2nd order differential transformation

[ARandomOWl]

Using the substitution y=x^(-2), transform:

[LATEX]2x \frac{d^2x}{dt^2}-6(\frac{dx}{dt})^2=x^2-3x^4...(1) [/LATEX]

into

[LATEX]\frac{d^2y}{dt^2}+y=3[/LATEX]

Attempted solution:

From y=x^(-2), y'=-2x^(-3), so:

[LATEX]\frac{dx}{dy}=-\frac{1}{2}x^3[/LATEX]
and therefore via the chain rule
[LATEX]\frac{dx}{dt}=-\frac{1}{2}x^3\frac{dy}{dt}...(2)[/LATEX]
and subsequently
[LATEX]\frac{d^2x}{dt^2}=-\frac{1}{2}x^3\frac{d^2y}{dt^2}-\frac{3}{2}x^2\frac{dy}{dt}...(3)[/LATEX]

Substituting (2) and (3) into (1) and cancelling some x gives
[LATEX]-x^2\frac{d^2y}{dt^2}-3x\frac{dy}{dt}-\frac{3}{2}x^4(\frac{dy}{dt})^2=1-3x^2[/LATEX]

Now I can't figure out what to do with the 2nd and 3rd term. I think they must cancel somehow in order to reach the answer. The (dy/dt)^2 term is the main problem. Any help is appreciated, thanks.

 

[MrAl]

Hi,

Quick spot: your [LATEX]\frac{d^2x}{dt^2}[/LATEX] is incorrect because the derivative of x^3 with respect to t is zero.
There's another trick too, but see if you can find it first.

 

[ARandomOWl]

Of course! So it's just:

[LATEX]\frac{d^2x}{dt^2}=-\frac{1}{2}x^3\frac{d^2y}{dt^2}[/LATEX]

I'll have another look at it and reply later. Does the trick have something to do with x=y^(-1/2) ?

 

[MrAl]

Hi,

By 'trick' i just meant that after you see this 'trick' which you've just solved, you should be able to see another such simplification which should get you to the final result with ease. I didnt want to come right out and state what it was but wanted to give you a chance to spot it on your own so next time you do a problem like this you'll be aware of these things. These arent really tricks, just noting some things about the equation(s). No problem if you dont spot it i'll mention what it is in another post. But this time go over all of your previous math carefully and i think you'll see it.
It was good that you noticed right away that the chain rule played a big part in solving this problem.

 

[ARandomOWl]

I am grateful of that, MrAl, thanks.
Ok, after substituting in my corrected (d^2x)/(dt^2) and multiplying by -1, I reach

[LATEX]x^2\frac{d^2y}{dt^2}+6(-\frac{1}{2}x^2\frac{dy}{dt})^2=3x^2-1[/LATEX]

Now if the 2nd term were to equal zero, I would reach the correct answer. I just cannot figure out what to do with the squared bracket.

 

[MrAl]

Hi again,

Ok, after substitutions i get:

3*(dy/dt)^2*x^2/2+1/x^2+d^2y/dt^2=3

So i was stuck with the dy/dt squared term too, but then reasoned that the first derivative of y with respect to t was zero, so that term goes to zero and we're left with:
1/x^2+d^2y/dt^2=3

and since 1/x^2 is equal to y, we get:
y+d^2y/dt^2=3

which rearranged comes out to the required solution.

But now you have me wondering if this is valid or not, so i'll have to take a better look at it as soon as possible. Right now im involved with a couple other things and in need of some rest from running around all morning. In the mean time, see what you think of this so far.

 

[ARandomOWl]

So i was stuck with the dy/dt squared term too, but then reasoned that the first derivative of y with respect to t was zero, so that term goes to zero and we're left with:
1/x^2+d^2y/dt^2=3

Can you please explain how you came to that conclusion?

 

[MrAl]

No. I'm too rusty. We're both overlooking something simple here.

[ADDED LATER]
Ok, we both did the parametric differentiation of the dx/dt incorrectly.
Since both x and y are functions of t, we must do the diff of x^3 like this:
d(x^3)/dt=3*x^2*dx/dt

So the derivative is not zero and it is not just 3*x^2, but is done as though x was a function of t (as it really is).

Once all the subs are done, we end up with an equation with just three terms and from there
we have to make one more subs by substituting y for 1/x^2 and simplify a little and we get the required new equation.
Im quite sure you'll be able to solve this now.

 

[ARandomOWl]

I'm afraid I don't follow. You are saying that my (d^2x/dt^2) is still wrong? From what you said:

[LATEX]\frac{d^2x}{dt^2}=-\frac{1}{2}x^3\frac{d^2y}{dt^2}-\frac{3}{2}x\frac{dx}{dt}\frac{dy}{dt}[/LATEX]

Which doesn't give me 3 terms after substitution nor seem to help at all. Which part am I mis-understanding?

 

[MrAl]

Hi,

He he, well i dont blame you for doubting, but you actually do have it right this time (in your last post). However, did you remember to FIRST substitute into d^2x/dt^2 the term(s) found for dx/dt itself? dx/dt comes up in the d^2x/dt^2 now because we have it (finally) correct, but in order to simplify completely we have to first substitute what we found earlier for dx/dt. The two inner terms then cancel completely, and we are left with d^2y/dt^2+y=3, but only after a little more manipulation where we subtract 1 from both sides, multiply both sides by -1, then divide by x^4, then substitute y in for 1/x^2.
The difference subst for dx/dt in dx^2/dt^2 makes is that we end up with a second term with the (dy/dt)^2 in it (that 'pesky' factor) but it's the negative of the other one, so they cancel nicely.

I promise this time if you still cant work it out i'll post the entire solution. I was pretty beat yesterday and even today had to run around a lot for some various reasons.

 

[ARandomOWl]

However, did you remember to FIRST substitute into d^2x/dt^2 the term(s) found for dx/dt itself? dx/dt comes up in the d^2x/dt^2 now because we have it (finally) correct, but in order to simplify completely we have to first substitute what we found earlier for dx/dt. The two inner terms then cancel completely, and we are left with d^2y/dt^2+y=3

Oh wow, I completely missed that bit Phew, now I have managed to do it Thanks very much for you help, MrAl.

[LATEX]2x \frac{d^2x}{dt^2}-6(\frac{dx}{dt})^2=x^2-3x^4...(1)[/LATEX]
[LATEX]y=\frac{1}{x^2}...(2)[/LATEX]

From 2:

[LATEX]\frac{dy}{dx}=-\frac{2}{x^3} \therefore \frac{dx}{dy}=-\frac{1}{2}x^3[/LATEX]

From the chain rule:

[LATEX]\frac{dx}{dt}=-\frac{1}{2}x^3\frac{dy}{dt}...(3)[/LATEX]

Subsequently:

[LATEX]\frac{d^2x}{dt^2}=-\frac{1}{2}x^3\frac{d^2y}{dt^2}-\frac{3}{2}x\frac{dx}{dt}\frac{dy}{dt}[/LATEX]

(This is the step I missed) Substituting in (3) gives:

[LATEX]\frac{d^2x}{dt^2}=\frac{3}{4}x^5(\frac{dy}{dt})^2-\frac{1}{2}x^3\frac{d^2y}{dt^2}...(4)[/LATEX]

Substituting (3) and (4) into (1):

[LATEX]2x(\frac{3}{4}x^5(\frac{dy}{dt})^2-\frac{1}{2}x^3\frac{d^2y}{dt^2})-6(-\frac{1}{2}x^3\frac{dy}{dt})^2=x^2-3x^4[/LATEX]

[LATEX]\frac{3}{2}x^6(\frac{dy}{dt})^2-x^4\frac{d^2y}{dt^2}-\frac{3}{2}x^6(\frac{dy}{dt})^2=x^2-3x^4[/LATEX]

Cancelling 1st and 3rd terms and dividing by -x^2:

[LATEX]x^2\frac{d^2y}{dt^2}=3x^2-1...(5)[/LATEX]

Finally from (2) we have:

[LATEX]x^2=\frac{1}{y}...(6)[/LATEX]

Substituting (6) into (5), multiplying through by y and re-arranging gives the final answer:

[LATEX]\frac{d^2y}{dt^2}+y=3[/LATEX]

Wow that took a while.

 

[MrAl]

Oh wow, I completely missed that bit Phew, now I have managed to do it Thanks very much for you help, MrAl.

Substituting (6) into (5), multiplying through by y and re-arranging gives the final answer:

[LATEX]\frac{d^2y}{dt^2}+y=3[/LATEX]

Wow that took a while.

Hi,

You're welcome. The only reason it took so long was because we both stumbled on taking a parametric derivative I dont use this kind of thing that much anymore but still like to review once in a while.