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Help with a simple Diff Eqn

  1. Dec 10, 2006 #1
    Here is the problem:

    Solve, (x+1)[tex]\frac{dy}{dx}[/tex] = x + 6

    Here is what I tried:

    I moved all the x's to one side and left the dy on the left of the equal sign to solve with the separation of variable method.

    I got, [tex]\int{dy}[/tex] = [tex]\int{\frac{(x+6)}{(x+1)}dx}[/tex]

    So here I just solve the integrals and I am done. I guess the real question is how do I go about solving the integral on the right? I seem to have forgotten some basic integral techniques.

    Thank you.
  2. jcsd
  3. Dec 10, 2006 #2
    I came accross another integral that I am not catching here that seems to be along the same line as this one. Is there a rule to dealing with these kinds of integrals? [tex]\int{\frac{x^2}{(1+x)}}dx[/tex]
  4. Dec 10, 2006 #3


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    For your first integral, you can write [tex]\frac{x+6}{x+1} = \frac{(x+1) + 5}{x+1} = 1 + \frac{5}{x+1}[/tex]which is easy to integrate.

    For the second one, you can change variables to u = 1 + x, so du = dx and your integrand becomes

    [tex]\frac{(u-1)^2}{u} = \frac{u^2 - 2u + 1}{u} = u - 2 + \frac{1}{u}[/tex],

    which again should be easy to integrate.
    Last edited: Dec 10, 2006
  5. Dec 10, 2006 #4
    Awesome, that was super easy once you look at it that way. Thanks, now I can look at other integrals and apply the same method. Life somehow just became much easier!! ^_^ Thank you!!!
  6. Dec 11, 2006 #5


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    No need for the substitution hinted.

    [tex] \frac{x^{2}}{x+1}=x-1+\frac{1}{x+1} [/tex]

  7. Dec 11, 2006 #6
    Oh cool, even better. Thanks!
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