nagle_differential_equations_ISM_Part12

nagle_differential_equations_ISM_Part12 - Exercises 2.4 18...

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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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nagle_differential_equations_ISM_Part12 - Exercises 2.4 18...

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