Question

answred

Answer 1

to find y any method

Answer 2

bernoulli equation or homogeneous

Answer 3

can you help me @.Sam.

Answer 4

\[xdx + ( y -2x)dy= 0\]
\[xdx=-(y-2x)dy \\ -\frac{x}{y-2x}=\frac{dy}{dx}\]
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Try using y=vx
Then
\[\frac{dy}{dx}=v+x\frac{dv}{dx}\]

Answer 5

Replace the y and \(\frac{dy}{dx}\) to v and the one given above
\[-\frac{x}{vx-2x}=v+x\frac{dv}{dx} \\ \frac{1}{2-v}=v+x\frac{dv}{dx}\]

Answer 6

yes then?

Answer 7

the numerator can break it for the denomentor?

Answer 8

Then set all v into the other side
\[\frac{1-2v+v^2}{(2-v)x}=\frac{dv}{dx} \\ \frac{(2-v)x}{1-2v+v^2}=\frac{dx}{dv} \\ \int\limits \frac{(2-v)}{1-2v+v^2} dx=\int\limits \frac{1}{x}dx\]

Answer 9

Should be dv
\[\int\limits\limits \frac{(2-v)}{1-2v+v^2} dv=\int\limits\limits \frac{1}{x}dx\]

Answer 10

i think you get stuck here though...no way to solve for "v" the integral gives a fraction and a log

Answer 11

i think we can use here synthetic division

Answer 12

my Im right?

Answer 13

@melmel , no that only works if the "2-v" was in denominator

Answer 14

http://www.wolframalpha.com/input/?i=integral+%282-v%29%2F%28v%5E2-2v%2B1%29

Answer 15

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

Answer 16

can you help me to solve and to prove that?

Answer 17

@.sam, already did but you cant go any further just sub in y/x for v

Answer 18

so what is my final answer?

Answer 19

i dunno depends on the instructions...ther is no "y =" answer

Answer 20

CAN YO HELP ME @zepdrix

Answer 21

coefficients by division?