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melmel

  • one year ago

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  1. melmel
    • one year ago
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    to find y any method

  2. melmel
    • one year ago
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    bernoulli equation or homogeneous

  3. melmel
    • one year ago
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    can you help me @.Sam.

  4. .Sam.
    • one year ago
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    \[xdx + ( y -2x)dy= 0\] \[xdx=-(y-2x)dy \\ -\frac{x}{y-2x}=\frac{dy}{dx}\] --------------------------------------- Try using y=vx Then \[\frac{dy}{dx}=v+x\frac{dv}{dx}\]

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

  6. melmel
    • one year ago
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    yes then?

  7. melmel
    • one year ago
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    the numerator can break it for the denomentor?

  8. .Sam.
    • one year ago
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    Then set all v into the other side \[\frac{1-2v+v^2}{(2-v)x}=\frac{dv}{dx} \\ \frac{(2-v)x}{1-2v+v^2}=\frac{dx}{dv} \\ \int\limits \frac{(2-v)}{1-2v+v^2} dx=\int\limits \frac{1}{x}dx\]

  9. .Sam.
    • one year ago
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    Should be dv \[\int\limits\limits \frac{(2-v)}{1-2v+v^2} dv=\int\limits\limits \frac{1}{x}dx\]

  10. dumbcow
    • one year ago
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    i think you get stuck here though...no way to solve for "v" the integral gives a fraction and a log

  11. melmel
    • one year ago
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    i think we can use here synthetic division

  12. melmel
    • one year ago
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    my Im right?

  13. dumbcow
    • one year ago
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    @melmel , no that only works if the "2-v" was in denominator

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

  15. dumbcow
    • one year ago
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    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

  16. melmel
    • one year ago
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    can you help me to solve and to prove that?

  17. dumbcow
    • one year ago
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    @.sam, already did but you cant go any further just sub in y/x for v

  18. melmel
    • one year ago
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    so what is my final answer?

  19. dumbcow
    • one year ago
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    i dunno depends on the instructions...ther is no "y =" answer

  20. melmel
    • one year ago
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    CAN YO HELP ME @zepdrix

  21. melmel
    • one year ago
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    coefficients by division?

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