Find differential equation by eliminating orbitary constant a,b, cfrom (x-a)^2+(y-b)^2+(z-c)^2=r^2 Where r is radius
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Let X=x-a, Y=y-b, Z=z-c, then X²+Y²+Z²=r².

Differentiating wrt to another variable t, 2XdX/dt + 2YdY/dt + 2ZdZ/dt = 0, so

XdX/dt + YdY/dt + ZdZ/dt = 0. Note that dX/dt can be replaced by dx/dt, etc.

Differentiating again:

(dX/dt)²+Xd²Xdt²+(dY/dt)²+Yd²Y/dt²+(dZ/dt)²+Zd²Z/dt²=0. Note that d²X/dt² can be replaced by d²x/dt², etc.

The original equation is that of a sphere centre (a,b,c), radius r. By transformation of axes so that the new origin is the centre of the sphere, a, b and c can be eliminated, and DEs created as above.

If dt=dx then we have:

X+Yy'+Zz'=0 (first degree)

1+Yy"+(y')²+Zz"+(z')²=0 (second degree)

The prime means d/dx. Double prime is d²/dx².

By further differentiation we arrive at three DEs containing three variables X, Y and Z and differentials involving x, y and z. From such a system of equations, X, Y and Z can be found in terms of differentials only. The algebra becomes very complicated. Once found, the variables can be replaced in X²+Y²+Z²=r². This will produce a rather complicated DE with no reference at all to a, b or c.

The third degree DE is:

Yy'''+3y'y"+Zz'''+3z'z"=0.

The second and third degrees DEs form a system of simultaneous equations containing only Y and Z, so both can found. These can then be substituted into the first degree DE to find X.

Finally, substitute X, Y and Z into X²+Y²+Z²=r² to arrive at a rather complicated equation involving x and derivatives wrt x, without any references to a, b or c.

To illustrate and simplify the process, I offer the following.

Yy"+Zz"=-A

Yy'''+Zz'''=-3B

where A=1+(y')²+(z')² and B=y'y"+z'z".

So, Y=(3Bz"-Az''')/(y"z'''-y'''z") and Z=(3By"-Ay''')/(y'''z"-y"z''').

And X=-(Yy'+Zz')=(3By"z'-3By'z"+Ay'z'''-Ay'''z')/(y"z'''-y'''z") and X²+Y²+Z²=r².

by Top Rated User (1.1m points)

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