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Verify solutions of differential equation

  1. Apr 21, 2006 #1
    Consider 2nd order differential equation:

    [itex]
    a_{2}y''(x) + a_{1}y'(x) + a_{0}y(x) = 0
    [/itex]

    Let b and c be the roots of the quadratic [itex]a_{2}x^2 + a_{1}x + a_{0}[/itex]

    Verify that when b = c then [itex]xe^{bx}[/itex] is a solution of the Differential equation.
    -----------------------------

    Ok so I differentiate y a couple times and get

    [itex]
    y = xe^{bx}
    [/itex]

    [itex]
    y' = xbe^{bx} + e^{bx}
    [/itex]

    [itex]
    y'' = xb^2e^{bx}+2be^{bx}
    [/itex]

    Substituting into the equation yields:

    [itex]
    a_{2}(xb^2e^{bx}+2be^{bx}) + a_{1}(xbe^{bx} + e^{bx}) + a_{0}(xe^{bx}) = 0
    [/itex]

    Now if I rearrange the quadratic equation for a0
    [itex]
    a_{0} = -a_{2}b^2 - a_{1}b
    [/itex]

    then substituting that in I get

    [itex]
    a_{2}(xb^2e^{bx}+2be^{bx}) + a_{1}(xbe^{bx} + e^{bx}) + (-a_{2}b^2 - a_{1}b)(xe^{bx}) = 0
    [/itex]

    [itex]
    = a_{2}2be^{bx} + a_{1}e^{bx} = 0
    [/itex]

    This is where I get stuck, there must be something ive missed which I can use to further simplify. I did the proof for [itex]y = e^{bx}[/itex] in the same way and it worked out easy. Can anybody give me a hint on what to do next? thanks
     
    Last edited: Apr 21, 2006
  2. jcsd
  3. Apr 21, 2006 #2

    nrqed

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    Doesn't that imply that [itex] a_0 = {a_1^2 \over 4 a_2}[/itex]? It does not seem to be what you used.

    Also, from th eway the question is posed, the b in the exponent has nothing to do with the b in "the roots are b and c", right?
     
  4. Apr 21, 2006 #3
    They are the same b and c as I take it.

    Where do you get [itex] a_0 = {a_1^2 \over 4 a_2}[/itex] from. I just assumed that it was trying to say [itex]a_{2}b^2 + a_{1}b + a_{0} = 0[/itex]
     
  5. Apr 21, 2006 #4

    nrqed

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    That's not the way I intrepreted the sentence, but I may be wrong.

    Consider [itex] a_2 x^2 + a_1 x + a_0 [/itex]. The two roots are [itex] x=-{a_1 \over 2} \pm {1 \over 2} {\sqrt{ a_1^2 - 4 a_0 a_2}} [/itex].
    Let's call these two roots b and c. If the two roots are equal, then the term under the square root must be zero.

    That's the only way I can make sense of "the two roots of the above quadratic are equal".

    But I may be completely off. Sorry.
     
  6. Apr 21, 2006 #5

    nrqed

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    I forgot to divide by a_2 in writing the two roots.

    I think this is what they meant to say , it seems to be working (if in addition I assume that the b equal to the unique root is the b appearing in the exponent)
     
  7. Apr 21, 2006 #6

    nrqed

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    assuming that the sqaure root vanishes, then the single root is -a_1 / 2 a_2. setting this equal to b, one gets a_1 = -2 b a_2.


    In addition, imposing a_0 = a_1^2/(4 a_2) gives a_0 = b^2 a_2.

    Using those two expressions for a_0 and a_1 in terms of a_2, you will get that everything cancels out.

    Patrick
     
  8. Apr 21, 2006 #7
    ahh i can see what you did now. I'll have a crack at it, thanks for your help
     
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