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Unformatted text preview: Exercises 2.4 18. Since M y = N x = 2 y 2 + sin( x + y ) , the equation is exact. We find F ( x,y ) = Z 2 x + y 2 cos( x + y ) dx = x 2 + xy 2 sin( x + y ) + g ( y ) , F y = 2 xy cos( x + y ) + g ( y ) = 2 xy cos( x + y ) e y g ( y ) = e y g ( y ) = e y . Therefore, F ( x,y ) = x 2 + xy 2 sin( x + y ) e y = C gives a general solution. 20. We find M y = y [ y cos( xy )] = cos( xy ) xy sin( xy ) , N x = x [ x cos( xy )] = cos( xy ) xy sin( xy ) . Therefore, the equation is exact and F ( x,y ) = Z ( x cos( xy ) y 1 / 3 ) dy = sin( xy ) 3 2 y 2 / 3 + h ( x ) F x = y cos( xy ) + h ( x ) = 2 1 x 2 + y cos( xy ) h ( x ) = 2 1 x 2 h ( x ) = 2 arcsin x, and a general solution is given by sin( xy ) 3 2 y 2 / 3 + 2 arcsin x = C. 22. In Problem 16, we found that a general solution to this equation is e xy xy 1 = C. Substituting the initial condition, y (1) = 1, yields e 1 = C . So, the answer is e xy xy 1 = e 1 . 51 Chapter 2 24. First, we check the given equation for exactness. dM dx = e t = N t . So, it is exact. We find F ( t,x ) = Z ( e t 1 ) dx = x ( e t 1 ) + g ( t ) , F t = xe t + g ( t ) = xe t + 1 g ( t ) = Z dt = t x ( e t 1 ) + t = C is a general solution. With x (1) = 1, we get (1) ( e 1) + 1 = C C = e, and the solution is given by...
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 Spring '08
 MAZMANI
 Equations

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