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DE - Solutions By Substitutions (algebraic manipulation problem)

  1. May 16, 2012 #1
    I can't for the life of me understand how the book went from step A to step B--even though I realize its just a pesky algebraic manipulation.

    Step A

    [tex] M(1,u) dx +N(1,u)[u dx+ x du] = 0 [/tex]

    Step B

    [tex] [M(1,u) + uN(1,u) ]dx + xN(1,u)du = 0[/tex]

    Step A to Step B? I tried factoring and tinkering around here and there but all was in vain.
     
  2. jcsd
  3. May 16, 2012 #2

    SammyS

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    You'll be embarrassed if we work it out for you !

    [itex] M(1,u) dx +N(1,u)[u dx+ x du] = 0 [/itex]  [itex] M(1,u) dx + u N(1,u) dx+ x N(1,u) du = 0 [/itex]
     
  4. May 16, 2012 #3
    To me it just seems that you wrote step B in an extremely similar form. I don't understand how you got to that step either. There must be something I'm misinterpreting because it seems like your missing at least a couple terms (if you factor).
     
  5. May 16, 2012 #4

    SammyS

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    Distribute the N(1,u):
    [itex]N(1,u)[u dx+ x du][/itex]​
     
  6. May 16, 2012 #5
    I always thought one must distribute everything? Given how we have two polynomials, shouldn't you always foil? Take for example:

    [tex] (x+2)(x-2) [/tex] It wouldn't make sense to only distribute the 2 in the first bracket and say the expression is equal to
    [tex]x+2x-4[/tex]
     
  7. May 17, 2012 #6

    SammyS

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    Actually, that's [itex]x^2+2x-4[/itex]

    but for your problem ...

    a + b(c+e) is a + bc + be . --- Right ?

    The a is M(1,u)dx ,

    the b is N(1,u) ,

    the c is u dx ,

    and the e is x du
     
  8. May 17, 2012 #7
    I know it is x^2+2x-4. But I was saying that by your method it would have been x+2x-4.

    Your logic is spot on. But something is still bothering me.

    a + b(c+e) is a + bc + be

    The a is x
    The b is +2
    The c is x
    The e is -2

    So then it would follow
    [tex]a+bc+be[/tex]
    [tex]x+2x-4[/tex]

    So What happened here? It doesn't come out to x^2+2x-4 like I was trying to say before.

    The correct answer would have been ac+bc+be. I know what you said is consistent but something is being mixed up here.

    Edit:

    Maybe what makes the difference is that there aren't brackets aroudn the first two terms. So that

    [tex] (a+b) (c+e) ≠ a+b(c+e)[/tex]

    Is this correct?
     
  9. May 17, 2012 #8

    SammyS

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    I misread what you said here.

    Yes, of course [itex] (x+2)(x-2) [/itex] is the same as [itex]x^2+2x-4\ .[/itex]

    What you have in this problem is more like [itex] x+2(x-2)\,, [/itex] which is indeed the same as [itex]x+2x-4\ .[/itex]

    In Step A) of your original post, the M(1,u) does not get distributed through anything.
     
    Last edited: May 17, 2012
  10. May 17, 2012 #9
    That's correct. Parentheses can make all the difference in an answer.
     
  11. May 17, 2012 #10
    Oh thanks!

    Wow, that is a really important algebraic property to know. I find it funny that I learn it now while in differential equations! :rofl:
     
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