Talk:General Solution to Chebyshev's Differential Equation

I don't understand this:

\(\ds 0\)

\(=\)

\(\ds \paren {1 - \sin^2 \theta} \frac {\d y} {\d x} \paren {\frac {\d y} {\d \theta} \frac {\d \theta} {\d x} } - \sin \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x}\)

substituting $x = \sin \theta$ and Chain Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \paren {1 - \sin^2 \theta} \frac {\d y} {\d \theta} \frac {\d \theta} {\d x} \paren {\frac {\d y} {\d \theta} \frac {\d \theta} {\d x} } - \sin \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x}\)

Chain Rule for Derivatives again

\(\ds \)

\(=\)

\(\ds \paren {\cos^2 \theta} \frac {\d y} {\d \theta} \frac 1 {\cos \theta} \paren {\frac {\d y} {\d \theta} \frac 1 {\cos \theta} } - \sin \theta \frac {\d y} {\d \theta} \frac 1 {\cos \theta}\)

Sum of Squares of Sine and Cosine and substituting $\dfrac {\d \theta} {\d x} = \dfrac 1 {\cos \theta}$

\(\ds \)

\(=\)

\(\ds \paren {\cos \theta} \paren {\frac {\d^2 y} {\d \theta^2} \frac 1 {\cos \theta} + \frac {\d y} {\d \theta} \frac {\sin \theta} {\cos^2 \theta} } - \sin \theta \frac {\d y} {\d \theta} \frac 1 {\cos \theta}\)

Product Rule for Derivatives, Derivative of Secant Function

Please excuse minor restructuring of comments; I'm trying to understand exactly what's going on.

I see this:

$0 = \paren {1 - \sin^2 \theta} \dfrac {\d y} {\d x} \paren {\dfrac {\d y} {\d \theta} \dfrac {\d \theta} {\d x} } - \sin \theta \dfrac {\d y} {\d \theta} \dfrac {\d \theta} {\d x}$

but I can't see why.

I think that $\dfrac {\d^2 y} {\d x^2}$ seems to should be $\map {\dfrac \d {\d x} } {\dfrac {\d y} {\d x} }$

which should be $\map {\dfrac \d {\d x} } {\dfrac {\d y} {\d \theta} \dfrac {\d \theta} {\d x} }$

which should be $\dfrac {\d^2 y} {\d x \d \theta} + \dfrac {\d y} {\d \theta} \dfrac {\d^2 \theta} {\d x^2}$.

using the product rule.

What this seems to be:

$\dfrac {\d y} {\d x} \paren {\dfrac {\d y} {\d \theta} \dfrac {\d \theta} {\d x} }$

and thence to:

$\dfrac {\d y} {\d \theta} \dfrac {\d \theta} {\d x} \paren {\dfrac {\d y} {\d \theta} \dfrac {\d \theta} {\d x} }$

looks like an expansion of $\paren {\dfrac {\d y} {\d x} }^2$ and not $\dfrac {\d^2 y} {\d x^2}$.

And in fact what I see is that in the last line of the above, a factor of $\cos \theta$ seems to have magically vanished.

Am I missing something obvious? --prime mover (talk) 18:13, 27 March 2025 (UTC)

$\quad \ds \paren {1 - x^2} \frac {\d^2 y} {\d x^2} - x \frac {\d y} {\d x} = 0$

$y = A + B \arcsin x $

$\ds \frac {\d y} {\d x} = \dfrac B {\sqrt{\paren {1 - x^2} } }$

$\ds \frac {\d^2 y} {\d x^2} = \dfrac {B x} {\sqrt{\paren {1 - x^2} } \paren {1 - x^2} } $

$\quad \ds \paren {1 - x^2} \dfrac {B x} {\sqrt{\paren {1 - x^2} } \paren {1 - x^2} } - x \dfrac B {\sqrt{\paren {1 - x^2} } } = 0$

Solution works. In fact, $\arccos x$ also works!

$\quad \ds \paren {1 - x^2} \frac {\d^2 y} {\d x^2} - x \frac {\d y} {\d x} = 0$

$y = A + B \arccos x $

$\ds \frac {\d y} {\d x} = -\dfrac B {\sqrt{\paren {1 - x^2} } }$

$\ds \frac {\d^2 y} {\d x^2} = -\dfrac {B x} {\sqrt{\paren {1 - x^2} } \paren {1 - x^2} } $

$\quad \ds \paren {1 - x^2} \paren {-\dfrac {B x} {\sqrt{\paren {1 - x^2} } \paren {1 - x^2} } } - \paren {-x \dfrac B {\sqrt{\paren {1 - x^2} } } } = 0$

Back to your question...the crux is that:

$\dfrac {\d y} {\d x} = \dfrac {\d y} {\d \theta} \dfrac {\d \theta} {\d x} $

We need to get out of the $x$ world and into the $\theta$ world.

Expanding on the above...

\(\ds 0\)

\(=\)

\(\ds \paren {1 - \sin^2 \theta} \frac {\d y} {\d x} \paren {\frac {\d y} {\d \theta} \frac {\d \theta} {\d x} } - \sin \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x}\)

substituting $x = \sin \theta$ and Chain Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \paren {1 - \sin^2 \theta} \frac {\d y} {\d \theta} \frac {\d \theta} {\d x} \paren {\frac {\d y} {\d \theta} \frac {\d \theta} {\d x} } - \sin \theta \frac {\d y} {\d \theta} \frac {\d \theta} {\d x}\)

Chain Rule for Derivatives again - notice that we changed $\dfrac {\d y} {\d x} = \dfrac {\d y} {\d \theta} \dfrac {\d \theta} {\d x} $

\(\ds \)

\(=\)

\(\ds \paren {\cos^2 \theta} \frac {\d y} {\d \theta} \frac 1 {\cos \theta} \paren {\frac {\d y} {\d \theta} \frac 1 {\cos \theta} } - \sin \theta \frac {\d y} {\d \theta} \frac 1 {\cos \theta}\)

Sum of Squares of Sine and Cosine and substituting $\dfrac {\d \theta} {\d x} = \dfrac 1 {\cos \theta}$

\(\ds \)

\(=\)

\(\ds \paren {\cos \theta} \frac {\d y} {\d \theta} \paren {\frac {\d y} {\d \theta} \frac 1 {\cos \theta} } - \frac {\d y} {\d \theta} \frac {\sin \theta } {\cos \theta}\)

new/clarifying step - cancel $\cos \theta$

\(\ds \)

\(=\)

\(\ds \paren {\cos \theta} \paren {\frac {\d^2 y} {\d \theta^2} \frac 1 {\cos \theta} + \frac {\d y} {\d \theta} \frac {\sin \theta} {\cos^2 \theta} } - \frac {\d y} {\d \theta} \frac {\sin \theta } {\cos \theta}\)

Product Rule for Derivatives, Derivative of Secant Function

\(\ds \)

\(=\)

\(\ds \paren {\frac {\d^2 y} {\d \theta^2} + \frac {\d y} {\d \theta} \frac {\sin \theta} {\cos \theta} } - \frac {\d y} {\d \theta} \frac {\sin \theta } {\cos \theta}\)

distributing $\cos \theta$

In short:

\(\ds y' '\)

\(=\)

\(\ds \dfrac {\d^2 y} {\d x^2}\)

Chain Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \frac {\d y} {\d x} \paren {\frac {\d y} {\d x} }\)

\(\ds \)

\(=\)

\(\ds \dfrac {\d y} {\d \theta} \dfrac {\d \theta} {\d x} \paren {\dfrac {\d y} {\d \theta} \dfrac {\d \theta} {\d x} }\)

\(\ds \)

\(=\)

\(\ds \dfrac {\d \theta} {\d x} \dfrac {\d y} {\d \theta} \paren {\dfrac {\d y} {\d \theta} \dfrac {\d \theta} {\d x} }\)

\(\ds \)

\(=\)

\(\ds \dfrac {\d \theta} {\d x} \paren {\frac {\d^2 y} {\d \theta^2} \dfrac {\d \theta} {\d x} + \frac {\d y} {\d \theta} \frac {\d^2 \theta} {\d x^2} }\)

Product Rule for Derivatives

Does this help? --Robkahn131 (talk) 19:21, 27 March 2025 (UTC)

I don't understand how $\dfrac {\d^2 y} {\d x^2} = \dfrac {\d y} {\d x} \paren {\dfrac {\d y} {\d x} }$

The first expression means the second derivative of $y$ wrt $x$ that is, $\map {\dfrac \d {\d x} } {\dfrac {\d y} {\d x} }$.

The second expression means the product of $\dfrac {\d y} {\d x}$ with $\dfrac {\d y} {\d x}$.

If $\dfrac {\d y} {\d x} \paren {\dfrac {\d y} {\d x} }$ is supposed to be interpreted as $\map {\dfrac \d {\d x} } {\dfrac {\d y} {\d x} }$, there's some serious abuse of notation going on.

Basically, $\dfrac {\d y} {\d x}$ is not an operator, it's the result of having operated on $y$ by the operator $\dfrac \d {\d x}$. --prime mover (talk) 20:46, 27 March 2025 (UTC)

You claim above:

$\dfrac {\d^2 y} {\d x^2} = \dfrac {\d^2 y} {\d x \d \theta} + \dfrac {\d y} {\d \theta} \dfrac {\d^2 \theta} {\d x^2}$.

I claim:

$\ds \dfrac {\d^2 y} {\d x^2} = \dfrac {\d \theta} {\d x} \paren {\frac {\d^2 y} {\d \theta^2} \dfrac {\d \theta} {\d x} + \frac {\d y} {\d \theta} \frac {\d^2 \theta} {\d x^2} }$.

I see it now! :)

Here is your comments with edits:

I think that $\dfrac {\d^2 y} {\d x^2}$ seems to should be $\map {\dfrac \d {\d x} } {\dfrac {\d y} {\d x} }$ - I agree.

which should be $\ds \dfrac {\d \theta} {\d x} \map {\dfrac \d {\d \theta} } {\dfrac {\d y} {\d \theta} \dfrac {\d \theta} {\d x} }$ - move to $\theta$ world.

which should be $\ds \dfrac {\d \theta} {\d x} \paren {\frac {\d^2 y} {\d \theta^2} \dfrac {\d \theta} {\d x} + \frac {\d y} {\d \theta} \frac {\d^2 \theta} {\d x^2} }$.

using the product rule. --Robkahn131 (talk) 21:29, 27 March 2025 (UTC)

Minor correction...

I think that $\dfrac {\d^2 y} {\d x^2}$ seems to should be $\map {\dfrac \d {\d x} } {\dfrac {\d y} {\d x} }$ - I agree.

which should be $\ds \dfrac {\d \theta} {\d x} \map {\dfrac \d {\d \theta} } {\dfrac {\d y} {\d \theta} \dfrac {\d \theta} {\d x} }$ - move to $\theta$ world.

which should be $\ds \dfrac {\d \theta} {\d x} \paren {\frac {\d^2 y} {\d \theta^2} \dfrac {\d \theta} {\d x} + \frac {\d y} {\d \theta} \frac {\d^2 \theta} {\d x \d \theta} }$.

using the product rule. --Robkahn131 (talk) 21:36, 27 March 2025 (UTC)

Let me think it through again when I've had a sleep --prime mover (talk) 00:47, 28 March 2025 (UTC)

... yes, that's the point I was trying to make, I just implemented it wrong.

I have restructured the comments slightly and merged one or two small steps. Hope this is all okay. --prime mover (talk) 08:47, 28 March 2025 (UTC)