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Unformatted text preview: Chapter 7 Solutions to §7.1 1. (Lf)(x) = (D – x)(2x – 3e 2x ) = D(2x – 3e 2x ) – x(2x – 3e 2x ) = (2 – 6e 2x ) – 2x 2 + 3xe 2x = 2(1 – x 2 ) + 3e 2x (x – 2). 3. (Lf)(x) = (D 3 – 2xD 2 )(2x – 3e 2x ) = D 3 (2x – 3e 2x ) – 2xD 2 (2x – 3e 2x ) = D 2 (2 – 6e 2x ) – 2xD(2 – 6e 2x ) = D(–12e 2x ) – 2x(–12e 2x ) = –24e 2x + 24xe 2x = 24e 2x [x – 1]. 5. Lf = (D 2 – x –1 D + 4x 2 )(sin(x 2 )) = D 2 (sin(x 2 )) – x –1 D(sin(x 2 )) + 4x 2 (sin(x 2 )) = D(2xcos(x 2 )) – x –1 (2xcos(x 2 )) + 4x 2 (sin(x 2 )) = 2[cos(x 2 ) – 2x 2 sin(x 2 )] – 2cos(x 2 ) + 4x 2 (sin(x 2 )) = 0. Thus, f ∈ Ker(L). 7. (L 1 L 2 )f = L 1 (f´ – 2x 2 f) = (xD + 1)(f´ – 2x 2 f) = xf´´ – 2x 3 f´ – 4x 2 f + f´ – 2x 2 f = xf´´ + (1 – 2x 3 )f´ – 6x 2 f, so that L 1 L 2 = xD 2 + (1 – 2x 3 )D – 6x 2 . (L 2 L 1 )f = L 2 (xD + 1)f = (D – 2x 2 )(xf´ + f) = xf´´ + f´ + f´ – 2x 2 (xf´ + f) = xf´´ + 2(1 – x 3 )f´ – 2x 2 f. Thus, L 2 L 1 = xD 2 + 2(1 – x 3 )D – 2x 2 , so that L 1 L 2 ≠ L 2 L 1 . 9. (L 1 L 2 ) = L 1 (D + b 1 )f = (D + a 1 )(f´ + b 1 f) = f´´ + [b 1 + a 1 ]f´ + (b 1 ´ + a 1 b 1 )f. Thus L 1 L 2 = D 2 + (b 1 + a 1 )D + (b 1 ´ + a 1 b 1 ). Similarly, L 2 L 1 = D 2 + (a 1 + b 1 )D + (a 1 ´ + b 1 a 1 ). Thus L 1 L 2 – L 2 L 1 = (b 1 ´ – a 1 ´), which is the zero operator if and only if b 1 ´ = a 1 ´. which can be integrated directly to obtain b 1 = a 1 + c 2 , where c 2 is an arbitrary constant. Consequently we must have: L 1 = D + a 1 (x), L 2 = D + [a 1 (x) + c 2 ]. 11. (D 2 + 4xD – 6x 2 )y = x 2 sin x and y´´ + 4xy´ – 6x 2 y = 0. 209 210 Ch. 7 Linear Differential Equations of Order n 13. Let a 1 , ..., a n be functions that are continuous on an interval I. Then, for any x in I, the initial value problem y (n) + a 1 (x)y (n – 1) + ... + a n – 1 (x)y´ + a n (x)y = 0, y(x ) = 0, y´(x ) = 0, ..., y (n – 1) (x ) = 0, has only the trivial solution y(x) = 0. Proof All of the conditions of the existence–uniqueness theorem are satisfied and y(x) = 0 is a solution; consequently, it is the only solution. 15. Substituting y(x) = e rx into the given DE yields e rx (r 3 + 3r 2 – 4r – 12) = 0, so that we will have a solution providing r satisfies r 3 + 3r 2 – 4r – 12 = 0, that is, (r – 2)(r + 2)(r + 3) = 0. Consequently, three solutions to the given DE are y 1 (x) = e 2x , y 2 (x) = e –2x , y 3 (x) = e –3x . Further, the Wronskian of these solutions is W[y 1 , y 2 , y 3 ](x) = e 2x e –2x e –3x 2e 2x –2e –2x –3e –3x 4e 2x 4e –2x 9e –3x = –20e –3x . Since the Wronskian is never zero, the solutions are LI on any interval. Hence the general solution to the DE is y(x) = c 1 e 2x + c 2 e –2x + c 3 e –3x ....
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 Spring '08
 Harris
 Linear Differential Equations

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