A5 Solutions

2 yx c1 e2zcoslnxc2 e2z sinz x x 0 2

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Unformatted text preview: 16 − 20 (b) Auxiliary = Roots: λ equation λ2 + 4λ + 5−2 0. i. = =± √ 2 −4 ± eq. 20 y z ) General solution to 16 −(2): =(−2= ci1 e−2z cos(z ) + c2 e−2z sin(z ). Roots: λ = ±. a) General solution to2eq. (1): y (x) = c1 e−2 ln(x) cos(ln(x)) + c2 e−2 ln(x) sin(ln(x)), i.e., General solution to eq. (2): y￿￿x) = c1 e−2zcos(ln(x))c2 e−2z sin(z ). x)), x > 0. ( 2 ( z ) = c x−2 cos(z ) + + c x−2 sin(ln( 2 (x + 1) y + (x 1+ 1) y ￿ − 2y = 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., 6. (i) + 1)2 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 yy = u(x)y1 = u(x)(x + 1). Then y2 = (x + 1)u￿ +...
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This note was uploaded on 05/29/2013 for the course AMATH 350 taught by Professor Noidea during the Fall '11 term at Waterloo.

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