2 yx c1 e 2 coslnxc2 e z sinz x x 0 z

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Unformatted text preview: 20 (b) Auxiliary = Roots: λ equation λ2 + 4λ + 5−2 0. i. = =± √ 2 −4 ±to 16 −(2): y (z ) = c e−2z cos(z ) + c e−2z sin(z ). 20 General = eq. 2 Roots: λsolution = −2 ± i1 −2 . a) General solution to2eq. (1): y (x) = c1 e−2zln(x) cos(ln(x))2+ c2 e−2 ln(x) sin(ln(x)), i.e., General solution to eq. (2): y￿￿x) = c1 e −2 cos(ln(x))c2 e− z sin(z ). x)), x > 0. (z ) c x cos(z ) + + x (x + 1)2 y ( +=x 1+ 1) y ￿ − 2y = c2. −2 sin(ln( ( 0 General solution to eq. (1): y (x) = c1 e−2 ln(x) cos(ln(x)) + c2 e−2 ln(x) sin(ln(x)), i.e., 2 ￿￿ ￿ 6. (i) + 1) y + 2(x + solution:= 0y (x) = c1 x−2 cos(ln(x)) + c2 x−2 sin(ln(x)), x > 0. x Check given 1)y − 2y ￿ ￿￿ 6. ((a) y1 2 y ￿￿x+ 2(x +y1)=￿ 1, 2y = 0 Substitute into DE: (x + 1)2 (0) + 2(x + 1)(1) − 2(x + 1) = 0. x + 1) = + 1 ⇒ 1 y − y1 = 0. So y1 is a solution of DE on −∞ < x < ∞. ￿ ￿￿ (a) y1 = x + 1 ⇒ y1 = 1, y1 = 0. Substitute into DE: (x + 1)2 (0) +￿￿2(x + 1)(1) − 2(x + 1) = 0. (b) So yy2is au(x)y1 = of(x)(xon 1). Then <￿ ∞.(x + 1)u...
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This homework help was uploaded on 02/22/2014 for the course AMATH 350 taught by Professor Davidhamsworth during the Fall '12 term at University of Waterloo, Waterloo.

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