Amath 350 assignment 7 solutions fall 2012 page 2 c

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Unformatted text preview: f n. AMATH 350 Assignment #7 Solutions - Fall 2012 Page 2 c) Use the variation of parameters method to solve the equation x2 y 00 +3xy 0 1 3y = . x Solution: We look for a solution of the form yp = u1 x 3 + u2 x. To find u1 and u2 , we must solve the system u01 x u1 3 3x (1) + u2 x = 0 4 + u2 = 1 x3 (2) 1 Multiplying (2) by x gives u1 ( 3x 3 + u2 x) = x2 , which we can subtract from (1) to get 40 1 u= 31 x x2 x ) u01 = 4 x2 ) u1 = . 8 ✓ ◆ 10 1 x 1 0 We also know, from (1), that u2 = u= = 3 . There41 4 x x 4 4x fore 1 u2 = . 8 x2 We have now determined that So y = C1 + C2 x x3 1 . 4x x2 x 8 3 = yp = 1 . 4x 1 x 8x 2 AMATH 350 Page 3 Assignment #7 Solutions - Fall 2012 These are two 2. y0 =Z Ay + F(x); separable DEs. Solving gives Z Z 1 dy1 Z 5x ✓ ◆ = 5 dt ) ln |y1 | = ◆ t + C ) y1 ✓ c1 e 5x . Similarly y2 = c2 e 5 1 1= 1 2 2 y1 21 4xe x 1 A= , and F(x) y1 (x) = .5x 5x 1 1 1 0 c1 e 5x . Thus the general solution of the2system is y = = y2 (x) 2 c2 e 2 2 1 1 2 =0 Thus the eigenvalues are: 2 2 ) . 5x 0 6....
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