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Vector of differential solutions

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi Guy's

    Let V be the set of functions [itex]f:R\rightarrowR[/itex]which solve the differential equation:

    [itex]\frac{d^2y}{dx^2}=y[/itex]

    Show that [itex]e_1:R\rightarrow R, x \rightarrow e^x[/itex] and [itex]e_2:R\rightarrow R, x\rightarrow cosh(x)[/itex]
    Comprise a basis for V.

    2. Relevant equations



    3. The attempt at a solution

    I have not done 2nd order differential equations and I was wondering if some one might give me a hint to a particular solution vector.

    I know I need to prove that [itex] e^x [/itex] and [itex] cosh(x) [/itex] are linear independant to be a basis, but I need to show that any solution vector can be generated by them.

    with the natural log we can't take the log of zero so when can the vector equal zero?
     
  2. jcsd
  3. Sep 14, 2009 #2

    HallsofIvy

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    Well, first of all, your [itex]e_1[/itex] and [itex]e_2[/itex] won't comprise a basis for the space of all functions- that's infinite dimensional. What you mean is that V is the set of all functions that satisfy that differential equation.

    That means you need to show the properties of a basis:
    1) They are independent. In this case that means that the only way you can have [itex]Ae^x+ Bcosh(x)= 0[/itex], for all x, is if A= B= 0.

    2) They span the space. That is, that if y is any function satifying the differential equation, then [itex]y= Ce^x+ Dcosh(x)[/itex] for some numbers C and D.
     
  4. Sep 14, 2009 #3


    Thankyou for your explaination.
    I don't know how I am going to prove this last statement. Do I need to have an explicit solution vector to show that it can be generated by [itex]y= Ce^x+ Dcosh(x)[/itex] ?
     
  5. Sep 15, 2009 #4
    when you are dealing with these 2nd order differential equations does it mean that there are two solutions?

    So if I prove that

    [itex]y= Ce^x+ Dcosh(x)= 0[/itex] if and only if the constants C and D are zero than that basis can generate the solution set of vectors V.
     
  6. Sep 15, 2009 #5
    As the functions inverses [itex]e^x[/itex] and [itex] cosh(x)[/itex]
    are not defined for 0 ie ln(0) is undifined etc is that enough to say that they must be independent as they can only equal zero if both constants are zero?
     
  7. Sep 15, 2009 #6

    HallsofIvy

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    No, whether they are indpendent or not has nothing to do with inverses.

    Suppose Aex+ Bcosh(x)= 0 for all x. In particular, Ae0+ B cosh(0)= A+ B= 0. Further, if Aex+ Bcosh(x)= 0, a constant, for all x, then its derivative must be 0 for all x also: Aex+ Bsinh(x)= 0 for all x. Now take x= 0.

    To show that these span the set of all solutions to this second order equation, suppose y is such a solution. Let A= y'(0) and B= y(0)- y'(0). Then, as before, z(x)= Aex+ B cosh(x) is a solution to the same differential equation (of course, you have shown that ex and cosh(x) satisfy the differential equation?) having z(0)= A+ B= y(0) and z'(0)= A= y'(0). Now use the "existence and uniqueness theorem".
     
    Last edited: Sep 15, 2009
  8. Sep 15, 2009 #7

    tiny-tim

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    Hi boneill3! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    Hint: put z = dy/dx + y :wink:
     
  9. Sep 16, 2009 #8
    Isn't the "existence and uniqueness theorem". mean that it is an solution in the space if the function and the first paritial derivative are continuous .

    You say Hint: put z = dy/dx + y

    we have
    [itex]\frac{d^2y}{dx^2}=y[/itex]
    so

    [itex]\frac{dy}{dx}=yx[/itex]

    so

    [itex]z = yx+y[/itex]
    and

    [itex]z' = y[/itex]

    which are both continuous proving that Z is in the solution space
     
  10. Sep 16, 2009 #9

    tiny-tim

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    nooo … you can't integrate y and get yx :redface:

    Differentiate z = dy/dx + y …

    what do you get? :smile:
     
    Last edited: Sep 16, 2009
  11. Sep 16, 2009 #10
    how do you differentiate dy/x ? what is its value?
     
  12. Sep 16, 2009 #11

    HallsofIvy

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    He didn't say "dy/x" he said "dy/dx". And its derivative if the second derivative, [itex]d^2y/dx^2[/itex]. The derivative of z= dy/dx+ y is [itex]d^2y/dx^2+ dy/dx[/itex].

    But I must admit, tiny tim, I don't see how that is related to this problem!
     
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