I am supposed to turn a linear differential equations with Euler coefficients into one with constant coefficients using the substitution $x = e^t$, but I'm not sure I've ever done a substitution for the independent variable before.

Taking the example $x^2 y'' + 3xy' + y = 0$, do I substitute only into the visible $x$'s yielding $e^{2t} \frac{d^2 y}{dt^2} + 3e^t \frac{dy}{dt} + y = 0$, or also into the variable the derivatives are with respect to, yielding $e^{2t} \frac{d^2 y}{d(e^t)^2} + 3e^t\frac{dy}{d(e^t)} + y = 0$, and in either case, how can I make the coefficients linear from here?

Such a sneaky disease

Don't write stuff like $d(e^t)$. Just use the chain rule properly.

But why does $O(|hu|^{m+1})$ assume $f$ needs to be $m+1$ times differentiable?

From the definition of big Oh, that is.

@schn: I'm tired of saying the same things over and over. I've given you the example to explain it. It's FALSE without the extra differentiability.

OK, @user10478, so $\dfrac{dy}{dt} = y'\dfrac{dx}{dt} = y' e^t = xy'$. Can you do the second derivative?

The time distance between onset of symptoms in the "infecter" and onset of symptoms in the infected person seems significantly smaller than the ~5-6 days average, with a long tail, for the actual incubation period

So the derivative of $xy'$ with respect to $t$, $x'y' + xy''$? Or does that assume the wrong chain of dependencies?

Don't use the prime for both x- and t-derivatives. That's horrendous.

I think thinking everything was with respect to $t$ but then maybe the answer is wrong.

@TedShifrin It's confusing with the exponent in $f(x)=|x|^{3/2}$. In the Taylor expansion I posted there are no fractions in the exponent of $hu$.

So after the substitution, $y$ is a function of $x$ which is a function of $t$, so ultimately everything should be with respect to $t$.

Nor should there be. I am telling you to do the first degree Taylor polynomial of $f$. That will be a regular polynomial. Calculate the value at $0$ and the first derivative at $0$ and write down the Taylor polynomial.

Right, so $\dfrac{d^2y}{dt^2} = \dfrac d{dt} (xy')$. You have to use chain rule again.

And product rule, of course.

Oh, I see what happened, your $y'$ with $xy'$ is respect to $x$.

Yes, to match the original problem.

I'm using prime for the original $x$-derivatives and $d/dt$ for $t$-derivatives. Otherwise it's hopeless.

@TedShifrin So $df/dx=(3 x)/(2 \sqrt(|x|))$ if $x\neq 0$.

That's not helpful. We're trying to compute the derivative AT $0$.

So I need to take $\frac{d}{dt}(x \frac{dy}{dx})$.

Yes. Product rule first, then chain rule.

@TedShifrin At 0 it is 0.

Correct. So the Taylor polynomial is $0+0\cdot (x-0) = 0$.

$\frac{dy}{dt} + x \frac{d^2 y}{dx^2}\frac{dx}{dt}$

Therefore, as I said ages ago, the error is all of $f(x)$, which is NOT $O(x^2)$.

@TedShifrin It doesn't have any error, or?

@user10478 The first term makes no sense to me. The second term, then (fixing your typo) is $x^2y''$.

@schn Come on. What's the definition of the error? It's $f(x)-P(x)$, where $P$ is the Taylor polynomial.

@TedShifrin Right :)

@TedShifrin So it is $O(|x|^{3/2})$?

The first term was $\frac{dx}{dt}\frac{dy}{dx}$. When $y$ is a function of $x$ which is a function of $t$, that's $\frac{dy}{dt}$ by the chain rule, right?

It is precisely $|x|^{3/2}$, which, as we have said several times, is NOT $O(x^2)$.

Oh, oh, you were too sneaky for me, @user10478. You're right.

So put all the terms together now.

I would have just written $dx/dt = e^t = x$ again, so we have $xy'$, which is, as we already had calculated $dy/dt$.

I'm not sure quite sure how you got $x^2 y''$ for the second term.

@TedShifrin And this is due to the fact that $|x|^{3/2}$ does not have a second derivative?

$d^2y/dx^2 = y''$, and $dx/dt = e^t = x$.

It's due to the fact that the exponent 3/2 is smaller than the exponent 2. But, ultimately, yes, you do not have the usual remainder formula because the function is not twice differentiable.

@TedShifrin How would you write a Taylor expansion with only finite terms, specifically the error?

I don't understand. We just wrote the Taylor polynomial. It is $0$. The error is the function, so it's $|x|^{3/2}$.

@TedShifrin Well, is there a way to use the big Oh notation although it isn't correct with m+1 in the exponent, or does one have to use the little oh notation?

The correct statement is little-oh unless you have more hypotheses. I've said this literally four times. I'm losing patience.

@TedShifrin If you know of a post or some text that deals with taylor expansions with finite differentiability of the function, I'd be very grateful. Thanks for your help though. In for instance Calculus by Adams this isn't treated, but maybe it is so basic :)

Most calculus books aren't going to mess with the subtlety. You could look at a mathematical treatment, like Spivak's Calculus.

Okay, I think I see what we did. We went backward, starting with the terms expected in the constant coefficient ODE, and tried to replace all $x$ with $t$ to arrive at the terms in the Euler ODE. So it would just be $\frac{d^2 y}{dt^2} + 3\frac{dy}{dt} + y = 0$.

@TedShifrin Is it treated in that book?

Yes, he discusses this very carefully. But he does NOT use little- or big-oh at all.

@user10478 Looks right.

Something doesn't seem quite right to me. The indical equation of the Euler ODE has $1$ root with algebraic multiplicity $2$, whereas the characteristic equation of the constant coefficient ODE has $2$ distinct roots. Is this a problem?

Oh, you made a mistake, didn't you?

Didn't you lose the extra $dy/dt$?

Should be one somewhere.

Okay I see it. So should the indical equation and characteristic equation always be identical, so in practice you can just immediately determine the constant coefficient ODE by the indical equation?

I haven't thought about it in generality, but since the change of variables doesn't change the qualitative nature of the equation, I would guess that's right.

Okay, thank you for the help.

Sure.