A second-order non-linear (but simple) DE

  • Thread starter Identity
  • Start date
  • Tags
    Non-linear
In summary, the equation looks simple, but integration can be a tricky process. The method that sv3t suggested is the best way to solve the equation.
  • #1
Identity
152
0
Can someone please give me a hint to solve

[tex]\frac{d^2x}{dt^2} = \frac{1}{x^2}[/tex]

It is not a homework problem, but something for fun. It looks simple but is the solution simple?

thanks
 
Physics news on Phys.org
  • #2
The solution appears simple.

[tex]\frac{d^2x}{dt^2} = \frac{1}{x^2}[/tex]
[tex]\int x^2 d^2x = \int dt^2[/tex]

Or am I making a mistake somewhere.

Cheers
 
  • #3
fatra2, what do you mean by [tex]dt^2[/tex]? How do you integrate that?

thanks
 
  • #4
You need to integrate over t twice.
 
  • #5
Just do:

[tex]
\begin{equation}
\frac{d^2x}{dt^2}=\frac{d\dot x}{dt}=\frac{d\dot x}{dx}\dot x=\frac{1}{2}\frac{d(\dot x^2)}{dx}=f(x)
\end{equation}
[/tex]
for your favourite [tex]f(x) [/tex]. I defined [tex] \dot x\equiv dx/dt[/tex] as usual.

Now solve for [tex] \dot x[/tex], and then solve for [tex]x[/tex].

This is a standard trick for these equations.
 
  • #6
Thanks sv3t!

fatra2 I tried that method and differentiated to check but I got a differetn answer, thanks though
 
  • #7
Identity said:
fatra2 I tried that method and differentiated to check but I got a differetn answer, thanks though

If you did, then you must have made a mistake in your integration. I did this integration, and the double-derivative brought me right back to the original equation. You might want to check your calculus.

Cheers
 
  • #8
This is what i did:

[tex]\int x^2 d^2x = \int dt^2[/tex]

[tex](\frac{x^3}{3}+C)dx = (t+D) dt[/tex]

[tex]\frac{x^4}{12}+Cx = \frac{t^2}{2}+Dt+F[/tex]

Now, implicitly differentiating with respect to t,

[tex]\frac{x^3}{3}\frac{dx}{dt}+C\frac{dx}{dt} = t+D[/tex]

And again,

[tex]x^2\frac{dx}{dt}+\frac{x^3}{3}\frac{d^2x}{dt^2}+C\frac{d^2x}{dt^2}=1[/tex]

Does not look like the original

Can you please show your working?
 
  • #9
Hi there,

Ok, you seem to complicate your life a little bit. Your different constants can be anything, so simplify your equation and make them equal to zero. You have the right to do so, and your equation will thank you for it.

Cheers
 
  • #10
You cannot integrate with respect to [tex]d^{2}x[/tex] as you integrate with [tex]dt^{2}[/tex]. The latter is a squared differential, so you just integrate twice, but the former is something completely different. Don't get confused.

The method that sv3t suggested is the best for solving this equation.
 
  • #11
Identity said:
This is what i did:

[tex]\int x^2 d^2x = \int dt^2[/tex]
No. You cannot separate second derivatives into differentials like that. (Reminding us of that is one reason for the odd placement of the "2"s in the second derivative.)

A first step might be to use "quadrature". Let y= dx/dt. Then [itex]d^2x/dt^2= dy/dt= (dy/dx)(dx/dt)= y dy/dx[/itex] so the equation becomes [itex]ydy/dx= 1/x^2[/itex]. That is, now, first order and separable: [itex]ydy= (1/x^2)dx[/itex]. Integrating both sides of that [itex}(1/2)y^2= -1/x + C[/itex].

Now, [itex]y= dx/dt= \sqrt{-2/x+ C/2}= \sqrt{(Cx- 4)/2x}[/itex] so [itex] dx/(\sqrt{(Cx-4)/2x}[/itex]. Now, unfortunately, you will find that integral cannot be done in terms of elementary functions.

Actually, the even simpler equation, dy/dx= 1/x, "Abel's equation", has solutions that must be written, if I remember correctly, in terms of Bessel functions.
 
  • #12
That's incorrect, Halls!

We have:
[tex]u=\sqrt{\frac{Cx-4}{2x}}\to{x}=\frac{4}{C-2u^{2}}\to{dx}=\frac{16udu}{(C-2u^{2})^{2}}[/tex]

Thus, you get the exactly solvable integral:
[tex]\int\frac{16du}{(C-2u^{2})^{2}}[/tex]
 
  • #13
For the OP:
Assuming C>0, we may rewrite this as:
[tex]\frac{16}{C^{2}}\int\frac{du}{(1-(u\sqrt{\frac{2}{C}})^{2})^{2}}[/tex]

Introducing:
[tex]s=u\sqrt{\frac{2}{C}}\to{du}=ds\sqrt{\frac{C}{2}}[/tex],
so that we essentially need to solve, with partial fractions decomposition:
[tex]\int\frac{ds}{(1-s^{2})^{2}}[/tex]
 
  • #14
If I have done my partial fractions decomposition correctly, we should have:
[tex]\frac{1}{(1-s^{2})^{2}}=\frac{1}{4}(\frac{1}{1+s}+\frac{1}{1-s}+\frac{1}{(1+s)^{2}}+\frac{1}{(1-s)^{2}})[/tex]

But don't count upon that, do it yourself!
 
  • #15
Thanks for the correction, Arildno.
 

Related to A second-order non-linear (but simple) DE

1. What is a second-order non-linear differential equation?

A second-order non-linear differential equation is a type of mathematical equation that involves two independent variables and their derivatives, as well as non-linear functions of these variables. This means that the equation cannot be written in a simple, linear form where the variables are raised to the power of one.

2. How does a second-order non-linear differential equation differ from a first-order linear equation?

A first-order linear differential equation involves only one independent variable and its derivative, and the equation can be written in a linear form. In contrast, a second-order non-linear differential equation involves two independent variables and their derivatives, and the equation cannot be written in a linear form due to the presence of non-linear functions.

3. What is the importance of second-order non-linear differential equations in science?

Second-order non-linear differential equations are important in many scientific fields, including physics, engineering, and biology. They can model complex systems and phenomena that cannot be accurately described by first-order linear equations, making them a powerful tool for understanding and predicting real-world phenomena.

4. Can a second-order non-linear differential equation be solved analytically?

In general, second-order non-linear differential equations cannot be solved analytically, meaning there is no explicit formula for the solution. However, in some special cases, such as when the equation is separable or exact, an analytical solution may be possible. In most cases, numerical methods must be used to approximate the solution.

5. How are second-order non-linear differential equations used in real-world applications?

Second-order non-linear differential equations are used in a wide range of real-world applications, such as modeling population growth, predicting the motion of objects under the influence of gravity, and understanding the behavior of electrical circuits. They are also used in the development of mathematical models for complex systems, such as climate change and economic dynamics.

Similar threads

Replies
6
Views
2K
Replies
3
Views
1K
Replies
2
Views
2K
  • Calculus
Replies
1
Views
933
Replies
8
Views
453
Replies
2
Views
1K
Replies
3
Views
1K
  • Differential Equations
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
898
Replies
3
Views
552
Back
Top