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I'm trying to find an integrating factor of the form $u(x,y)$ for the equation $$(a\cos xy-y\sin xy)~dx+(b\cos xy-x\sin xy)~dy=0$$ using an approach suggested in the comments for this post: https://math.stackexchange.com/a/139160/170231

I'm wondering if this "greedy" method allows for me to try to find a function $C(x,y)$ such that I can eliminate the appropriate variable when solving for either $F(x)$ or $G(y)$.

Work thus far: $$\begin{align*} F(x)(b\cos xy-x\sin xy)\quad&\\ -G(y)(a\cos xy-y\sin xy)&=(-ax\sin xy-\sin xy-xy\cos xy)\\&\quad-(-by\sin xy-\sin xy-xy\cos xy)\\\\ F(x)(b\cos xy-x\sin xy)\quad&\\ -G(y)(a\cos xy-y\sin xy)&=(by-ax)\sin xy\\\\ F(x)&=\frac{(by-ax)\sin xy+G(y)(a\cos xy-y\sin xy)}{b\cos xy-x\sin xy} \end{align*}$$ and so, based on that post, I would try to look for a constant or function $C$ that, when setting $$(by-ax)\sin xy+G(y)(a\cos xy-y\sin xy)=C(b\cos xy-x\sin xy)$$ let's me do a proper simplification. Is it possible?

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user170231
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  • I think I might have a solution. I tried setting $F(x)=\alpha x^\gamma$ and $G(y)=\beta y^\delta$, and I find that $\alpha=a,~\beta=b,~\gamma=\delta=0$, which gives an integrating factor $u(x,y)=\exp(ax+by)$, which seems to work! If anyone else has any suggestions, please feel free to leave them. – user170231 Oct 28 '14 at 06:30

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A very simple way to solve the ODE is : $$(a+b y')\cos(xy)=(y+x y')\sin(xy)=-\big(cos(xy)\big)'$$ $$a+b y'+\frac{\big(cos(xy)\big)'}{cos(xy)}=0$$ $$a x+by+ln(cos(xy))=C$$

In addition :

With the method that you are trying, you got to: $$F(x)(b \cos xy-x \sin xy)-G(y)(a \cos xy-y \sin xy)= (by-ax)\sin xy $$ $$(bF-aG)\cos xy-(xF-yG-by+ax)\sin xy= 0$$ $$bF-aG=0$$ $$xF-yG-by+ax=0$$ leading to : $$F=a$$ $$G=b$$ and the integrating factor is $$e^{ax+by}$$ The initial equation becomes : $$e^{ax+by}(a\cos xy-y\sin xy)~dx+e^{ax+by}(b\cos xy-x\sin xy)~dy=0$$ which is the total differential of $e^{ax+by}\cos(xy)=c$ So, the solution on implicit form is : $$e^{ax+by}\cos(xy)=c$$ or : $$ax+by+ln(\cos xy)=ln(c)=C$$

JJacquelin
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