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Differential equation

  1. Jul 7, 2004 #1
    if y = e^(2x) . ( ax + b ), when a and b is constant find dy/dx and dee 2 y dee x squared.

    i already find the answer for this part and it is correct.
    dy/dx = e^(2x) . (2ax + 2b + a )
    dee two y dee x squared is equal to
    4e^(2x) . (ax + b + a )

    this is the second part of the question which im not able to solve. tried substitute the derivative value in to the equation below, but the equation is getting more complex. any hint?

    find values of constants p adn q such that
    dee two y dee x squared + p ( dy / dx ) + qy = 0
    for all values of a and b

    any hint?
    Last edited: Jul 7, 2004
  2. jcsd
  3. Jul 7, 2004 #2
    try to use m_1=m_2 and you'll get p=-4 and q=4, I'm not sure how do you say that in english, but in my language people call it roots of multiplicity.
  4. Jul 7, 2004 #3
    your answer is absolutely correct.
    but i never learnt the method u mentioned.
  5. Jul 7, 2004 #4
    double checked...
    no typo.
  6. Jul 7, 2004 #5


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    Are you trying to say



    It's pronounced "d squared y ..." not "d two y ..."

    Anyways, did you try to do anything with the equation once you substituted?
  7. Jul 7, 2004 #6
    once i substituted, i try to factorise - stuck in the middle.
    i try to expand it - found it no use..

    bout the d2y/dx2, im sorry. but the book here im using teach me to pronounce it that way.
  8. Jul 7, 2004 #7


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    Do you know what you are trying to do with the equation? For instance, if I told you that [itex]m x e^x + n e^x = 0[/itex] for all x, what could you tell me about m and n?

    Oh, and just to let you know, I think most people would interpret

    "d 2 y d x squared" as saying:

    Last edited: Jul 7, 2004
  9. Jul 7, 2004 #8
    well, will that mean m = 0 and n = 0 ?
    im not sure.
  10. Jul 7, 2004 #9


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    Right; in my example, the equation can only be true if m = n = 0.

    Can you apply this idea to your problem?
  11. Jul 8, 2004 #10
    sorry, hurkyl. i need more guidance.

    do u mean i should do this?
    4 ( ax + b + a ) = 0
    p ( 2ax + 2b + a ) = 0
    q ( ax + b ) = 0

  12. Jul 8, 2004 #11
    i still cannot get it.
  13. Jul 8, 2004 #12


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    Insert the derivatives in (*), omit e^(2x), you get a first order polynomial of x that should be identically zero. That can be valid only if both the coefficient of x and the constant term are zero. That means two eguations for p and q in terms of the constants a and b.
  14. Jul 8, 2004 #13
    alrite. i solved the question. thank you Hurkyl, ehild and wisky!
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