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Unformatted text preview: Exercises 2.4
Diﬀerentiating F (x, y ) with respect to y gives Fy (x, y ) = x2 + g (y ) = N (x, y ) = x2 − 1. From this we see that g = −1. (As a partial check we note that g (y ) does not involve x.) Integrating gives g (y ) = (−1) dy = −y. Since the constant of integration will be incorporated into the parameter of the solution, it is not written here. Substituting this expression for g (y ) into the expression that we found for F (x, y ) yields F (x, y ) = x2 y + 3x − y. Therefore, the solution of the diﬀerential equation is x2 y + 3x − y = C ⇒ y= C − 3x . x2 − 1 The given equation could be solved by the method of grouping. To see this, express the diﬀerential equation in the form (2xy dx + x2 dy ) + (3 dx − dy ) = 0. The ﬁrst term of the left-hand side we recognize as the total diﬀerential of x2 y . The second term is the total diﬀerential of (3x − y ). Thus we again ﬁnd that F (x, y ) = x2 y + 3x − y and, again, the solution is x2 y + 3x − y = C . 11. Computing partial derivatives of M (x, y ) = cos x cos y + 2x and N (x, y ) = −(sin x sin y + 2y ), we obtain ∂ ∂M = (cos x cos y + 2x) = − cos x sin y , ∂y ∂y ∂N ∂ = [− (sin x sin y + 2y )] = − cos x sin y , ∂x ∂x ∂M ∂N ⇒ = , ∂y ∂x and the equation is exact. 61 ...
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