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gerryliyana Group Title

Solve dy/dx = (x + 4y -1)^2

  • one year ago
  • one year ago

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  1. gerryliyana Group Title
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    solve \(\frac{ dy }{ dx } = (x + 4y -1)^{2}\)

    • one year ago
  2. carson889 Group Title
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    It is 2x + 8y - 2.

    • one year ago
  3. carson889 Group Title
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    Got it by expanding the brackets: x^2 + 4yx - x + 4xy + 16y^2 -4y -x - 4y + 1. Which simplifies to: x^2 + 8xy + 16y^2 - 2x -8y + 1. Taking the derivative of that with respect to x: 2x + 8y + 0 - 2 - 0 + 0 Which simplifies to 2x + 8y -2 or 2(x+4y-1)

    • one year ago
  4. carson889 Group Title
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    Or using chain rule: \[2(x+4y-1)^{1}(\frac{ d }{ dx }(x+4y-1))\] \[\frac{ d }{ dx } = 1\] Therfore, \[2(x+4y-1)(1) = 2x + 8y - 2\]

    • one year ago
  5. Idealist Group Title
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    2(x+4y-1)=2x+8y-2.

    • one year ago
  6. gerryliyana Group Title
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    ok thank all a lots i appreciate it :)

    • one year ago
  7. Idealist Group Title
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    No Problem.

    • one year ago
  8. gerryliyana Group Title
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    @carson889: wait i mean \[\frac{ dy }{ dx } = (x+4y - 1)^{}\] \((x+4y -1)^{2}\) is the result of \(\frac{ dy }{ dx }\) it isn't \[\frac{ d }{ dx} (x +4y -1)^{2}\] any idea???

    • one year ago
  9. LolWolf Group Title
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    You need to use the chain rule, as this is implicit differentiation, not partial.

    • one year ago
  10. gerryliyana Group Title
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    what did u get it??

    • one year ago
  11. LolWolf Group Title
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    So, are we trying to find \(y\) ?

    • one year ago
  12. LolWolf Group Title
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    (As a function of \(x\) )

    • one year ago
  13. gerryliyana Group Title
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    it seems like http://openstudy.com/study#/updates/50a9b0e8e4b064039cbd16a2

    • one year ago
  14. LolWolf Group Title
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    Yeah, so it's an actually differential equation which you have to solve. Here:

    • one year ago
  15. LolWolf Group Title
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    Define (all in terms of x): \[ v=4y+x\implies v'=4y'+1 \]So: \[ \frac{v'-1}{4}=(v-1)^2\implies\\ \frac{v'}{4v^2-8v+5}=1 \]Integrating both sides: \[ \int \frac{v'}{4v^2-8v+5}\;dx=\int 1\;dx=\\ \frac{\tan^{-1}(2v-2)}{2}=x+C \]Solving for \(v\): \[ v=\frac{\tan(2(x+C))+2}{2} \]Substituting \(y\): \[ y=\frac{\tan(2(x+C))-2x+2}{8} \] That took forever.

    • one year ago
  16. gerryliyana Group Title
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    awesome..., thank u @LolWolf

    • one year ago
  17. LolWolf Group Title
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    Sure thing

    • one year ago
  18. gerryliyana Group Title
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    @LolWolf

    • one year ago
  19. gerryliyana Group Title
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    i tried to calculate on Maple 13, but, look at the picture below

    • one year ago
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  20. gerryliyana Group Title
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    is not equal to (x+4y-1)^2

    • one year ago
  21. LolWolf Group Title
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    This is the equation I get, when solving the differential with Mathematica:

    • one year ago
  22. LolWolf Group Title
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    In real form:

    • one year ago
  23. gerryliyana Group Title
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    ok...,

    • one year ago
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  24. gerryliyana Group Title
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    it is right, isn't it?

    • one year ago
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