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(a) Solution: Try u = x + y. Then This is separable Thence 1 du dx u² = x² + C 1+ dy dx dy dx = = udu X - - Y x + y 2x 1+ 2xdx - U U 2x U (x + y)² = 2(x² + C) y(x) = ± √2(x² + C) = x.

(a) Solution: Try u = x + y. Then This is separable Thence 1 du dx u² = x² + C 1+ dy dx dy dx = = udu X - - Y x + y 2x 1+ 2xdx - U U 2x U (x + y)² = 2(x² + C) y(x) = ± √2(x² + C) = x.

Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter12: Quadratic Functions
Section12.4: Graphs Of Quadratic Equations: The Discriminant
Problem 7OE
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how am I getting the 1 + (2x-u)/u here?

(a) Solution: Try u = x + y. Then du d.x This is separable Thence = 1² u² = x² + C 2 dy dx dy 1+ = dx X Y x + y udu = - = 1+ 2x 2xdx - U ՂԱ 2x U (x + y)² = 2(x² + C) y(x) = ±√√2(x² + C) − x.
Transcribed Image Text:(a) Solution: Try u = x + y. Then du d.x This is separable Thence = 1² u² = x² + C 2 dy dx dy 1+ = dx X Y x + y udu = - = 1+ 2x 2xdx - U ՂԱ 2x U (x + y)² = 2(x² + C) y(x) = ±√√2(x² + C) − x.
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