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Best_Mathematician

find general solution of this differential equation dy/dx = y-2

  • one year ago
  • one year ago

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  1. iambehindyou
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    oh, its easily variable separable, dy/ (y-2) =dx did you try integrating this ?

    • one year ago
  2. Best_Mathematician
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    no...i got confused if i need to subtract it or what

    • one year ago
  3. iambehindyou
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    no subtract. dy/ (y-2) =dx got this ? can you integrate ?

    • one year ago
  4. Best_Mathematician
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    oh ya...is the answer |dw:1364766331963:dw|

    • one year ago
  5. iambehindyou
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    yes, now do you need to isolate 'y' also ? if not, thats your answer.

    • one year ago
  6. Best_Mathematician
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    ya i need to get y out

    • one year ago
  7. Best_Mathematician
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    as i need to find c

    • one year ago
  8. iambehindyou
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    write c as ln c'

    • one year ago
  9. iambehindyou
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    then use log properties.

    • one year ago
  10. Best_Mathematician
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    lets say we getta find c at x=1, y=1 how?

    • one year ago
  11. iambehindyou
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    you are making this up ? general solution means we need 'c' nevertheless, y can't be 1 here.

    • one year ago
  12. Best_Mathematician
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    |dw:1364766554143:dw|

    • one year ago
  13. Best_Mathematician
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    ya i am making up...now lets find particular solution

    • one year ago
  14. iambehindyou
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    y must be >=2 so that log term is defined...

    • one year ago
  15. Best_Mathematician
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    no it is not

    • one year ago
  16. Best_Mathematician
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    it is 1 and 1

    • one year ago
  17. iambehindyou
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    ln of negative numbers are not defined in reals.

    • one year ago
  18. Best_Mathematician
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    then we can't find c?

    • one year ago
  19. iambehindyou
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    do we need to find 'c'? and yes, if x=y=1, then c is complex number...

    • one year ago
  20. Best_Mathematician
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    yes we need to find particular solution

    • one year ago
  21. Best_Mathematician
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    at y(1) = 1

    • one year ago
  22. cwrw238
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    integrating gives ln (y - 2) = x + c so y - 2 = e^(x + c) = e^x * e*c = Ae^x y = Ae^x + 2

    • one year ago
  23. iambehindyou
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    oh, good....now we don't have to actually worry about c being complex..

    • one year ago
  24. Best_Mathematician
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    whats Ae^x

    • one year ago
  25. Best_Mathematician
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    no we dont its calculus

    • one year ago
  26. iambehindyou
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    you didn't get this ? : ln (y - 2) = x + c so y - 2 = e^(x + c) = e^x * e*c = Ae^x y = Ae^x + 2 what you have doubt with ?

    • one year ago
  27. iambehindyou
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    just plug in x=1, y=1 there,and you'll get A

    • one year ago
  28. Best_Mathematician
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    ok now the initial condition is y(0)=1

    • one year ago
  29. iambehindyou
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    so, put x=0 y=1 this should not be any problem...

    • one year ago
  30. Best_Mathematician
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    |dw:1364767623910:dw|

    • one year ago
  31. Best_Mathematician
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    |dw:1364767697343:dw|

    • one year ago
  32. iambehindyou
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    no, put e^c = A you can keep the general form in A and A=-2 is valid.

    • one year ago
  33. iambehindyou
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    you have to find a constant, not specifically 'c' A can be used, as valid constant.

    • one year ago
  34. Best_Mathematician
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    can 1 put c1 and c2 instead....i really dont know what is A

    • one year ago
  35. iambehindyou
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    sure, you can even put 'Best_Mathematician' instead Best_Mathematician = -2 :P

    • one year ago
  36. Best_Mathematician
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    ya whatever lol...so A is just a constant right

    • one year ago
  37. iambehindyou
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    yup.

    • one year ago
  38. Best_Mathematician
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    haha thanks...do you want going further in this equation

    • one year ago
  39. iambehindyou
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    this is your question...do you wanna go ?

    • one year ago
  40. Best_Mathematician
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    ya lol..... Use a technological tool and Euler's method with ten steps of size 0.1 to estimate y(1). Make certain to include supporting screen captures.

    • one year ago
  41. iambehindyou
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    sorry... that was latin to me :P

    • one year ago
  42. Best_Mathematician
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    ok

    • one year ago
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