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

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.Sam.
  • .Sam.
\[xdx + ( y -2x)dy= 0\] \[xdx=-(y-2x)dy \\ -\frac{x}{y-2x}=\frac{dy}{dx}\] --------------------------------------- Try using y=vx Then \[\frac{dy}{dx}=v+x\frac{dv}{dx}\]
.Sam.
  • .Sam.
Replace the y and \(\frac{dy}{dx}\) to v and the one given above \[-\frac{x}{vx-2x}=v+x\frac{dv}{dx} \\ \frac{1}{2-v}=v+x\frac{dv}{dx}\]
anonymous
  • anonymous
yes then?
anonymous
  • anonymous
the numerator can break it for the denomentor?
.Sam.
  • .Sam.
Then set all v into the other side \[\frac{1-2v+v^2}{(2-v)x}=\frac{dv}{dx} \\ \frac{(2-v)x}{1-2v+v^2}=\frac{dx}{dv} \\ \int\limits \frac{(2-v)}{1-2v+v^2} dx=\int\limits \frac{1}{x}dx\]
.Sam.
  • .Sam.
Should be dv \[\int\limits\limits \frac{(2-v)}{1-2v+v^2} dv=\int\limits\limits \frac{1}{x}dx\]
dumbcow
  • dumbcow
i think you get stuck here though...no way to solve for "v" the integral gives a fraction and a log
anonymous
  • anonymous
i think we can use here synthetic division
anonymous
  • anonymous
my Im right?
dumbcow
  • dumbcow
@melmel , no that only works if the "2-v" was in denominator
dumbcow
  • dumbcow
http://www.wolframalpha.com/input/?i=integral+%282-v%29%2F%28v%5E2-2v%2B1%29
dumbcow
  • dumbcow
turns out you can't solve for "y" directly http://www.wolframalpha.com/input/?i=y%27%28x%29+%3D+x%2F%282x-y%28x%29%29
anonymous
  • anonymous
can you help me to solve and to prove that?
dumbcow
  • dumbcow
@.sam, already did but you cant go any further just sub in y/x for v
anonymous
  • anonymous
so what is my final answer?
dumbcow
  • dumbcow
i dunno depends on the instructions...ther is no "y =" answer
anonymous
  • anonymous
CAN YO HELP ME @zepdrix
anonymous
  • anonymous
coefficients by division?

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