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gerryliyana

solve dy/dx= (x + 4y - 1)^{2}

  • one year ago
  • one year ago

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  1. gerryliyana
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    \[\frac{ dy }{ dx } = (x + 4y - 1)^{2}\]

    • one year ago
  2. gerryliyana
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    @jim_thompson5910 can u help me?

    • one year ago
  3. ksaimouli
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    |dw:1353368437237:dw|

    • one year ago
  4. gerryliyana
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    i'm sorry @ksaimouli , but i cannot see clearly

    • one year ago
  5. ksaimouli
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    do u want to solve for dy/dx

    • one year ago
  6. ksaimouli
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    use chain rule

    • one year ago
  7. gerryliyana
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    how to find y ??

    • one year ago
  8. ksaimouli
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    then how do u find x ?

    • one year ago
  9. ksaimouli
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    nothing but 1 dx/dxso y =1 dy/dx

    • one year ago
  10. gerryliyana
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    ok.., is it use separable equation or linear?

    • one year ago
  11. ksaimouli
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    it is linear no need to find y until they mentioned y=something u will find it in optimization

    • one year ago
  12. ksaimouli
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    hold on is this integration

    • one year ago
  13. gerryliyana
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    ok i'll try., :)

    • one year ago
  14. saifoo.khan
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    You have to differentiate it? Right?

    • one year ago
  15. gerryliyana
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    i have to find the solution of differential equations...,

    • one year ago
  16. saifoo.khan
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    Meaning you have to differentiate that equation?

    • one year ago
  17. gerryliyana
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    yes

    • one year ago
  18. hartnn
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    dy/dx= (x + 4y - 1)^{2} u = x+4y-1 du/dx = 1 + 4dy/dx so, dy/dx = (1/4)*(du/dx -1) (1/4)*(du/dx -1) = u^2 du/dx = 4u^2 +1 now variables are easily seperable.

    • one year ago
  19. gerryliyana
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    @hartnn

    • one year ago
  20. gerryliyana
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    \[\frac{ du }{ dx } = 4u^{2} + 1\] \[\int\limits \frac{ du }{ 4u^{2}+1 } = \int\limits dx\] \[\int\limits \frac{ du }{ (2u)^{2}+1 } = \int\limits dx\] \[\tan^{-1} (2u) = x\] \[\tan^{-1} (x+4y-1) = x\] \[\tan x = x +4y-1\] \[4y = \tan x - x + 1\] \[y = \frac{ \tan x - x + 1 }{ 4 }\] am i right @hartnn ??

    • one year ago
  21. gerryliyana
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    oh i'm sorry, i made a mistake...,

    • one year ago
  22. hartnn
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    tan^-1 2u /2 this ?

    • one year ago
  23. gerryliyana
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    \[\tan^{-1} (2(x+4y-1)) = x\] \[\tan^{-1} (2x+8y-2)) = x\] \[\tan x = 2x + 8y - 2\] \[y = \frac{ \tan x - 2x + 2 }{ 8 }\]

    • one year ago
  24. hartnn
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    the co-efficient is divided, not multiplied.

    • one year ago
  25. hartnn
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    (tann^-1.... )/2

    • one year ago
  26. hartnn
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    tan2x = x+4y-1 + c don't forget +c

    • one year ago
  27. gerryliyana
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    ok \[\tan 2x = x + 4y -1 + C\] \[y = \frac{ \tan 2x - x + 1 - C }{ 4 }\] is it right ?

    • one year ago
  28. hartnn
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    yes, seems correct now :)

    • one year ago
  29. gerryliyana
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    yes i got it.., Thanks a lot. I appreciate all your help :)

    • one year ago
  30. hartnn
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    welcome ^_^

    • one year ago
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