Integration problem

  • Thread starter tomwilliam
  • Start date
  • #1
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Homework Statement


v(dv/dx)= k(2v+1)
Find an explicit solution for x in terms of v.

Homework Equations


v=speed, x=distance, k=constant


The Attempt at a Solution


I can solve this by dividing by (2v+1), then integrating with respect to x using the chain rule.
I believe there is another method, though, involving substitution of 2v+1=u
so:
v(dv/dx)= ku
Now I'm not sure how (or even if it's possible) to integrate ku with respect to x, as it is a function of v, not x. Can someone point me in the right direction?
 

Answers and Replies

  • #2
35
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Write the numerator v as 1/2(2v - 1 + 1) and separate the terms properly to easily solve this integral :)
 
  • #3
135
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Thanks
Isn't that still the separation of variables method though?
 
  • #4
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Indeed it is...which other method do you wish to use? It's a variable-separable equation..
 
  • #5
135
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It's just that I can solve it ok using the sep of variables method. I'm asked to use the substitution 2v+1=u to solve it, but I can't seem to work out how that approach works.
 
  • #6
35
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Ok you mean -:
2v + 1 = u and dv = du, v = (u - 1)/2.
Now try it!
 
  • #7
135
2
Thanks, that's exactly what I'm trying to do:

So I make the substitions you suggested, and I get:

((u-1)/2) (du/dx) = ku

Using the chain rule on the LHS I can rewrite:

∫(u-1)/2 du = k ∫ u dx

now the LHS is easy, but is the right-hand side just kux+C?

(u^2)/4 - (u/2) = kux+C

After I reinsert what I've substituted, this isn't the same as the answer I get with separation of variables. What did I do wrong?
I'm afraid I don't really understand this substitution business yet...
Thanks for your time, much appreciated by the way.
 
  • #8
135
2
Scrap that last post, I've worked a few things out:

I understand that once you make the substitution u=1+2v then you can rewrite v as v=(u-1)/2, so the equation becomes

((u-1)/2) (dv/dx) = ku

Then I understand that du/dv = 2, so we can multiply the equation by du/dv to get:

((u-1)/2) (dv/dx)(du/dv) = 2ku

And then by the chain rule simplify to

∫ (u-1)/2 du = ∫ 2ku dx

is this right? I'm still not sure how to integrate u (or 1+2v) with respect to x, but it looks slightly closer to the right answer.
Thanks again
 
  • #9
35
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Did they say that you can't use the variable-separable method after substitution? I think you have to apply the substitution and then use the variable-separable method...meaning all u terms on the LHS.
 
  • #10
135
2
I see now! Thanks very much for your time.
I think you're right, I'm supposed to make the substitution, then bring everything to the lhs. It's not too different from just separating the variables first, actually.

Just one question. When making the substitution, I understand all the stages except the justification for making dv=du. If we have said that u=1+2v, then what is the argument for making dv=du? I know it proves necessary, but if it comes up again in a different question I won't necessarily know what the consequences are of making a substitution like this.
Thanks
 
  • #11
169
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That was a typo, I suppose. It's du = 2 dv, actually. Just like you already calculated.
 
  • #12
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Ah yes pardon me...it is a typo as earl pointed out!
 

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