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Unformatted text preview: linear equations, we nd an integrating factor ( x ) = 1 /x and multiply the equation by ( x ) to get 1 x du dx 1 x 2 u = x d dx 1 x u = x 1 x u = Z ( x ) dx = 1 2 x 2 + C 1 u = 1 2 x 3 + C 1 x y = 1 x 3 / 2 + C 1 x = 2 Cx x 3 . Also, y 0 is a solution which was lost when we multiplied the equation by u 2 (in terms of y , divided by y 2 ) to obtain a linear equation. 83...
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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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