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**Unformatted text preview: **¿ and N ( tx ,ty )=( tx ) 2 −( tx ) ( ty )= t 2 x 2 − t 2 ( xy )= t 2 ( x 2 − xy ) ¿ t 2 N ( x , y ) both of coefficients are homogeneous functions of degree 2. If we let y = ux ( u = y x ) then dy = du∙ x + udx ( x 2 + u 2 x 2 ) dx + ( x 2 − x 2 u ) [ du∙ x + u dx ]= ( 1 + u 2 ) dx +( 1 − u ) [ du∙ x + u dx ]= ( 1 + u 2 ) dx +( 1 − u ) udx +( 1 − u ) xdu = ( 1 + u 2 + u − u 2 ) dx +( 1 − u ) udx = ( 1 + u ) dx +( 1 − u ) udx = dx x + ( 1 − u ) du ( 1 + u ) = dx x + ( − 1 + 2 ( 1 + u ) ) du = After integrating the last line gives ln | x |− u + 2ln | 1 + u |= ln ∨ C ∨ ¿ ln | x |− y x + 2ln | 1 + y x | = ln ∨ C ∨ ¿ y x = ln | x C | + ln | ( x + y ) x 2 2 | y x = ln | ( x + y ) x 2 2 x C | , y x = ln | ( x + y ) Cx 2 | or y = x ln | ( x + y ) Cx 2 | ....

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- Fall '15
- Linear Algebra, Algebra, Equations, Trigraph, dx, homogeneous functions