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Choice of basis

  1. Nov 21, 2005 #1
    for the problem:
    y``+y=secx

    why can't 0 and e^(-1x) be choices for the basis?
     
  2. jcsd
  3. Nov 21, 2005 #2

    HallsofIvy

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    Be precise! What you mean is "why can't 0 and e-x be a basis for the solution space of the corresponding homogeneous equation".

    I am sorely tempted to ask why in the world you think it should be. For one thing, the "0" vector (here, the function that is identically 0) is never a member of a basis- any set of vectors including 0 cannot be independent. But then I also wonder where you got "e-x[/sup}"!
    What is the characteristic equation for y"= y= 0? What are its solutions?
     
  4. Nov 22, 2005 #3
    i figured that if you wanted to solve the homogenous equation, then suppose that y=e^(namda)x
    and subsitute that into the equation to find namda which equals 0 and
    (-1)...
    so i thought that since those two are the solutions to the homogeneous question, then couldn't they be the basis?
     
  5. Nov 22, 2005 #4

    matt grime

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    When did e^{namdax} (lambda?) become zero when namda is zero?

    And the qustion still remains: a basis of what?

    You haven't solved y''+y=0 correctly anyway. e^0 is constant, and y=const is not a solution of that equation, nor for that matter is e^{-x}
     
    Last edited: Nov 22, 2005
  6. Nov 22, 2005 #5

    HallsofIvy

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    First, it's "lambda", not "nambda":smile:. Yes, you "suppose that [itex]y= e^{\lambda x}[/itex] and substitute that into the equation to find lamba which equals..." No, it does not equal 0 and -1! I asked before, what is the characteristic equation? What equation do you get when you substitute [itex]y= e^{\lambda x}[/itex]?
     
  7. Nov 24, 2005 #6
    sorry about the spelling! thanks for correcting that!
    the characteristic equation is (lambda)^2+(lambda)=0
    => [(lamda)((lamda)+1)]=0
    =>lamda=0, -1 ?
     
  8. Nov 24, 2005 #7

    matt grime

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    no it isn't, not unless you made a typo in the original post. you're characteristic equation is for y'' + y', when you wrote y'' + y
     
  9. Nov 24, 2005 #8

    HallsofIvy

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    The differential equation was y"+ y= 0. If [itex]y= e^{\lambda x}[/itex] then [itex]y'= \lambdae^{\lambda x}[/itex] and [itex]y"= \lambda^2e^{\lambda x}[/itex]. The characteristic equation is [/itex]\lambda^2+ 1= 0[/itex] which has roots i and -i. A basis for the solution space is {sin x, cos x}.
    If the original equation was y"+ y'= 0, then the characteristic equation is y"+ y= 0 which has roots 0 and -1. In that case a basis for the solution space would {e0x= 1, e-x}
     
  10. Nov 25, 2005 #9
    opps... thank you for correcting me!!!
     
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