Chem Differential Eq HW Solutions Fall 2011 182

Chem Differential Eq HW Solutions Fall 2011 182 - A182...

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A182 Appendix A Ordinary Differential Equations: Review of Concepts and Methods To evaluate the last integral, we use the partial fractions decomposition 1 (1 - x 2 ) x 2 = 1 (1 - x )(1 + x ) x 2 = A (1 - x ) + B (1 - x ) + C x + D x 2 ; = A (1 + x ) x 2 + B (1 - x ) x 2 + C (1 - x 2 ) x + D (1 - x 2 ) (1 - x )(1 + x ) x 2 1= A (1 + x ) x 2 + B (1 - x ) x 2 + C (1 - x 2 ) x + D (1 - x 2 ) . Take x =0 D =1 . x 1=2 A, A = 1 2 . x = - 1 B, B = 1 2 . Checking the coefficient of x 3 , we Fnd C = 0. Thus 1 (1 - x 2 ) x 2 = 1 2(1 - x ) + 1 2(1 + x ) ++ 1 x 2 ± 1 (1 - x 2 ) x 2 dx = - 1 2 ln(1 - x )+ 1 2 ln(1 + x ) - 1 x = 1 2 ln ² 1+ x 1 - x ³ - 1 x . So y 2 = x ´ 1 2 ln ² x 1 - x ³ - 1 x µ = x 2 ln ² x 1 - x ³ - 1 . Hence the general solution y = c 1 x + c 2 ´ x 2 ln ² x 1 - x ³ - 1 µ . 13. Put the equation in standard form: x 2 y ±± + xy ±
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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