Help with a simple Diff Eqn

  • Thread starter prace
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  • #1
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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.
 

Answers and Replies

  • #2
102
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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]
 
  • #3
Mute
Homework Helper
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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:
  • #4
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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!!!
 
  • #5
dextercioby
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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]

No need for the substitution hinted.

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

Daniel.
 
  • #6
102
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Oh cool, even better. Thanks!
 

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