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Differential Equation Help

  1. May 20, 2007 #1
    1. The problem statement, all variables and given/known data
    Find constants A, B, and C such that the function y = Ax^2+Bx+C satisfies the differential equation y''+y'-2y=x^2

    2. The attempt at a solution

    d/dx (y) = d/dx (Ax^2+Bx+C) = 2Ax+B
    y' = 2Ax+B

    d/dx (y') = d/dx (2Ax+B) = 2A
    y'' = 2A

    Now subbing back into the differential equation given:
    (2A) + (2Ax+B) - 2y = x^2
    2A + (2Ax+B) - 2(Ax^2+Bx+C) = x^2
    2A + 2Ax + B - 2Ax^2 - 2Bx - 2C = x^2

    Moving the left hand side around:
    (2A-2C+B) + (2A-2B)x - 2Ax^2 = x^2

    However, now I have to solve for the constants and I'm not exactly sure how to figure that out. Any help would be great! Thanks guys :D
    Last edited: May 20, 2007
  2. jcsd
  3. May 20, 2007 #2
    Well, in the final equation, you have an x^2 term, an x term and a constant on the left, while you have only an x^2 term on the right. Now, based on this, what can you conclude about the coefficients on the left?
  4. May 20, 2007 #3
    Well the coefficients on the left hand side if we were to rearrange it to
    -2Ax^2 + (2A-2B)x + (2A-2C+B)
    would be -2, 1, 1 which in a way is kind of similar to the equation:
    -2x^2 + x+ 1 which is a polynomial. However, I'm not too sure if I should be moving the x^2 term from the right hand side over to the left hand side and solving for 0. Haha I'm sorry if I didn't really answer your question.
  5. May 20, 2007 #4
    Okay, let me put in a different way...

    Suppose ax^2+bx+1 = 3x^2+cx+d (a,b,c,d constants)

    What can you say about a, b, c and d?
  6. May 20, 2007 #5
    Hmmm I must not be seeing the logic behind this at all, and usually I would. Maybe I'm just having an off day or something. 12 years of schooling and 1 year of University is failing me today. The only thing I can think of would be that you are moving the right hand side over and subtracting it to be:
    ax^2+bx+1 - 3x^2 - cx - d = 0
    (a-3)x^2 + (b-c)x - d + 1 = 0

    I feel really silly right now for not understanding because I know it is more likely really simple because the same principle for what you are asking will apply to my above question.
  7. May 20, 2007 #6

    D H

    Staff: Mentor

    The way to solve a problem like

    [tex]a_nx^n + \cdots + a_1x + a_0 = A_nx^n + \cdots + A_1x + A_0[/tex]

    is to split this one equation into n equations by grouping like powers of x:

    [tex]a_nx^n = A_nx^n[/tex]
    [tex]a_1x = A_1x[/tex]
    [tex]a_0 = a_0[/tex]

    from which [itex]a_k = A_k[/itex].

    How does this work? Evaluate the expression at x=0. The constant terms are all that are left; they must be equal. The higher-order powers must be equal as well (prove recursively). You don't need the proof to solve the problem. Simply equate like terms.
    Last edited: May 20, 2007
  8. May 20, 2007 #7
    OHHH alright! Thank you so much both of you. That last explanation I think helped the most. I knew it could not be a hardship to figure it out but I made it seem harder than it should be. The keywords that helped me the most was "equate like terms" which I know is what neutrino was trying to state too.

    Going back to my question, I'm finding that A = -1/2, B = -1/2 and C = -3/4 I believe. I'm hoping that is right anyway. :D
  9. May 20, 2007 #8

    D H

    Staff: Mentor


  10. May 20, 2007 #9
    Excellent, yeah I just went back and subbed into my formula the answers I just got:
    - 2Ax^2 + (2A-2B)x + (2A-2C+B) = x^2

    Which ends up equaling x^2 on both sides, so the identity must be true.
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