100_pdfsam_math 54 differential equation solutions odd

100_pdfsam_math 54 differential equation solutions odd - xy...

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Chapter 2 23. Given equation is homogeneous because dy dx = x y x + y = 1 ( y/x ) 1+( y/x ) . Therefore, substituting u = y/x , we obtain a separable equation. u + x du dx = 1 u 1+ u x du dx = u 2 2 u +1 1+ u u +1 u 2 +2 u 1 du = dx x Z 1+ u u 2 +2 u 1 du = Z dx x 1 2 ln | u 2 +2 u 1 | = ln | x | + C 1 ( u 2 +2 u 1 ) x 2 = C, and, substituting back u = y/x , after some algebra we get a general solution y 2 +2 xy x 2 = C . 25. In this diFerential form, M ( x, y )= y ( x y 2) and N ( x, y )= x ( y x + 4). Therefore, ∂M ∂y = x 2 y 2 , ∂N ∂x = y 2 x +4 ∂N/∂x ∂M/∂y M = ( y 2 x +4) ( x 2 y 2) y ( x y 2) = 3( x y 2) y ( x y 2) = 3 y , which is a function of y alone. Therefore, the equation has a special integrating factor µ ( y ). We use formula (9) on page 70 of the text to ±nd that µ ( y )= y 3 . Multiplying the equation by µ ( y ) yields y 2 ( x y 2) dx
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Unformatted text preview: + xy − 3 ( y − x + 4) dy = 0 ⇒ F ( x, y ) = Z y − 2 ( x − y − 2) dx = y − 2 x 2 2 − ( y − 1 + 2 y − 2 ) x + g ( y ) ⇒ ∂F ∂y = − y − 3 x 2 − ( − y − 2 − 4 y − 3 ) x + g ( y ) = N ( x, y ) = xy − 3 ( y − x + 4) ⇒ g ( x ) = 0 ⇒ g ( y ) ≡ , and so F ( x, y ) = y − 2 x 2 2 − x ( y − 1 + 2 y − 2 ) = C 1 ⇒ x 2 y − 2 − 2 xy − 1 − 4 xy − 2 = C is a general solution. In addition, y ≡ 0 is a solution that we lost when multiplied the equation by µ ( y ) = y − 3 (i.e., divided by y 3 ). 96...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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