Unformatted text preview: 2 √ x + y = x + C which, on solving for y , yields y = ´ x 2 + C 2 µ 2 − x. Thus, we have y = ( x + C ) 2 4 − x. 19. The righthand side of this equation has the form G ( x − y ) with G ( t ) = ( t + 5) 2 . Thus we substitute t = x − y ⇒ y = x − t ⇒ y = 1 − t , 82...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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