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Homework Help: Ordinary Diff Eq. - Help with Calculus

  1. Apr 15, 2010 #1
    1. The problem statement, all variables and given/known data
    Hey all. I'm having issues with a few Diff. Eq. problems. I think I'm screwing up at the point of u-substitution, but I do not know for sure.

    2. Relevant equations
    I'm evaluating: (x2 + x3)*integral (x4/(x4+2x5+x6)

    3. The attempt at a solution
    I eliminate x4 inside the integral and evaluate:
    x*(x+1)*integral(dx/(x+1)2)
    = x2*(x+1)/-(x+1). Cancel out (x+1), and my answer is
    = -x2.

    However, the back of the book says the answer should be x2 (positive). Can anyone point out as to what I'm doing wrong? I've gotten a few of these wrong, usually because they are in the denominator and I do u-substitution.

    P.S.: If anyone has a link on how to use Latex Reference, it'd be greatly appreciated. I tried a few things and could not get it to work, for the life of me.
    Latex text: [tex]y_2[/tex]

    Many thanks.
     
    Last edited: Apr 15, 2010
  2. jcsd
  3. Apr 15, 2010 #2

    Mark44

    Staff: Mentor

    First problem is in the line above. x2 + x3 = x2(1 + x).
    Next problem is above. You are treating (x + 1) as if it were a constant, but it isn't. You can't cancel an (x + 1) factor outside the integral with one in the denominator of the integrand.

    You need to evaluate
    [tex]x^2(1 + x) \int \frac{dx}{(x + 1)^2}[/tex]

    A simple substitution will work here.

     
  4. Apr 15, 2010 #3

    Cyosis

    User Avatar
    Homework Helper

  5. Apr 15, 2010 #4

    Mark44

    Staff: Mentor

    I agree with Cyosis that the answer you got is correct.
    [tex]x^2(1 + x) \int \frac{dx}{(x + 1)^2} = -x^2 + C[/tex]

    I might have misunderstood what you cancelled, thinking that you cancelled something inside the integrand with something outside it.
     
  6. Apr 15, 2010 #5
    Hey guys. Sorry but it's difficult for me to format the problem correctly. I'm not being lazy intentionally, I'm just very frustrated from not getting a very cool feature to work right. After one hour I've figured out I need to clear my cache each time I need to preview (before all I was getting was "2y"). Now onto the problem.

    The problem is an ODE - Constructing a 2nd solution from a known solution, where if you have an equation in standard form: [tex]y^{''} + P(x)y^{'} + Q(x)y = 0[/tex], and we know a solution [tex]y_1[/tex], a second solution [tex]y_2[/tex] is given by:

    [tex]y_2 = y_1 \int \frac{e^{- \int {P(x)dx}}}{y_1^2}dx[/tex].

    The equation I am given is:
    [tex]x^2y^{''} - 4xy^{'} + 6y = 0[/tex]
    In standard form:
    [tex]y^{''} - \frac {4{y^{'}}}{x} + \frac{6y}{x^2} = 0 [/tex]

    Evaluate [tex]e^{-\int {P(x)dx}} = e^{- \int {\frac {-4}{x}}dx} = e^{4 ln x} = x^4[/tex]

    Now to get y2:
    [tex]y_2 = (x^2 + x^3) \int {\frac {x^4}{\left({x^2 + x^3}\right)^2}}dx[/tex]
    I expand the denominator inside the integral:
    [tex]y2 = (x^2 + x^3) \int {\frac {x^4 dx}{x^6 + 2x^{5} + x^{4}}}[/tex]
    factoring out [tex]x^4[/tex] in order to cancel:
    [tex]y2 = (x^2 + x^3) \int {\frac {x^4 dx}{x^4 \left(x^2 + 2x + 1 \right)}}[/tex]
    and the expression simplifies to:
    [tex]y_2 = (x^2 + x^3) \int {\frac {dx}{\left({x+1}\right)^2}}[/tex]
    which evaluates to:
    [tex]y_2 = \frac {x^2 \left (x+1 \right)}{- \left(x+1 \right)}[/tex]
    and this yields:
    [tex]y_2 = -x^2[/tex]

    Again, my issue is that the answer in the back of the book is the negative of this. This is the third answer or so that evalutes to the negative of the real answer, so I figure I must be doing some systematic error.

    Your feedback is greatly appreciated.
     
    Last edited: Apr 15, 2010
  7. Apr 15, 2010 #6

    Mark44

    Staff: Mentor

    One way to check is to see if your solution satisfies the d.e., which is something you probably don't need to do if your answer agrees with the one in the book, but you should do if the two answers don't agree.

    In this case, both your answer and the one in the book satisfy the d.e. The reason is that for y2, any multiple of x2 (including the -1 multiple) is a solution, and this is due in part to a constant of integration that wasn't included somewhere, and also because the d.e. is homogeneous.
     
  8. Apr 15, 2010 #7
    substituting the negative and positive into the original equation, both are solutions.

    don't 2nd order differential equations have solutions of the form,

    y = c_1 * y_1(x) + c_2 * y_2(x)

    if y_1 and y_2 are solutions?

    edit: ahhh, Mark44 beat me to it!
     
    Last edited: Apr 15, 2010
  9. Apr 15, 2010 #8
    Guys thanks for the help. I had forgotten all about the general solution, so now I see that they just chose to write the solutions as a positive multiple of [tex]y_2[/tex].

    Thanks!!
     
  10. Apr 15, 2010 #9

    Mark44

    Staff: Mentor

    BTW, I forgot to mention that the site has been experiencing problems of not rendering correctly when you click the Preview button. I'm pretty sure people are working on fixing the problem.
     
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