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Best_Mathematician

  • one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  15. Best_Mathematician
    • one year ago
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    no it is not

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

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

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

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

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

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

  22. cwrw238
    • one year ago
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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

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

  24. Best_Mathematician
    • one year ago
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    whats Ae^x

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

  26. iambehindyou
    • one year ago
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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 ?

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

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

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

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

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

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

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

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

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

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

  37. iambehindyou
    • one year ago
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    yup.

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

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

  40. Best_Mathematician
    • one year ago
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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.

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

  42. Best_Mathematician
    • one year ago
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    ok

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