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I want to solve the differential equation $(3xy+y^2)+(x^2+xy)y'=0$. If I use the integrating factor $\mu (x)=x$ so that the original differential equation becomes exact, then the general solution that I get does include the the particular solution $y=-2x$. But if I use the integrating factor $\mu (x)=\frac{1}{xy(2x+y)}$ (so that the original differential equation becomes exact), then the general solution that I get does not include the particular solution $y=-2x$. Why? What should I do so that an integrating factor that I choose will give me the general solution that includes all particular solutions of a differential equation that I want to solve?

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What I get with your second integrating factor is a general (implicit) solution expressed as $$ \ln(x) + \dfrac{1}{2} \ln(2x+y) + \dfrac{1}{2} \ln(y) = c $$ The particular solution $y = -2x$ is not really "missing": it is a limiting case as $c \to -\infty$. If you multiply by $2$ and take exponential of both sides, it becomes $$ x^2 y (2x+y) = e^{2c} $$ which with $e^{2c}$ renamed is the same general solution as I get with the first integrating factor.

I don't think there's any way to avoid all cases where particular solutions occur only as limiting cases of the general solution.

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  • $\begingroup$ Can we declare that both of the integrating factors lead to the same set of solutions? $\endgroup$ – shuxue Jun 17 '16 at 17:55

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