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anonymous

  • 5 years ago

All Calculus students dream :) Find all pair of non constant functions f, g such that (fg)'=f'g' and (fg)''=f''g''

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  1. anonymous
    • 5 years ago
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    i saw this yesterday:)

  2. anonymous
    • 5 years ago
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    oh yeah

  3. anonymous
    • 5 years ago
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    watchmath had answered:)

  4. anonymous
    • 5 years ago
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    no he didnt give A,B,C

  5. anonymous
    • 5 years ago
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    i forgot his solution, but

  6. anonymous
    • 5 years ago
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    lets do it now then..:)

  7. anonymous
    • 5 years ago
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    we had 2 eqns. (f/f')+(g/g')=1 and (f'/f")+(g'/g")=1

  8. anonymous
    • 5 years ago
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    its impossible

  9. anonymous
    • 5 years ago
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    we have f'g + fg' = f'g'

  10. anonymous
    • 5 years ago
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    and f''g + 2f'g' + fg'' = f''g''

  11. anonymous
    • 5 years ago
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    ya so dvide by f'g' both sides...tats same!

  12. nowhereman
    • 5 years ago
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    It's easy, just choose functions with disjoint support. Then either f or g and thus f' or g' likewise are zero.

  13. anonymous
    • 5 years ago
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    whats disjoint support

  14. nowhereman
    • 5 years ago
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    Support is the set where they are non-zero. So that just means at every point either of the functions is zero.

  15. anonymous
    • 5 years ago
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    f'g + fg' = f'g', f' ( g' - g ) /g' = f

  16. anonymous
    • 5 years ago
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    we have to find nonconstant solutions nowhereman!!

  17. anonymous
    • 5 years ago
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    f = fg' / ( g' - g)

  18. nowhereman
    • 5 years ago
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    Does f'g' have to be non-constant too?

  19. anonymous
    • 5 years ago
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    my bad f' = fg' / ( g' - g)

  20. anonymous
    • 5 years ago
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    now, take the derivative of this

  21. anonymous
    • 5 years ago
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    cantorset wat r u tryin t say? its f/f'+g/g'=1

  22. anonymous
    • 5 years ago
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    now do you get that?

  23. anonymous
    • 5 years ago
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    what is the problem the 2 eqns. (f/f')+(g/g')=1 and (f'/f")+(g'/g")=1

  24. anonymous
    • 5 years ago
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    saub, we know that (fg)' = f'g + fg' , correct ? so f'g + fg' = f'g'

  25. anonymous
    • 5 years ago
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    yes so why are my equations impossible? (f/f')+(g/g')=1 and (f'/f")+(g'/g")=1

  26. anonymous
    • 5 years ago
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    (g/g')+(f/f')=1 fg'' + 2g'f' + f''g=f''g'' saubhik: how do you get this? (f'/f")+(g'/g")=1

  27. nowhereman
    • 5 years ago
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    Well I would substitute \(F = \ln f\) and \(G = \ln g\) so that you get \(1 = F'(1 - 1/G')\) and thus \(G' = \int \frac{1}{1-\frac{1}{G'}}\)

  28. anonymous
    • 5 years ago
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    k...amogh u got me... (f/f')+(g/g')=1 is correct and u have to differentiate this and use the 2nd result...

  29. anonymous
    • 5 years ago
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    Yes and I get this: fg'' + 2g'f' + f''g = f''g''

  30. anonymous
    • 5 years ago
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    ok..so we have to solve: (g/g')+(f/f')=1 and fg'' + 2g'f' + f''g = f''g''

  31. anonymous
    • 5 years ago
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    PS:If this question were my dream it wud be my nightmare :P lol

  32. anonymous
    • 5 years ago
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    Yea but how did you get this? (f'/f")+(g'/g")=1, sorry if its stupid.

  33. anonymous
    • 5 years ago
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    hmmm, ln g(x) + ln f(x) = integral 1 ?

  34. anonymous
    • 5 years ago
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    f/f'+g/g'=1, if you take integral of both sides

  35. anonymous
    • 5 years ago
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    no tats not right i think!

  36. anonymous
    • 5 years ago
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    oh, thats f'/f , woops

  37. anonymous
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    integral of f/f' is not ln f!

  38. anonymous
    • 5 years ago
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    yes..amogh is correct

  39. anonymous
    • 5 years ago
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    Solving these will require deep math cantorset!!

  40. anonymous
    • 5 years ago
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    unless and until u start guessing:P

  41. anonymous
    • 5 years ago
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    it will be ugly to start solving there will be innumerable veriables and integrals...we have to use the exactness condition..

  42. anonymous
    • 5 years ago
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    f=e^{3x/2},g=e^{3x} works . (source:watchmath)

  43. anonymous
    • 5 years ago
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    haha

  44. anonymous
    • 5 years ago
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    i remember that solution

  45. anonymous
    • 5 years ago
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    so can we generalize...

  46. anonymous
    • 5 years ago
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    Can you send me the link to the question you posted earlier yesterday?

  47. anonymous
    • 5 years ago
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    http://openstudy.com/groups/mathematics/updates/4dda92c95d7a8b0b4fb71986

  48. anonymous
    • 5 years ago
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    Thanks

  49. nowhereman
    • 5 years ago
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    I already told you how to do it. Use the substitution G = ln g and F = ln f or in other terms g = e^G and f = e^F, then for every G you can find an F so that f'g' = (fg)' because the condition can be reformulated to say F' = 1/(1-1/G') which can be solved by integration.

  50. anonymous
    • 5 years ago
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    youre missing a G there

  51. anonymous
    • 5 years ago
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    what is integral(G'/(G'-1)) nowhereman?

  52. anonymous
    • 5 years ago
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    nowhereman is not missing a G cantorset..he used 1/(1-1/G') but how to do this integral???

  53. anonymous
    • 5 years ago
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    , his G is not the same as g

  54. anonymous
    • 5 years ago
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    G = ln g and F = ln f

  55. anonymous
    • 5 years ago
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    yups..i know.....but he is right cantorset u r a calculus tutor..how do we do this integral? any ideas?

  56. anonymous
    • 5 years ago
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    F'=G'/(G'-1) is correct.

  57. anonymous
    • 5 years ago
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    it is integral G ' / ( 1- G' ) ?

  58. anonymous
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    F'=G'/(G'-1) i mean

  59. anonymous
    • 5 years ago
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    yups..integrate both sides u get F=integral(G'/(G'-1)) what will be the RHS?

  60. anonymous
    • 5 years ago
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    first divide that

  61. anonymous
    • 5 years ago
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    k. so 1/(1-1/G') then?

  62. anonymous
    • 5 years ago
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    G'/(G'-1) = 1 + 1/ G' -1

  63. anonymous
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    G'/(G'-1) = 1 + 1/ (G' -1)

  64. anonymous
    • 5 years ago
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    how will u do the later? its the same problem :P

  65. anonymous
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    hmmm, with respect to x ?

  66. anonymous
    • 5 years ago
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    its a differential equation...we can differentiate with respect to the variables of the function..i.e. G(x) wrt x and F(x) wrt x...

  67. anonymous
    • 5 years ago
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    ok...lets go to the main problem after making nowhereman's substitutions we get F'G'=F'+G'..how will u solve this diff eqn?

  68. anonymous
    • 5 years ago
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    dunno

  69. anonymous
    • 5 years ago
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    please dont make fun of me...see that F'G'=F'+G' implies integral(dF(x)*dG(x))=F(x)+G(x) thats an integration with respect to 2 variables!!!

  70. anonymous
    • 5 years ago
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    we need multiple integration here? :)

  71. anonymous
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    am here cantorset..

  72. anonymous
    • 5 years ago
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    ok , im lost, how did you get that , capital F' and G'

  73. anonymous
    • 5 years ago
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    see F=lnf and G=lng then use f'g'=(fg)'

  74. anonymous
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    itll give F'G'=F'+G'

  75. anonymous
    • 5 years ago
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    then integrate but i dont know the integral of LHS..

  76. watchmath
    • 5 years ago
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    f'g+fg'=f'g' (f'-f)g'-f'g=0 g'- f'g/(f'-f)=0 Now you can use integrating factor method. The answer will be still in terms of f and f'. But that is ok, we will use the second requirement.

  77. anonymous
    • 5 years ago
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    yah...

  78. anonymous
    • 5 years ago
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    ok , multiply both sides by e ^ int ( f' / (f' - f)

  79. watchmath
    • 5 years ago
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    yes :)

  80. anonymous
    • 5 years ago
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    what is int ( f' / (f' - f)

  81. anonymous
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    keep it like this? we need to simplify...

  82. watchmath
    • 5 years ago
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    just keep it like that

  83. anonymous
    • 5 years ago
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    g.e ^ int ( f' / (f' - f) =0 is the answer

  84. watchmath
    • 5 years ago
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    =constant

  85. anonymous
    • 5 years ago
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    yess..

  86. anonymous
    • 5 years ago
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    how will u use the 2nd fact watchmath?

  87. anonymous
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    this is pretty ugly to differentiate..

  88. watchmath
    • 5 years ago
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    just plugin the g. You need fundamental calculus theorem to compute g'.

  89. anonymous
    • 5 years ago
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    website keeps frezing on me

  90. anonymous
    • 5 years ago
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    ok im a little slow, one sec

  91. anonymous
    • 5 years ago
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    watchmath how will u extract the g?

  92. anonymous
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    g'- f'/(f'-f)*g =0

  93. anonymous
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    k...lot of work though..

  94. anonymous
    • 5 years ago
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    but process is clear..

  95. anonymous
    • 5 years ago
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    now we know , dy/dx + p(x) *y = q(x), the solution is y = [int ( int e^p(x) )*q(x) dx + c ] / e^int p(x)

  96. anonymous
    • 5 years ago
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    g.e^[f'/(f'-f)]=constant.

  97. anonymous
    • 5 years ago
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    now since q(x) is zero, that simplifies the numerator, so its g = C / ( e^ int ( f

  98. anonymous
    • 5 years ago
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    ln|g|+[f'/(f'-f)]=constant

  99. anonymous
    • 5 years ago
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    now cantorset..how will u get g?

  100. anonymous
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    , looks like you made a mistake watchman

  101. anonymous
    • 5 years ago
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    how ill we use (fg)"=f"g"?

  102. anonymous
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    f ' g + f g' - f ' g ' = 0 g' ( f - f ' ) + f ' g = 0 g ' + f' / ( f- f ' ) g = 0

  103. anonymous
    • 5 years ago
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    cantorset the integral of what u obtained is ln|g|+[f'/(f'-f)]=constant

  104. anonymous
    • 5 years ago
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    g = C / e^ int ( f ' / ( f - f' )

  105. anonymous
    • 5 years ago
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    does that make sense so far?

  106. anonymous
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    g = C e^ int ( f ' / f' - f)

  107. anonymous
    • 5 years ago
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    ln g = C + int f' / ( f ' - f )

  108. anonymous
    • 5 years ago
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    saubik im not sure how you got that

  109. anonymous
    • 5 years ago
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    watchman, im sure that g = C / e^ int ( f ' / ( f - f' )

  110. anonymous
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    or g = C e^ int ( f ' / f' - f)

  111. anonymous
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    ya ur right

  112. anonymous
    • 5 years ago
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    saubhik, can you show me your reasoning

  113. anonymous
    • 5 years ago
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    yep but i simplified it...:) its ln|g|+[f'/(f-f')]=constant

  114. anonymous
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    its ok..we are getting the same results..

  115. anonymous
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    now how will u plug in g from (fg)"=f"g"

  116. anonymous
    • 5 years ago
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    saubhik, our answers dont match, you are missing an integral sign

  117. anonymous
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    oops..i meant it sorry :)

  118. anonymous
    • 5 years ago
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    oh ok , thats better

  119. anonymous
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    ln|g|+int[f'/(f-f')]=constant

  120. anonymous
    • 5 years ago
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    i guess you plug this g into the second equation, ?

  121. anonymous
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    f'' g + 2 f ' g' + f g'' - f '' g'' = 0

  122. anonymous
    • 5 years ago
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    yeah..but its madness...i guess who had this as "dream"!! :)

  123. anonymous
    • 5 years ago
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    i think this should be minus

  124. anonymous
    • 5 years ago
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    ln|g| - int[f'/(f-f')]=constant

  125. anonymous
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    no..i had interchange f and f' so minus is taken care of..

  126. anonymous
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    oh ok , i did it a little different

  127. anonymous
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    f'' g + 2 f ' g' + f g'' - f '' g'' = 0

  128. anonymous
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    this is not a very nice equation to plug g :( <<i need to go now.. cantorset...will come after smtym.bye>>

  129. anonymous
    • 5 years ago
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    nowhere man...how can we integrate F'G'=F'+G' ? using ur substitutions?

  130. nowhereman
    • 5 years ago
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    I don't think that the integration can be done for a general G. But to find examples this is a good approach. For example choose F = x^2.

  131. anonymous
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    well this is bad:(

  132. nowhereman
    • 5 years ago
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    No, I mean you can't find the integral of the general G itself. So why should you be able to find it of a term involving the general G. In fact you can choose G always in such a way, that the integrand is arbitrary. Set G = integral 1/(1-1/H) and you will get H = 1/(1-1/G').

  133. watchmath
    • 5 years ago
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    express g' and g'' in terms of g. Then you will have an equation in terms of f,f' and f''. It is ugly but you will be able to factor it to get (f'' - (f')^2 )(2f'-f '')=0. From here we have f=Ae^(Bx) or f=a+be^(cx) (since f andg symmetric, g also of the same form). Then we need to adjust the constants to make it right.

  134. watchmath
    • 5 years ago
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    Actually after I replace f,f',f'' by a,b,c it isn't too bad. Also actually we just need algebra to get g'=f'g/(f'-f) (no need invoking integral factor)

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