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Can i apply the same technique as I did with real solutions? D.E. 2nd order

  1. Feb 20, 2006 #1
    Hello everyone!
    I had a question, I solved a problem very similar to this one but it has 2 real soltions which took the form of:
    y (x) = Ae^rx + Be^(rx)

    But now i have repeated roots:
    y(x) = e^(rx)*(A + Bx);

    HEre is the problem:
    Find y as a function of x if
    x^2 y'' - 11 x y' + 36 y = 0,

    y(1) = -2, y'(1) = -3.

    Ivey told me i could should change e^(rx) to just x^r in my last problem which worked great!

    Now if i have a repeated root case how would I do this? would it be:
    y(x) = x^r(A+Bx)?
  2. jcsd
  3. Feb 20, 2006 #2


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    Remember what I said about making the substitution x= et? Then t= ln x so dy/dx= (dy/dt)(dt/dx)= (1/x)dy/dt. Repeating that,
    [tex]\frac{d^2y}{dx^2}= (1/x^2)\frac{d^2y}{dx^2}- (1/x^2)dy/dx[/tex]

    Your differential equation becomes
    [tex]\frac{d^2y}{dt^2}- 12\frac{dy}{dx}+ 36y= 0[/tex].
    That's a de with constant coefficients that has the same characteristic equation as you got: r2- 12r+ 36= (r- 6)2= 0.

    Okay that equation has solutions, as you know well,
    [tex]y(t)= Ae^{6t}+ Bte^{6t}[/tex]. Now put t= ln x:
    Of course, [itex]6t= 6 ln x= ln x^6[/itex] so that [itex]e^{6t}= e^{ln x^6}= x^6[/itex].
    [tex]y(t)= Ax^6+ Bln(x)x^6[/tex].
    You see how that t became ln x?
    Last edited by a moderator: Feb 20, 2006
  4. Feb 20, 2006 #3
    Thanks ivey! i'm not sure if i totally fallowed what u wrote but i'm assuming u ment to tell me that:
    y = Ax^6 + Bln(x)*x^6 ?
    y' = 6Ax^5 +6Bln(x)^x^5 + Bx^5
    Well i used that and with the intial conditions:
    y(1) = -2
    y'(1) = -3
    and i found
    A and B.
    A = -2
    B = -1

    So i wrote:
    y = -2x^6-ln(x)*x^6
    but it didn't like it! Some parts of ur latex arn't showing up, i don't know if its my browser or what, i'm on Linux but still using firefox

  5. Feb 21, 2006 #4


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    A is -2 but B is NOT -1! Recalculate that. Be careful with your differentiation.
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