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Vector spaces problem -linear algebra

  1. Jul 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi guys , I have this problem ,well actually I don't understand the solution they provide , Here's the problem statement and the solution .
    linear6.JPG
    May someone please explain the solution to me?? Thanks so much, Sorry for my bad english


    2. Relevant equations
    1.I understand that f'-af=0 and the kernel is the space of the solutions that satisfy that equation but I don't get what they do after that...why do they divide f(t) by e^(at)?
    2. why do they conclude that exists a constant c such that f(t)=ce^at??
     
  2. jcsd
  3. Jul 3, 2013 #2

    HallsofIvy

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    The problem is this: we have the vector space of all infinitely differentiable functions, D is the differentiation operator.
    a) Find the kernel of the linear operator D- I.
    b) Find the kernel of the linear operator D- aI.

    It is simplest to solve (b) first, then take a= 1 to solve (a).
    If f is any function in the kernel of D- aI then, by definition of "kernel" we must have f'- af= 0.
    That is the same as df/dx= af which is a separable equation: [tex]df/f= adx[/tex]. Integrating both sides, ln(f)= ax+ d where c is the constant of integration. Taking the exponential of both sides [itex]f(x)= e^{ax+ c}= Ce^{ax}[/itex] where [itex]C= e^c[/itex].

    That is how I would have solved the problem. I suspect that your text, knowing that f must be an exponential, started from that:
    [tex]\left(\frac{f(x)}{e^{ax}}\right)'= \frac{f'(x)e^{ax}- f(x)ae^{ax}}{e^{2ax}}= \frac{(f'(x)- af(x))e^{ax}}{e^{2ax}}[/tex]
    by the product rule.

    And, because [itex]f'(x)- af(x)= 0[/itex], the right side is 0, the derivative of [itex]f(x)/e^{ax}[/itex] is 0 so that [itex]f(x)/e^{ax}[/itex] is a constant: [itex]f(x)/e^{ax}= C[/itex] so [itex]f(x)= Ce^{ax}[/itex].

    (And your English is excellent. Far better than my (put language of your choice here).
     
  4. Jul 3, 2013 #3
    Thank you so much HallsofIvy , I understood all your explanation . It's cool to know that there's still good people who like to help others .Greetings from Chile
     
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