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Let us show that in such a case the present method does not work. From (6), uϭ Ύ M dx ϩ k(y) ϭ Ϫxy ϩ k(y), hence dk 0u ϭ Ϫx ϩ . 0y dy Now, 0u> 0y should equal N ϭ x, by (4b). However, this is impossible because k(y) can depend only on y. Try ᭿ (6*); it will also fail. Solve the equation by another method that we have discussed. Reduction to Exact Form. Integrating Factors The ODE in Example 3 is Ϫy dx ϩ x dy ϭ 0. It is not exact. ], (11) Ϫy dx ϩ x dy x 2 ϭϪ y x 2 dx ϩ y 1 dy ϭ d a b ϭ 0. x x Integration of (11) then gives the general solution y>x ϭ c ϭ const.

Such a function F(x, y) is then called an integrating factor of (12). EXAMPLE 4 Integrating Factor The integrating factor in (11) is F ϭ 1>x 2. Hence in this case the exact equation (13) is FP dx ϩ FQ dy ϭ Ϫy dx ϩ x dy x 2 y ϭ d a b ϭ 0. x Solution y x ϭ c. These are straight lines y ϭ cx through the origin. ) It is remarkable that we can readily find other integrating factors for the equation Ϫy dx ϩ x dy ϭ 0, namely, 1>y 2, 1>(xy), and 1>(x 2 ϩ y 2), because (14) Ϫy dx ϩ x dy y 2 x ϭ d a b, y Ϫy dx ϩ x dy xy x ϭ Ϫd aln b , y Ϫy dx ϩ x dy x ϩy 2 2 y ϭ d aarctan b .

We can find this factor F(x) by Theorem 1 in the previous section or we can proceed directly, as follows. We multiply (1) by F(x), obtaining (1*) Fy r ϩ pFy ϭ rF. The left side is the derivative (Fy) r ϭ F r y ϩ Fy r of the product Fy if pFy ϭ F r y, pF ϭ F r . thus By separating variables, dF>F ϭ p dx. By integration, writing h ϭ ͐ p dx, ln ƒ F ƒ ϭ h ϭ Ύ p dx, F ϭ eh. thus With this F and h r ϭ p, Eq. (1*) becomes ehy r ϩ h r ehy ϭ ehy r ϩ (eh) r y ϭ (ehy) r ϭ reh. By integration, ehy ϭ Ύ e r dx ϩ c.

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