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.