68_pdfsam_math 54 differential equation solutions odd

68_pdfsam_math 54 differential equation solutions odd -...

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Chapter 2 19. Taking partial derivatives of M ( x, y )=2 x + y/ (1 + x 2 y 2 )and N ( x, y )= 2 y + x/ (1 + x 2 y 2 ) with respect to y and x , respectively, we get ∂M ∂y = ∂y ± 2 x + y 1+ x 2 y 2 ² = (1)(1 + x 2 y 2 ) yx 2 (2 y ) (1 + x 2 y 2 ) 2 = 1 x 2 y 2 (1 + x 2 y 2 ) 2 , ∂N ∂x = ∂x ± 2 y + x 1+ x 2 y 2 ² = (1)(1 + x 2 y 2 ) xy 2 (2 x ) (1 + x 2 y 2 ) 2 = 1 x 2 y 2 (1 + x 2 y 2 ) 2 . Therefore, the equation is exact. F ( x, y )= Z ± 2 x + y 1+ x 2 y 2 ² dx = x 2 + Z d ( xy ) 1+( xy ) 2 = x 2 + arctan( xy )+ g ( y ) ∂F ∂y = ∂y ³ x 2 + arctan( xy )+ g ( y ) ´ = x 1+( xy ) 2 + g 0 ( y )= 2 y + x 1+ x 2 y 2 g 0 ( y )= 2 y g ( y )= y 2 F ( x, y )= x 2 y 2 + arctan( xy ) and a general solution then is given implicitly by x 2 y 2 + arctan( xy )= c . 21. We check the equation for exactness. We have M ( x, y )=1 /x +2 y 2 x , N ( x, y )=2 yx 2 cos y , ∂M ∂y = ∂y ± 1 x +2 y 2 x ² =4 yx, ∂N ∂x = ∂x ( 2 yx 2 cos y ) =4 yx. Thus ∂M/∂y =
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Unformatted text preview: ∂N/∂x . Integrating M ( x, y ) with respect to x yields F ( x, y ) = Z ± 1 x + 2 y 2 x ² dx = ln | x | + x 2 y 2 + g ( y ) . Therefore, ∂F ∂y = ∂ ∂y ³ ln | x | + x 2 y 2 + g ( y ) ´ = 2 x 2 y + g ( y ) = N ( x, y ) = 2 yx 2 − cos y ⇒ g ( y ) = − cos y ⇒ g ( y ) = Z ( − cos y ) dy = − sin y ⇒ F ( x, y ) = ln | x | + x 2 y 2 − sin y, and a general solution to the given diFerential equation is ln | x | + x 2 y 2 − sin y = c. 64...
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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 University of California, Berkeley.

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