MATH 263 Lecture Notes - Integrating Factor, Partial Fraction Decomposition, Final Examination
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Solutions of final examination: (a) (8 points) solve the di erential equation. 2xyy(cid:48) + x2 = 3y2 . (b) (2 points) give another method that we have learned of solving this equation. You may only brie y outline the second method without carrying it out in detail. Re-arranging, we have x2 3y2 + 2xyy(cid:48) = 0 which is not exact since my = 6y and nx = 2y, and these are not equal. = 4 x which is a function of x alone. Thus we nd an integrating factor (x) by solving. Thus multiplication of the original de by this integrating factor gives the. 1 x2 3 y2 x4 + 2 y x3 y(cid:48) = 0 which is exact (as is easily checked). Thus we seek f (x, y) such that. Fy = 2 y x3 , fx = Thus f (x, y) = y2 x3 1 x and so the solutions are given implicitly by y2 x3 1 x.