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Y'' + py'+ qy = 0 explain why the value of y''(a) is determined by the values of y(a)

  1. Oct 7, 2011 #1
    Indicate why we can impose only n initial conditions on a solution of nth order linear differential equation.

    A) Given the equation y'' + py'+ qy = 0
    explain why the value of y''(a) is determined by the values of y(a) and y'(a).

    B) Prove that the equation y'' - 2y' -5y =0
    has the solution satisfying the conditions y(0) = 1, y'(0) = 0, and y''(0) = C
    if and only if C = 5.
     
  2. jcsd
  3. Oct 7, 2011 #2
    Re: Y'' + py'+ qy = 0 explain why the value of y''(a) is determined by the values of

    The DE is linear so it must have a general solution which is a linear combination of two linearly independent solutions y1(x) and y2(x).

    y(x)=c1y1(x) + c2y2(x) .

    c1 and c2 can be determined uniquely from the given initial conditions. So the result can be deduce from here.
     
  4. Oct 8, 2011 #3
    Re: Y'' + py'+ qy = 0 explain why the value of y''(a) is determined by the values of

    Do you need to the dimension of the solution set to a second order system is two dimensional?
     
  5. Oct 8, 2011 #4

    AlephZero

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    Re: Y'' + py'+ qy = 0 explain why the value of y''(a) is determined by the values of

    Because y''(a) = -py'(a) - qy(a).

    Move along, please, there's nothing to explain here...
     
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