48_pdfsam_math 54 differential equation solutions odd

# 48_pdfsam_math 54 differential equation solutions odd - ( x...

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Chapter 2 Here P ( x )= x/ ( x 2 +1),so Z P ( x ) dx = Z x x 2 +1 dx = 1 2 ln( x 2 +1) . Thus the integrating factor is µ ( x )= e (1 / 2) ln( x 2 +1) = e ln [ ( x 2 +1) 1 / 2 ] =( x 2 +1) 1 / 2 . Multiplying equation (2.9) by µ ( x ) yields ( x 2 +1) 1 / 2 dy dx + x ( x 2 +1) 1 / 2 y = x ( x 2 +1) 1 / 2 , which becomes d dx ± ( x 2 +1) 1 / 2 y ² = x ( x 2 +1) 1 / 2 . Now we integrate both sides and solve for y to Fnd ( x 2 +1) 1 / 2 y =( x 2 +1) 1 / 2 + C y =1+ C ( x 2 +1) 1 / 2 . This solution is valid for all x since P ( x )and Q ( x ) are continuous for all x . 17. This is a linear equation with P ( x )= 1 /x and Q ( x )= xe x which is continuous on any interval not containing 0. Therefore, the integrating factor is given by
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Unformatted text preview: ( x ) = exp Z 1 x dx = e ln x = 1 x , for x &gt; . Multiplying the equation by this integrating factor yields 1 x dy dx y x 2 = e x D x y x = e x . Integrating gives y x = e x + C y = xe x + Cx. Now applying the initial condition, y (1) = e 1, we have e 1 = e + C C = 1 . 44...
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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.

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