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Differential Equations
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to find y any method
bernoulli equation or homogeneous
can you help me @.Sam.

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Other answers:

\[xdx + ( y -2x)dy= 0\] \[xdx=-(y-2x)dy \\ -\frac{x}{y-2x}=\frac{dy}{dx}\] --------------------------------------- Try using y=vx Then \[\frac{dy}{dx}=v+x\frac{dv}{dx}\]
Replace the y and \(\frac{dy}{dx}\) to v and the one given above \[-\frac{x}{vx-2x}=v+x\frac{dv}{dx} \\ \frac{1}{2-v}=v+x\frac{dv}{dx}\]
yes then?
the numerator can break it for the denomentor?
Then set all v into the other side \[\frac{1-2v+v^2}{(2-v)x}=\frac{dv}{dx} \\ \frac{(2-v)x}{1-2v+v^2}=\frac{dx}{dv} \\ \int\limits \frac{(2-v)}{1-2v+v^2} dx=\int\limits \frac{1}{x}dx\]
Should be dv \[\int\limits\limits \frac{(2-v)}{1-2v+v^2} dv=\int\limits\limits \frac{1}{x}dx\]
i think you get stuck here way to solve for "v" the integral gives a fraction and a log
i think we can use here synthetic division
my Im right?
@melmel , no that only works if the "2-v" was in denominator
turns out you can't solve for "y" directly
can you help me to solve and to prove that?
@.sam, already did but you cant go any further just sub in y/x for v
so what is my final answer?
i dunno depends on the instructions...ther is no "y =" answer
CAN YO HELP ME @zepdrix
coefficients by division?

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