Wikipedia:Reference desk archive/Mathematics/2006 August 27

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Linear Differential Equation

${\displaystyle {\frac {dy}{dx}}+y=y^{3}}$

I know how to solve linear differential equations, but the y3 on the right side has totally confused me. Here y=y(x).--Patchouli 08:57, 27 August 2006 (UTC)

See the section "Linear differential equations with variable coefficients" on Linear differential equation. That has the method for solving such a DE. Dysprosia 09:03, 27 August 2006 (UTC)
I had looked at that already. It doesn't help me eliminate y3.--Patchouli 09:07, 27 August 2006 (UTC)
That's not a linear differential equation because of the y3 term. Did you try separation of variables? We have an article about it (see separation of variables), but you're probably better of looking it up in a textbook or lecture notes. -- Jitse Niesen (talk) 09:22, 27 August 2006 (UTC)
Oh, whoops. Dysprosia 11:37, 27 August 2006 (UTC)

You should have directed me to Bernoulli differential equation.

${\displaystyle u=y^{1-3}=y^{-2}}$.

Equation${\displaystyle {\frac {du}{dx}}-2u=-2}$

Integrating factor is ${\displaystyle e^{-2x}}$.

u=1+c×e2×x.

${\displaystyle y^{2}={\frac {1}{1+c\times e^{2x}}}.}$--Patchouli 09:38, 27 August 2006 (UTC)

While that method works here, so does separation of variables, which in this case is quite easy:
${\displaystyle {\frac {dy}{dx}}+y=y^{3}}$
${\displaystyle {\frac {dy}{dx}}=y^{3}-y}$
${\displaystyle {\frac {dx}{dy}}={\frac {1}{y^{3}-y}}}$
${\displaystyle x=\int {\frac {dy}{y^{3}-y}}={\frac {1}{2}}{\log \left(1-{\frac {1}{y^{2}}}\right)}+c}$
${\displaystyle y^{2}={\frac {1}{1-\exp {2(x-c)}}}}$
--LambiamTalk 10:31, 27 August 2006 (UTC)
• I wasn't thinking of partial fractions for integrating::${\displaystyle x=\int {\frac {dy}{y^{3}-y}}={\frac {1}{2}}{\log \left(1-{\frac {1}{y^{2}}}\right)}+c}$.

Your solution is imaginative.--Patchouli 11:03, 27 August 2006 (UTC)

Continuous functions

I was recently reading a (non-WP) article that uses the notation ${\displaystyle C^{(k,\alpha )}}$ with ${\displaystyle 0<\alpha <1}$ and k an integer to denote some sort of continuity. The paper failed to describe the meaning of this notation. From the context, it was clear that it was kind-of-like ${\displaystyle C^{k}}$ viz. the set of k-times differentiable functions. For a while, I thought this might refer to Sobolev space, but the ${\displaystyle 0<\alpha <1}$ condition makes this seem unlikely. Any hints about what this notation might be? FWIW, this is a survey article, so the author was clearly assuming that this is a basic notation. FWIW, this is the article: Potential theory, the notation shows up just a little more than 1/3 of the way in. Its mostly just plain-old stock, ordinary integrals and derivatives on plain-old flat space, nothing fancy, nothing topological, or otherwise whiz-bang, nothing beyond multivariate calculus. linas 15:31, 27 August 2006 (UTC)

Could it be functions with Holder continuous k-th derivative? See Holder condition (Igny 16:16, 27 August 2006 (UTC))
Why yes, of course, that's exactly it. Thanks. linas 03:05, 28 August 2006 (UTC)