Solve the given differential equation

In summary, substitution of x for 1+x in the equation x(1+x^2)^{-\frac{1}{5}}dx gives the same result as substitution of x for 1+x^2 in the equation x(1+x^2)^{-\frac{1}{2}}dx.
  • #1
chwala
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Homework Statement
See attached
Relevant Equations
separation of variables
My interest is only on the highlighted part, i can clearly see that they made use of chain rule i.e
by letting ##u=1+x^2## we shall have ##du=2x dx## from there the integration bit and working to solution is straightforward. I always look at such questions as being 'convenient' questions.

Now to my question, supposing we had ##x(1+x^2)^{-\frac{1}{5}} dx## in the place of ##x(1+x^2)^{-\frac{1}{2}} dx##
How would we handle this?

1691325775186.png
 
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  • #2
chwala said:
Homework Statement: See attached
Relevant Equations: separation of variables

My interest is only on the highlighted part, i can clearly see that they made use of chain rule i.e
by letting ##u=1+x^2## we shall have ##du=2x dx## from there the integration bit and working to solution is straightforward. I always look at such questions as being 'convenient' questions.

Now to my question, supposing we had ##x(1+x^2)^{-\frac{1}{5}} dx## in the place of ##x(1+x^2)^{-\frac{1}{2}} dx##
How would we handle this?

View attachment 330199
You would handle it the same way that I would suppose the original integration was done: Sub ##u = 1 + x^2## and go from there.

-Dan
 
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  • #3
chwala said:
Homework Statement: See attached
Relevant Equations: separation of variables

My interest is only on the highlighted part, i can clearly see that they made use of chain rule i.e
by letting ##u=1+x^2## we shall have ##du=2x dx## from there the integration bit and working to solution is straightforward. I always look at such questions as being 'convenient' questions.

Now to my question, supposing we had ##x(1+x^2)^{-\frac{1}{5}} dx## in the place of ##x(1+x^2)^{-\frac{1}{2}} dx##
How would we handle this?

View attachment 330199
Where did you get ##\dfrac{1}{5}## from?
 
  • #4
fresh_42 said:
Where did you get ##\dfrac{1}{5}## from?
I just came up with that part of the question. I will amend my question and re-post it again. (post 5)
 
  • #5
topsquark said:
You would handle it the same way that I would suppose the original integration was done: Sub ##u = 1 + x^2## and go from there.

-Dan
I will amend the question as it is still 'convenient' as in it is easy to solve.

Now to my question, supposing we had say ##x(1+x^{\frac{1}{2}})^{-\frac{1}{5}} dx## in the place of ##x(1+x^2)^{-\frac{1}{2}} dx##
How would we handle this?
 
  • #6
chwala said:
I just came up with that part of the question. I will amend my question and re-post it again. (post 5)
I do not see any ##5##. So where exactly do you see it?
 
  • #7
chwala said:
I will amend the question as it is still 'convenient' as in it is easy to solve.

Now to my question, supposing we had say ##x(1+x^{\frac{1}{2}})^{-\frac{1}{5}} dx## in the place of ##x(1+x^2)^{-\frac{1}{2}} dx##
How would we handle this?
I don't see that a simple substitution would work for this revised problem. For your 2nd example above, the expression in parentheses is ##1 + x^2##. Outside the parentheses you have something that is almost the differential of ##1 + x^2## (differing only by a constant multiplier). For a similar kind of substitution for your first example you would need the differential (or a constant multiple of it) outside the parentheses.
 
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  • #8
fresh_42 said:
Where did you get ##\dfrac{1}{5}## from?
He's asking about a different, but related, problem.
chwala said:
Now to my question, supposing we had ##x(1+x^2)^{−1/5}dx## in the place of ##x(1+x^2)^{−1/2}dx##
How would we handle this?
Which was answered by @topsquark.
topsquark said:
You would handle it the same way that I would suppose the original integration was done: Sub ##u=1+x^2## and go from there.
 
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  • #9
chwala said:
I will amend the question as it is still 'convenient' as in it is easy to solve.

Now to my question, supposing we had say ##x(1+x^{\frac{1}{2}})^{-\frac{1}{5}} dx## in the place of ##x(1+x^2)^{-\frac{1}{2}} dx##
How would we handle this?
The easiest way to see it is to do it in two steps:
1. Let ##u = x^{1/2}##.

2. Then let ##v = u + 1##.

But you can come up with all sorts of ugly examples, and many of them do not have an actual solution. If your question is about integration techniques of
##\displaystyle \int \dfrac{x \, dx }{(1 + x^a)^b}##

you really need to put it in it's own thread.

-Dan
 
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  • #10
chwala said:
Homework Statement: See attached
Relevant Equations: separation of variables

My interest is only on the highlighted part, i can clearly see that they made use of chain rule i.e
by letting ##u=1+x^2## we shall have ##du=2x dx## from there the integration bit and working to solution is straightforward. I always look at such questions as being 'convenient' questions.

Now to my question, supposing we had ##x(1+x^2)^{-\frac{1}{5}} dx## in the place of ##x(1+x^2)^{-\frac{1}{2}} dx##
How would we handle this?

View attachment 330199
What happened with the y' in the separation process( from top line to 2nd line)?
 
  • #11
WWGD said:
What happened with the y' in the separation process( from top line to 2nd line)?
Check post ##8##. My interest is solely on the highlighted part of post ##1## in red.
 
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