4-3. Homogeneous equations(separation of variable)

# 4-3. Homogeneous equations(separation of variable) -...

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Homogeneous equations Definition 1. If a function f possesses the property f ( tx,ty ) = t α f ( x , y ) for some real number α , the f is said to be homogeneous function of degree α . For example, f ( x , y ) = x 3 + y 3 is homogeneous function of degree 3 , because f ( tx ,ty ) =( tx ) 3 +( ty ) 3 = t 3 x 3 + t 3 y 3 = t 3 ( x 3 + y 3 ) = t 3 f ( x , y ) . A function f ( x , y ) = x 3 + y 3 + 1 is not homogeneous function. A first order DE in differential form M ( x , y ) dx + N ( x, y ) dy = 0 is said to homogeneous if both of coefficient M ( x , y ) and N ( x , y ) are homogeneous functions of the same degree. That is M ( tx ,ty ) = t α M ( x, y ) and N ( tx ,ty ) = t α N ( x , y ) . If M ( x , y ) and N ( x , y ) are homogeneous functions of degree α , we can write also M ( x , y ) = x α M ( 1, u ) and N ( x , y ) = x α N ( 1, u ) , where u = y x and M ( x , y ) = y α M ( v , 1 ) and N ( x , y ) = y α N ( v , 1 ) , where v = x y . Then a homogeneous equation M ( x , y ) dx + N ( x, y ) dy = 0 can be written as x α M ( 1, u ) dx + x α N ( 1, u ) dy = 0 or M ( 1, u ) dx + N ( 1, u ) dy = 0 .

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Since u = y x then y = ux and dy = du∙ x + udx . Then M ( 1, u ) dx + N ( 1, u ) [ du∙ x + udx ]= 0 M ( 1, u ) dx + N ( 1, u ) xdu + N ( 1, u ) udx = 0 [ M ( 1, u ) + N ( 1, u ) u ] dx + N ( 1, u ) xdu = 0 dx x + N ( 1, u ) du M ( 1, u ) + N ( 1, u ) u = 0 Thus, we get a separable DE in variables u and x . Example . Solve ( x 2 + y 2 ) dx + ( x 2 xy ) dy = 0 . Solution. In this case M ( x , y ) = x 2 + y 2 and N ( x , y ) = x 2 xy . Since y t ¿ ¿ M ( tx ,ty ) =( tx ) 2 + ¿ and N ( tx,ty ) =( tx )
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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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