General Conic

[winged trout]

How do I find the differential equation corresponding to the conic section ax^2 + by^2 + cxy + dx + ey + f = 0?

[steep tulip]

Start with
ax² + by² + cxy + dx + ey + f = 0

[winged trout]

then...

[steep tulip]

differenciate once 2ax + 2by y' + c(xy' + y) + d + ey' = 0

do it again, this is harder and messy but still. 2a + 2b[(y')² + y y''] + c(xy'' + 2y') + ey'' = 0

Now you’ve got 2 equations (from 1st & 2nd derivatives) plus the original. Eliminate the constants a, b, c, d, e, f, that gives you the differential equation of the conic.

[winged trout]

Oh so you mean something like this?

Ax^2 + By^2 + Cx + Dy + Exy + F = 0
=> (x^2 + B' y^2 + C' x + D' y + E' xy + F' = 0
=> 2x + 2B' y y' + C' + D' y' + E' (y + xy') = 0
=> x + B' y y' + C'' + D'' y' + E'' (y + x y') = 0
=> 1 + B' (y'^2 + y y'') + D'' y'' + E'' (y' + y' + x y'') = 0
=> 1/B + y'^2 + y y'' + D''' y'' + 2E''' y' + E''' x y'' = 0
=> 2 y' y'' + y' y'' + y y''' + D''' y''' + 2E''' y'' + E''' y'' + E''' x y''' = 0
=> 3 y' y'' + y y''' + D''' y''' + 3 E''' y'' + E''' x y''' = 0
=> x = - (3y' y'' + y y''' + D''' y''' + 3 E''' y'')/E'''y'''
=> -(E'''y''')^2 = [3 (y'')^2 + 3 y' y''' + y'y''' + y y'''' + D''' y'''' + 3 E''' y'''][E'''y'''] - [3y' y'' + y y''' + D''' y''' + 3 E''' y''][E'''y'''']
=> -E''' * (y''')^2 = y'''[3 (y'')^2 + 4 y' y''' + y y'''' + D''' y'''' + 3 E ''' y'''] - y''''[3 y'y'' + y y''' + D''' y''' + 3 E''' y'']
=> D''' (y''' y'''' - y''' y'''') = -E''' * (y''')^2 - y'''[3 (y'')^2 + 4 y' y''' + y y'''' + 3 E''' y'''] + y''''[3 y' y'' + y y''' + 3 E''' y'']
=> 0 = -E''' * (y''')^2 - y'''[3 (y'')^2 + 4 y' y''' + y y'''' + 3 E''' y'''] + y''''[3 y' y'' + y y''' + 3 E''' y'']
=> E''' (-(y''')^2 - 3(y''')^2 + 3 y'' y'''') = y'''[3(y'')^2 + 4 y' y''' + y y''''] - y''''[3y'y'' + y y''']
=> E''' (3 y'' y'''' - 4 (y''')^2) = y'''[3(y'')^2 + 4 y' y''' + y y''''] - y''''[3y'y'' + y y''']
=> E''' = [y'''[3(y'')^2 + 4 y' y''' + y y''''] - y''''[3y'y'' + y y''']]/[(3 y'' y'''' - 4 (y''')^2)]
=> E''' = [3 (y'')^2 y''' + 4 y' (y''')^2 + y y''' y'''' - 3 y' y'' y'''' - y y''' y'''']/[3 y'' y'''' - 4 (y''')^2]
=> E''' = [ 3 (y'')^2 y''' + 4 y' (y''')^2 - 3 y' y'' y'''' ] / [ 3 y'' y'''' - 4 (y''')^2 ]
=> 0 = [ 6 y'' (y''')^2 + 3 (y'')^2 y'''' + 4 y'' (y''')^2 + 8 y' y''' y'''' - 3 (y'')^2 y'''' - 3 y' y''' y'''' - 3 y' y'' y''''' ][ 3 y'' y'''' - 4 (y''')^2 ] - [ 3 (y'')^2 y''' + 4 y' (y''')^2 - 3 y' y'' y'''' ][ 3 y''' y'''' + 3 y'' y''''' - 8 y''' y'''' ]

=> [ 10 y'' (y''')^2 + 5 y' y''' y'''' - 3 y' y'' y''''' ][ 3 y'' y'''' - 4 (y''')^2 ] - [ 3 (y'')^2 y''' + 4 y' (y''')^2 - 3 y' y'' y'''' ][ 3 y'' y''''' - 5 y''' y'''' ] = 0
=> 30 (y'')^2 (y''')^2 y'''' + 15 y' y'' y''' (y'''')^2 - 9 y' (y'')^2 y'''' y''''' - 40 y'' (y''')^4 - 20 y' (y''')^3 y'''' + 12 y' y'' (y''')^2 y''''' - 9 (y'')^3 y''' y''''' - 12 y' y'' (y''')^2 y''''' + 9 y' (y'')^2 y'''' y''''' + 15 (y'')^2 (y''')^2 y'''' + 20 y' (y''')^3 y'''' - 15 y' y'' y''' (y'''')^2 = 0
=> 45 (y'')^2 (y''')^2 y'''' - 40 y'' (y''')^4 - 9 (y'')^3 y''' y''''' = 0
=> 45 y'' (y''')^2 y'''' - 40 (y''')^4 - 9 (y'')^2 y''' y''''' = 0
=> 45 y'' y''' y'''' - 40 (y''')^3 - 9 (y'')^2 y''''' = 0
=> 9 y''''' (y'')^2 - 45 y'' y''' y'''' + 40 (y''')^3 = 0

[steep tulip]

ending in a 2nd order DE involving y, y', y'' only.

[winged trout]

It's 5th order apparently.

[steep tulip]

this is too much, for a conic you only need up to 2nd derivative, not 5th

like this

[winged trout]

That's only for the usual conics, this was the general conic.

[light umbra]

X_X

[winged trout]

Even for a circle, if we eliminate x (and h, k, r), then it's still 3rd order:

3y'(y'')² - (y')²y''' - y''' = 0

[steep tulip]

not rlly your mistake is thinking “6 constants --need like 5 derivatives”, that’s not how this one works because the constants appear linearly, so they can be eliminated much earlier.

[winged trout]

The answer is actually correct so...

[steep tulip]

ye, that was js and easier approach.

an

[winged trout]

Can you try doing it to see if that works?

[steep tulip]

kk, give me a sec

[light umbra]

Summary of Properties
Order: 5 (because there are 5 independent arbitrary constants).
Degree: 2 (due to the

term).
Geometric Meaning: This equation represents all possible parabolas, ellipses, and hyperbolas in a plane. If you were looking for a specific type (like just circles), the order would be lower (order 3) because circles have fewer independent constants.

[winged trout]

Is this an AI?

[light umbra]

nope

my friend sait it

said*

[steep tulip]

Ax² + By² + Cxy + Dx + Ey + F = 0 ...(1)

diff once:
2Ax + 2By y' + C(xy' + y) + D + Ey' = 0 ...(2)

diff twice:
2A + 2B[(y')² + y y''] + C(2y' + x y'') + Ey'' = 0 ...(3)

now eliminate D from (2):
D = -2Ax - 2By y' - C(xy' + y) - Ey'

plug into stuff and start cancelling terms

then use (1) to eliminate F

basically keep eliminating A, B, C, D, E step by step

eventually everything falls into something with just y, y', y'' yes ur correct

[light umbra]

wow

[steep tulip]

?

[winged trout]

Alright, thanks for verifying!