# Solve the given differential equation

• chwala
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.
chwala
Gold Member
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?

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

fresh_42
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?

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)

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?

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?

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.

topsquark and chwala
fresh_42 said:
Where did you get ##\dfrac{1}{5}## from?
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?
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.

chwala, topsquark and fresh_42
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

chwala
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)?

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.

WWGD

## What is the general solution to the given differential equation?

The general solution to a differential equation is a family of functions that includes all possible solutions. It usually involves one or more arbitrary constants. For example, the general solution to the differential equation $$\frac{dy}{dx} = ky$$ is $$y = Ce^{kx}$$, where $$C$$ is an arbitrary constant.

## How do you determine the particular solution to a differential equation?

The particular solution to a differential equation is found by applying initial conditions or boundary conditions to the general solution. For instance, if the general solution is $$y = Ce^{kx}$$ and you are given $$y(0) = y_0$$, you can solve for $$C$$ to find the particular solution. In this case, $$y_0 = Ce^{k \cdot 0}$$, so $$C = y_0$$, giving the particular solution $$y = y_0e^{kx}$$.

## What methods can be used to solve a first-order differential equation?

Several methods can be used to solve first-order differential equations, including separation of variables, integrating factors, and exact equations. For example, the separation of variables method involves rearranging the equation so that each variable appears on a different side of the equation and then integrating both sides.

## What is an initial value problem in the context of differential equations?

An initial value problem (IVP) is a differential equation accompanied by specified values of the unknown function and its derivatives at a given point. These values are used to find a unique solution to the differential equation. For example, the IVP for $$\frac{dy}{dx} = 3y$$ with $$y(0) = 2$$ involves finding the specific solution that satisfies both the differential equation and the initial condition.

## How do you solve a second-order differential equation?

Second-order differential equations can be solved using various methods depending on their form. Common methods include the characteristic equation for linear homogeneous equations with constant coefficients, undetermined coefficients, and variation of parameters for non-homogeneous equations. For example, the characteristic equation for $$ay'' + by' + cy = 0$$ is $$ar^2 + br + c = 0$$, and the roots of this equation determine the form of the general solution.

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