Solve the given differential equation

[chwala]

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?

 

[topsquark]

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

 

[fresh_42]

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?

 

[chwala]

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)

 

[chwala]

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?

 

[fresh_42]

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?

 

[Mark44]

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.

 

[Mark44]

Where did you get $\dfrac{1}{5}$ from?

He's asking about a different, but related, problem.

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.

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.

 

[topsquark]

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

 

[WWGD]

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)?

 

[chwala]

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.