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

[Identity]

Can someone please give me a hint to solve

$$\frac{d^2x}{dt^2} = \frac{1}{x^2}$$

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

thanks

 

[fatra2]

The solution appears simple.

$$\frac{d^2x}{dt^2} = \frac{1}{x^2}$$
$$\int x^2 d^2x = \int dt^2$$

Or am I making a mistake somewhere.

Cheers

 

[Identity]

fatra2, what do you mean by $$dt^2$$? How do you integrate that?

thanks

 

[fatra2]

You need to integrate over t twice.

 

[sv3t]

Just do:

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

Now solve for $$\dot x$$, and then solve for $$x$$.

This is a standard trick for these equations.

 

[Identity]

Thanks sv3t!

fatra2 I tried that method and differentiated to check but I got a differetn answer, thanks though

 

[fatra2]

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

 

[Identity]

This is what i did:

$$\int x^2 d^2x = \int dt^2$$

$$(\frac{x^3}{3}+C)dx = (t+D) dt$$

$$\frac{x^4}{12}+Cx = \frac{t^2}{2}+Dt+F$$

Now, implicitly differentiating with respect to t,

$$\frac{x^3}{3}\frac{dx}{dt}+C\frac{dx}{dt} = t+D$$

And again,

$$x^2\frac{dx}{dt}+\frac{x^3}{3}\frac{d^2x}{dt^2}+C\frac{d^2x}{dt^2}=1$$

Does not look like the original

Can you please show your working?

 

[fatra2]

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

 

[elibj123]

You cannot integrate with respect to $$d^{2}x$$ as you integrate with $$dt^{2}$$. 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.

 

[HallsofIvy]

This is what i did:

$$\int x^2 d^2x = \int dt^2$$

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 $d^2x/dt^2= dy/dt= (dy/dx)(dx/dt)= y dy/dx$ so the equation becomes $ydy/dx= 1/x^2$. That is, now, first order and separable: $ydy= (1/x^2)dx$. Integrating both sides of that [itex}(1/2)y^2= -1/x + C[/itex].

Now, $y= dx/dt= \sqrt{-2/x+ C/2}= \sqrt{(Cx- 4)/2x}$ so $dx/(\sqrt{(Cx-4)/2x}$. 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.

 

[arildno]

That's incorrect, Halls!

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

Thus, you get the exactly solvable integral:
$$\int\frac{16du}{(C-2u^{2})^{2}}$$

 

[arildno]

For the OP:
Assuming C>0, we may rewrite this as:
$$\frac{16}{C^{2}}\int\frac{du}{(1-(u\sqrt{\frac{2}{C}})^{2})^{2}}$$

Introducing:
$$s=u\sqrt{\frac{2}{C}}\to{du}=ds\sqrt{\frac{C}{2}}$$,
so that we essentially need to solve, with partial fractions decomposition:
$$\int\frac{ds}{(1-s^{2})^{2}}$$

 

[arildno]

If I have done my partial fractions decomposition correctly, we should have:
$$\frac{1}{(1-s^{2})^{2}}=\frac{1}{4}(\frac{1}{1+s}+\frac{1}{1-s}+\frac{1}{(1+s)^{2}}+\frac{1}{(1-s)^{2}})$$

But don't count upon that, do it yourself!

 

[HallsofIvy]

Thanks for the correction, Arildno.