Explanation Given: ( 2 x − y ) d x + ( x + y − 3 ) d y = 0 , y ( 0 ) = 2 Approach: ( a 1 x + b 1 y + c 1 ) d x + ( a 2 x + b 2 y + c 2 ) d y = 0 For a 1 b 2 ≠ a 2 b 1 , let x = X + h and y = Y + k , where h and k satisfy a 1 h + b 1 k + c 1 = 0 a 2 h + b 2 k + c 2 = 0 Then the equation will become homogeneous equation with u and v variables. Calculation: ( 2 x − y ) d x + ( x + y − 3 ) d y = 0 ( x + y − 3 ) d y = − ( 2 x − y ) d x d y d x = − 2 x − y x + y − 3 Let x = X + h and y = Y + k Therefore, d Y d X = − 2 ( X + h ) − ( Y + k ) X + h + Y + k − 3 d Y d X = − 2 X + 2 h − Y − k X + h + Y + k − 3 d Y d X = − ( 2 X − Y ) + ( 2 h − k ) ( X + Y ) + ( h + k − 3 ) ... ( 1 ) 2 h − k = 0 h + k − 3 = 0 Then h = 1 and k = 2 Equation (1) becomes d Y d X = − ( 2 X − Y ) ( X + Y ) ... ( 2 ) Let V = Y X Y = V X d Y d X = X d V d X + V Equation (2) becomes X d V d X + V = − 2 X − V X X + V X X d V d X + V = − 2 − V 1 + V X d V d X = − 2 − V 1 + V − V X d V d X = − V 2 + 2 1 + V 1 + V V 2 + 2 d V = − 1 X d X ( 1 V 2 + 2 + V V 2 + 2 ) d V = − 1 X d X Integrate on both sides

7th Edition

ISBN: 9780321977175

Author: Nagle

Publisher: PEARSON CO

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Differential Equations

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Solution Summary: The author explains how the equation becomes a homogeneous equation with u and v variables.

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