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Unformatted text preview: MATH 138 Physicsbased Section 007 Assignment 6 Solutions 1. Boundary Value Problems A boundary value problem is a differential equation with conditions on the solution that occur at distinct values of the independent variable. Specifically, given the DE ay 00 ( t ) by ( t ) + cy ( t ) = 0 An initial value problem would arise from constraints of the form y (0) = A , y (0) = B , while constraints of the form y ( t 1 ) = A , y ( t 2 ) = B specify a boundary value problem. Initial value problems of this type are always soluble, but boundary value problems may admit no solutions. In the following two cases, solve the boundary value problem, if possible, or show that no solution exists. a) y 00 ( t ) 3 y ( t ) + 2 y ( t ) = 0 , y (0) = 1, y (3) = 0 b) y 00 ( t ) + 4 y ( t ) = 0 , y (0) = 1, y ( π ) = 0 Solution: a) The characteristic equation is r 2 3 r + 2 = 0 which we can write as ( r 2)( r 1) = 0. Thus the general solution is y ( t ) = C 1 e 2 t + C 2 e t . The boundary conditions then read y (0) = C 1 + C 2 = 1 y (3) = C 1 e 6 + C 2 e 3 = 0 The first equation yields C 1 = 1 C 2 and substituting this into the second, we find (1 C 2 ) e 6 + C 2 e 3 = i.e. (1 C 2 ) e 3 + C 2 = giving (1 e 3 ) C 2 = e 3 so that C 2 = e 3 e 3 1 = 1 1 e 3 , while C 1 = e 3 1 e 3 . The solution is then y ( t ) = e 3 1 e 3 e 2 t + 1 1 e 3 e t b) In this case the characteristic equation is r 2 + 4 = 0 which has purely imaginary solutions r = ± 2 i . The general solution is then y ( t ) = C 1 sin(2 t ) + C 2 cos(2 t ) . The boundary conditions then read y (0) = C 2 = 1 y ( π ) = C 2 = 0 Clearly these cannot both be satisfied. We conclude that this boundary value problem has no solutions. 2. Repeated Roots Recall that when the parameters in the 2nd order DE ay 00 ( t ) + by ( t ) + cy ( t ) = 0 satisfy b 2 4 ac = 0, there is only one solution of the form e rt , namely e b 2 a t . The second independent solution is y ( t ) = te b 2 a t , as can be verified from the differential equation. The form of this second solution can be developed as follows. Suppose the roots of the characteristic equation are close to one another, but are not equal. Let them be r and r + ε , so that a general solution takes the form y ( t ) = C 1 e rt +...
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This note was uploaded on 01/06/2011 for the course MATH 138 taught by Professor Anoymous during the Winter '07 term at Waterloo.
 Winter '07
 Anoymous
 Math

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