Wikipedia:Reference desk/Archives/Mathematics/2006 August 27

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Linear Differential Equation

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

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)[reply]

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)[reply]

I had looked at that already. It doesn't help me eliminate y³.--Patchouli 09:07, 27 August 2006 (UTC)[reply]

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)[reply]

Oh, whoops. Dysprosia 11:37, 27 August 2006 (UTC)[reply]

You should have directed me to Bernoulli differential equation.

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

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

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

u=1+c×e^2×x.

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

While that method works here, so does separation of variables, which in this case is quite easy:

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

{\frac {dy}{dx}}=y^{3}-y

{\frac {dx}{dy}}={\frac {1}{y^{3}-y}}

x=\int {\frac {dy}{y^{3}-y}}={\frac {1}{2}}{\log \left(1-{\frac {1}{y^{2}}}\right)}+c

y^{2}={\frac {1}{1-\exp {2(x-c)}}}

--Lambiam ^Talk 10:31, 27 August 2006 (UTC)[reply]

I wasn't thinking of partial fractions for integrating:: $x=\int {\frac {dy}{y^{3}-y}}={\frac {1}{2}}{\log \left(1-{\frac {1}{y^{2}}}\right)}+c$ .

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

Continuous functions

I was recently reading a (non-WP) article that uses the notation $C^{(k,\alpha )}$ with $0<\alpha <1$ 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 $C^{k}$ viz. the set of k-times differentiable functions. For a while, I thought this might refer to Sobolev space, but the $0<\alpha <1$ 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)[reply]

Could it be functions with Holder continuous k-th derivative? See Holder condition (Igny 16:16, 27 August 2006 (UTC))[reply]

Why yes, of course, that's exactly it. Thanks. linas 03:05, 28 August 2006 (UTC)[reply]