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Unclear concept on simple 2nd order differential eq

  1. Nov 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Hi,I am learning to solve 2nd-order differential eq.
    Suppose I have a equation
    dy/dx - 3x = 0...........(1)
    Then dy/dx = 3x -----> x = 3(x^2)/2
    Now if I have a 2nd order ODE such that:
    d^2y/dx^2 = 3.............(2)
    Then it could be solved by integrating both sides wrt x twice,which yields
    y = 3(x^2)/2 + Ax + B

    Now,consider the case:
    d^2y/dx^2 -6dy/dx + 9 = 0.........(3)
    I know it could be solved by using the idea of auxiliary equation(Putting y = Ae^(sx) into the initial eq)

    But,why it's not possible to solve (3) by using direct integration on both sides as in (1) and (2)?
    I can illustrate it here:
    d^2y/dx^2 = 6dy/dx - 9,then I integrate both sides wrt x
    dy/dx = 6 -9x + A
    y = 6x -9(x^2)/2 + Ax + B
    It's obviously incorrect,the (6+A)x term vanishes in the 2nd order derivative.Can someone tell me what's wrong?(I mean on the idea)

    Thx
    2. Relevant equations


    3. The attempt at a solution
    I have illustrated it in the problem statement.
     
  2. jcsd
  3. Nov 10, 2015 #2

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    Didn't you lose a ##y## here?
    As in: ##y'=6y-9x+A##.
     
  4. Nov 10, 2015 #3
    Thx so much^^,a silly mistake
     
  5. Nov 10, 2015 #4

    Mark44

    Staff: Mentor

    As already noted by @samy A, you lost a factor of y in the second line of your work.

    Starting with y'' = 6y' - 9, if we integrate (w. respect to x), we get
    y' = 6y - 9x + A

    If you try the same trick again, you will have y on the left side, but will have ##\int y dx## on the right side. Since you don't know what y is in terms of x, the integral can't be calculated.
     
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