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Unformatted text preview: y dx dy + 2 x = 5 y 2 ⇒ dx dy + 2 y x = 5 y 2 . Therefore, P ( y ) = 2 /y and the integrating factor, µ ( y ), is µ ( y ) = exp ³Z 2 y dy ´ = exp (2 ln  y  ) =  y  2 = y 2 . Multiplying the equation (in standard form) by y 2 and integrating yield y 2 dx dy + 2 y x = 5 y 4 ⇒ d dy ( y 2 x ) = 5 y 4 ⇒ y 2 x = Z 5 y 4 dy = y 5 + C ⇒ x = y − 2 ( y 5 + C ) = y 3 + Cy − 2 . 15. To put this linear equation in standard form, we divide by ( x 2 + 1) to obtain dy dx + x x 2 + 1 y = x x 2 + 1 . (2.9) 43...
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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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