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gerryliyana

  • 2 years ago

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

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

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

  3. ksaimouli
    • 2 years ago
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    |dw:1353368437237:dw|

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

  5. ksaimouli
    • 2 years ago
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    do u want to solve for dy/dx

  6. ksaimouli
    • 2 years ago
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    use chain rule

  7. gerryliyana
    • 2 years ago
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    how to find y ??

  8. ksaimouli
    • 2 years ago
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    then how do u find x ?

  9. ksaimouli
    • 2 years ago
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    nothing but 1 dx/dxso y =1 dy/dx

  10. gerryliyana
    • 2 years ago
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    ok.., is it use separable equation or linear?

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

  12. ksaimouli
    • 2 years ago
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    hold on is this integration

  13. gerryliyana
    • 2 years ago
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    ok i'll try., :)

  14. saifoo.khan
    • 2 years ago
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    You have to differentiate it? Right?

  15. gerryliyana
    • 2 years ago
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    i have to find the solution of differential equations...,

  16. saifoo.khan
    • 2 years ago
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    Meaning you have to differentiate that equation?

  17. gerryliyana
    • 2 years ago
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    yes

  18. hartnn
    • 2 years ago
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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.

  19. gerryliyana
    • 2 years ago
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    @hartnn

  20. gerryliyana
    • 2 years ago
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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 ??

  21. gerryliyana
    • 2 years ago
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    oh i'm sorry, i made a mistake...,

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

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

  24. hartnn
    • 2 years ago
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    the co-efficient is divided, not multiplied.

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

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

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

  28. hartnn
    • 2 years ago
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    yes, seems correct now :)

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

  30. hartnn
    • 2 years ago
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    welcome ^_^

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