AMath350.F13.A5.Sol

# Thus y2 x 1 3 x 1 x 1 2 is a solution for x

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Unformatted text preview: . Thus y2 = (x + 1) 3 (x + 1) = (x + 1) 2 is a solution for x > 1. y1 and y2 are linearly independent on x > 1 by Proposition 3.31. So the general solution of DE is yh = c1 (x + 1) + c2 (x + 1) 2 . b) xy 00 y 0 + 4 x3 = 0 4 0 00 i) y1 = sin x2 =) y1 = 2x cos x2 , y1 = 2 cos x2 Substituting this into the DE, we have xy 00 y 0 +4x3 y = 2x cos x2 4x3 sin x2 4x2 sin x2 . 2x cos x2 +4x3 sin x2 = 0 ii) Let y2 = u(x) sin (x2 ). 0 Then y2 = u0 sin (x2 ) + 2ux cos (x2 ), 00 and y2 = u00 sin (x2 ) + 2u0 x cos (x2 ) + 2u0 x cos (x2 ) + 2u cos (x2 ) 2 4ux sin (x2 ). Substituting this into the DE, we obtain u00 x sin x2 + 4u0 x2 cos x2 + 2ux cos x2 u0 sin x2 =) 4ux3 sin x2 2ux cos x2 + 4ux3 sin x2 = 0 u00 x sin x2 + u0 4x2 cos x2 sin x2 = 0. This is a ﬁrst-order DE for u0 , so let v = u0 , and rewrite it as ✓ ◆ dv (sin (x2 ) 2x2 cos (x2 )) v 1 2 = = 4x cot x v dx x sin (x2 ) x ln v = ln x 2 ln sin x2 ✓ ◆ x = ln sin2 (x2 ) =) =) =) v = x csc2 x2 u= 1 cot x2 2 1 Therefore a second solution is sin x2 cot x2 2 The general solution of the DE can be written as = 1 cos x2 ....
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## This note was uploaded on 02/08/2014 for the course AMATH 350 taught by Professor Davidhamsworth during the Fall '12 term at University of Waterloo, Waterloo.

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