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It's been quite a while since I last dealed with DE's. I'd appreciate if you could help me with the official, or usual, classification of the next DE's and/or if there are some definite methods to solve them. Hints will also be welcome:


$$(2)\;\;\;\;\;\;\;\;y'x+y\left(\ln^2x+\ln^2y-2\ln x\ln y\right)=0\;\;,\;x,y>0$$

I'm guessing here one could write

$$\ln^2x+\ln^2y-2\ln x\ln y=\left(\ln x-\ln y\right)^2=\ln^2\frac{x}{y}$$


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Are you sure you copied (1) correctly? Apart from the $4x+8y-6$, the right side depends only on $y+2x$. If you changed $4x+8y-6$ to $4y+8x-6$, things would be much simpler. – Robert Israel Nov 29 '12 at 0:20
Of course, you're right! – DonAntonio Nov 29 '12 at 0:28
I am fond of ODEs. Nice Don. I didn't know you posted a question in this area. + – Babak S. Mar 11 '13 at 18:47
up vote 3 down vote accepted

That guess is a good one. Further hint: $y'$ depends only on $y/x$.

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Thanks, I see your point: $$y'=-\frac{y}{x}\,\log^2\frac{y}{x}$$ Putting now $$z=\frac{y}{x}\Longrightarrow \frac{dy}{dx}=z+x\frac{dz}{dx}$$ so the eq. becomes $$z+x\frac{dz}{dx}=-z\log^2z\Longrightarrow-\frac{dz}{z(\log^2z+1)}=\frac{dx}{x}‌​\Longrightarrow$$ $$-\arctan\log z=\log x+C\Longrightarrow\log z=\tan(-\log z)=\tan\log x+C'$$ I wonder whether this is enough... – DonAntonio Nov 29 '12 at 0:27

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