This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 2a Practical, Fall 2008: Solutions to Homework 4 You are bound by the honor code to not look at this file if you did not take Math 2a in the Fall of 2008 1. Solve the following equations or initial value problems. (20pts) (a) y 00 2 y + y = e t 1+ t 2 , y (0) = 0, y (0) = 1 Solution: First we solve the corresponding homogeneous equation y 00 2 y + y = 0. The characteristic equation is r 2 2 r + 1 = ( r 1) 2 , which has root 1 with multiplicity 2. Thus the general solution of the homogeneous equation is c 1 e t + c 2 te t . Now we need to find a particular solution of the nonhomogeneous equation. Using variation of parameters, well try to find a solution of the form Y ( t ) = u ( t ) e t + v ( t ) te t . Computing derivatives, we see that this would mean Y ( t ) = u ( t ) e t + u ( t ) e t + v ( t ) te t + v ( t ) e t + v ( t ) te t Y 00 ( t ) = u 00 ( t ) e t + 2 u ( t ) e t + u ( t ) e t + v 00 ( t ) te t + 2 v ( t ) te t + 2 v ( t ) e t + 2 v ( t ) e t + v ( t ) te t . Plugging this into the nonhomogeneous equation gives u 00 ( t ) e t + v 00 ( t ) e t + 2 v ( t ) e t = e t 1 + t 2 . Thus we want to have u 00 ( t ) + v 00 ( t ) + 2 v ( t ) = 1 1 + t 2 . Letting v ( t ) = 0, we obtain u 00 ( t ) = 1 1+ t 2 . To get this, we can take u ( t ) = tan 1 ( t ), and u ( t ) = Z tan 1 ( t ) dt = t tan 1 ( t ) Z t 1 + t 2 dt = t tan 1 ( t ) 1 2 log(1 + t 2 ). Note that we may omit constants since we only need to find one such u . Thus a particular solution of the original equation is Y ( t ) = t tan 1 ( t ) 1 2 log(1 + t 2 ) e t . Hence the general solution to the original equation is given by y ( t ) = c 1 e t + c 2 te t + t tan 1 ( t ) 1 2 log(1 + t 2 ) e t . Plugging in y (0) = 0, we get c 1 1 2 log(1) = 0, so thus c 1 = 0. We also have y ( t ) = c 1 e t + c 2 e t + c 2 te t + t tan 1 ( t ) 1 2 log(1 + t 2 ) e t + t 1 + t 2 + tan 1 ( t ) t 1 + t 2 e t , 1 and plugging in y (0) = 1 we get c 1 + c 2 = 1, so thus c 2 = 1. Hence we have found that the solution to the initial value problem is y ( t ) = te t + t tan 1 ( t ) 1 2 log(1 + t 2 ) e t . (b) t 2 y 00 ty 3 y = t 3 + t Solution: Letting P ( t ) = t 2 , Q ( t ) = t , R ( t ) = 3 we see that P 00 ( t ) Q ( t ) + R ( t ) = 0 and thus the underlying homogeneous equation is exact(see problem 4). Thus, we can write it as ( Py ) + ( fy ) = 0. See problem 4 for a proof that f = Q P , which here is...
View
Full
Document
This note was uploaded on 01/08/2011 for the course MA 2a taught by Professor Makarov,n during the Fall '08 term at Caltech.
 Fall '08
 Makarov,N
 Differential Equations, Statistics, Equations, Probability

Click to edit the document details