ARandomOWl
New Member
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.
[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.