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Unformatted text preview: ( x ) is found to be ( x ) = exp Z 2 3 x dx = exp 2 3 ln  x  = x 2 / 3 . Multiplying equation (2.23) by x 2 / 3 gives x 2 / 3 v + 2 3 x 1 / 3 v = 2 3 x 5 / 3 ( x 2 / 3 v ) = 2 3 x 5 / 3 . We now integrate both sides and solve for v to nd x 2 / 3 v = Z 2 3 x 5 / 3 dx = 1 4 x 8 / 3 + C v = 1 4 x 2 + Cx 2 / 3 . Substituting v = y 2 gives the solution y 2 = 1 4 x 2 + Cx 2 / 3 ( x 2 + 4 y 2 ) x 2 / 3 = 4 C or, cubing both sides, ( x 2 + 4 y 2 ) 3 x 2 = C 1 , where C 1 := (4 C ) 3 is an arbitrary constant. 92...
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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 University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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