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Unformatted text preview: Chapter 2 32. To obtain a homogeneous equation, we make a substitution x = u + h , y = v + k with h and k satisfying 2 h + k + 4 = 0 h 2 k 2 = 0 h = 6 5 , k = 8 5 . This substitution yields (2 u + v ) du + ( u 2 v ) dv = 0 dv du = v + 2 u 2 v u = ( v/u ) + 2 2( v/u ) 1 . We now let z = v/u (so, v = z + uz ) and conclude that z + u dz du = z + 2 2 z 1 u dz du = z + 2 2 z 1 z = 2 z 2 + 2 z + 2 2 z 1 Z (2 z 1) dz z 2 z 1 = 2 Z du u ln z 2 z 1 = 2 ln  u  + C 1 ln u 2 z 2 u 2 z u 2 = C 1 ln v 2 uv u 2 = C 1 ln y + 8 5 2 y + 8 5 x + 6 5 x + 6 5 2 = C 1 (5 y + 8) 2 (5 y + 8) (5 x + 6) (5 x + 6) 2 = C, where C = 25 e C 1 6 = 0 is any constant. Separating variables, we lost two constant solutions z ( u ), which are the zeros of the polynomial z 2 z 1. They can be included in the above formula by taking C = 0. Therefore, a general solution is given by (5 y + 8) 2 (5 y + 8) (5 x + 6) (5 x + 6) 2 = C, where C is an arbitrary constant. 34. In Problem 2, we found that the given equation can be written as a Bernoulli equation, dx dt 1 2 t x = t 2 x 1 . Thus, 2 x dx dt 1 t x 2 = t ( v = x 2 ) dv dt 1 t v = t....
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.
 Spring '08
 MAZMANI
 Equations

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