C = a + 4 5 . 21c) To find the critical value, we consider the limit as t → ∞ , and find the value of a where the limiting behavior of the solutions changes. Note that as t → ∞ , sin ( t ) and cos ( t ) are bounded by 1, so in passing to the limit we have lim t →∞ y ( t ) = lim t →∞ ( a + 4 5 ) e t 2 , and so we see that the limit is + ∞ if a > 4 5 , and the limit is∞ if a < 4 5 , so a = 4 5 is our critical value. 1 Section 2.3: 2) This problem is of the general type y = rate in  rate out, so to set up the problem we simply need to find each of the terms. Letting y ( t ) be the amount of salt in the tank at time t , we have that the rate of salt going into the tank is ratein = 2 γ g L , since the amount of salt going in is the...
View
Full Document
 Spring '10
 Edgars
 Constant of integration, Boundary value problem

Click to edit the document details