arrow_back Show that if $(\partial N / \partial x-\partial M / \partial y) /(x M-y N)$ depends only on the product $x y$, that is

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Show that if $(\partial N / \partial x-\partial M / \partial y) /(x M-y N)$ depends only on the product $x y$, that is
$$\frac{\partial N / \partial x-\partial M / \partial y}{x M-y N}=H(x y)$$
then the equation $M(x, y) d x+N(x, y) d y=0$ has an integrating factor of the form $\mu=\mu(x y)$. Give the general formula for $\mu(x y)$.

Solution We look for an integrating factor of the form $\mu=\mu(x y)$. Multplying through by $\mu$ gives
$$\mu(x y) M(x, y) d x+\mu(x y) N(x, y) d y=0$$
which is exact if $\frac{\partial}{\partial y}(\mu(x y) M(x, y))=\frac{\partial}{\partial x}(\mu(x y) N(x, y))$. Expanding yields
$$\mu(x y) \frac{\partial M}{\partial y}+x \mu^{\prime}(x y) M=\mu(x y) \frac{\partial N}{\partial x}+y \mu^{\prime}(x y) N$$
which upon rearrangement results in
$$\frac{\mu^{\prime}(x y)}{\mu(x y)}=\frac{\partial N / \partial x-\partial M / \partial y}{x M-y N}$$
Thus, we see that an integrating factor of the form $\mu=\mu(x y)$ exists provided that the R.H.S. of (5) satisfies
$$\frac{\partial N / \partial x-\partial M / \partial y}{x M-y N}=H(x y)$$
for some function $H$. The formula for the integrating factor is then given by
$$\mu(t)=e^{\int H(t) d t}$$
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