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homework set 17-Solutions - BILKENT UNIVERSITY Department...

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B ˙ ILKENT UNIVERSITY Department of Mathematics MATH 225, LINEAR ALGEBRA and DIFFERENTIAL EQUATIONS, Solution of Homework set 1 # 17 U. Mu˘gan July 16, 2008 1) a) Substituting y = e rx into the D.E., we get the C.E. r 2 - 2 r + 1 = 0 which gives r 1 = r 2 = 1. Then two L.I. solutions y 1 = e x and y 2 = xe x . So the general solution is y ( x ) = c 1 e x + c 2 xe x . Or, we can use the method of reduction of order if y 1 = e x , then the second L.I. solution is of the form y 2 = y 1 v ( x ) = e x v ( x ). D.E yields v 00 = 0. Solving the D.E. for v gives v ( x ) = c 1 + c 2 x, c 1 , c 2 = constant. The general solution of the given D.E. can be obtained from y = e x v ( x ) . b) Substituting y = e rx into the D.E. we get the C.E. 25 r 2 - 20 r + 4 = 0 which may be written as (5 r - 2) 2 = 0. The roots of the C.E. are r 1 = r 2 = 2 / 5. Thus, y ( x ) = c 1 e 2 x/ 5 + c 2 xe 2 x/ 5 , c 1 , c 2 = constant. c) Substituting y = e rx into the D.E. we get the C.E. r 2 - 2 r + 10 = 0 and hence the roots are r 1 = 1 + 3 i, r 2 = 1 - 3 i . Thus, y ( x ) = c 1 e x cos 3 x + c 2 e x sin 3 x, c 1 , c 2 = constant. 2) a) The C.E. is r 2 + 4 r + 4 = ( r + 2) 2 = 0, which has the repeated root r 1 = r 2 = r = - 2. Then the general solution is y ( x ) = c 1 e - 2 x + c 2 xe - 2 x , c 1 , c 2 = constant. Since the I.C. are given at x = - 1, we can write the general solution as y ( x ) = c 1 e - 2( x +1) + c 2 xe - 2( x +1) . Then y 0 ( x ) = - 2 c 1 e - 2( x +1) + 2 c 2 e - 2( x +1) - 2 c 2 xe - 2( x +1) and hence c 1 - c 2 = 2 and - c 1 + 3 c 2 = 1 which yields c 1 = 7 and c 2 = 5. Thus the general solution is y ( x ) = 7 e - 2 x + 5 xe - 2 x , and decaying exponentially as x → ∞ . b) C.E. is 9 r 2 - 12 r + 4 = 0 and its roots are r 1 = r 2 = 2 / 3. Therefore, the general solution of the D.E. is y = c 1 e 2 x/ 3 + c 2 xe 2 x/ 3 , c 1 , c 2 = constant The initial conditions yield c 1 = 2 and c 2 = - 7 / 3. Since r > 0, y → ∞ as x → ∞ . c) C.E. is 9 r 2 + 6 r + 82 = 0 and its roots are r 1 , 2 = - (1 / 3) ± 3 i . Therefore, the general solution of the D.E. is y = c 1 e - x/ 3 cos 3 x + c 2 e - x/ 3 sin 3 x, c 1 , c 2 = constant The initial conditions yield c 1 = - 1 and c 2 = 5 / 9. Since the real part of the roots are negative, y 0 as x → ∞ .
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