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Integration by parts question

  1. Feb 19, 2008 #1
    [SOLVED] integration by parts question

    1. The problem statement, all variables and given/known data
    I think my first problem is at integration by parts, but let me know if you see a different error in my work.

    Edit: I shoud mention that this is Calculus 2 and we just learned first order differential equations today. So I don't have any fancy methods available.

    Find the solution of
    [tex]x^3 y'+ax^2 y = 2x^{-a+4} \sin(x)[/tex]
    with
    [tex]y(\pi) = 0[/tex]


    2. Relevant equations
    Format of first order differential equations:
    [tex]y' + P(x) y = q(x)[/tex]

    [tex]H(x) = \int P(x) dx[/tex]


    3. The attempt at a solution

    [tex]x^3 y'+ax^2 y = 2x^{-a+4} \sin(x)[/tex]

    [tex]y' + \frac{a y}{x} = \frac{2 x^{-a + 4} \sin(x)}{x^3}[/tex]

    [tex]y' + \frac{a y}{x} = 2 x^{-a + 1} \sin(x)[/tex]

    [tex]H(x) = \int \frac{a}{x} dx[/tex]

    [tex]H(x) = a \ln{x}[/tex]

    Take [tex]e^{H(x)} = x^a[/tex] and multiply it to both sides

    [tex]x^a y' + \frac{a y}{x} x^a = x^a 2 x^{-a + 1} \sin(x)[/tex]

    [tex](x^a y)' = 2 x^{-a^2 + a} \sin(x)[/tex]

    [tex] x^a y = \int 2 x^{-a^2 + a} \sin(x) dx[/tex]

    This is where I am stuck. The only way I can think of to solve that integral is integration by parts, but the only way I can see that working is by repeating it enough times for the part of the integral before sin(x) to disappear. So I think I went wrong somewhere? But I don't see where.
     
    Last edited: Feb 19, 2008
  2. jcsd
  3. Feb 19, 2008 #2
    leads to

    [tex](x^\alpha\,y)'=2\,x\,\sin x[/tex]
     
  4. Feb 19, 2008 #3
    Ohh... Well that was easy. I just forgot how to multiply! Thanks.
     
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