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Equation data
Category:3. Linear Partial Differential Equations
Subcategory:3.3. Second-Order Elliptic Equations
Equation(s):$\displaystyle x^a\frac{\partial ^2w}{\partial y^2}+b\frac{\partial^2w}{\partial x^2}=0$.
Solution(s),
Transformation(s),
Integral(s)
:
1. Particular solutions:\hfill\break
$w_0(x,y)=1$,\\
$w_1(x,y)=y$,\\
$\displaystyle w_2(x,y)=y^2-\frac {2x^{a+2}}{b(a+1)(a+2)}$,\\
$\displaystyle w_3(x,y)=y^3-\frac {6yx^{a+2}}{b(a+1)(a+2)}$,\\
$\displaystyle w_4(x,y)=y^4-\frac {12y^2x^{a+2}}{b(a+1)(a+2)}+\frac {24x^{2a+4}}{b^22!\left(\frac{a+1}{a+2}\right)_2(a+2)^4}$,\\
$\displaystyle w_5(x,y)=y^5-\frac {20y^3x^{a+2}}{b(a+1)(a+2)}+\frac {120yx^{2a+4}}{b^22!\left(\frac{a+1}{a+2}\right)_2(a+2)^4}$,\\
$\dots$,\\
$\displaystyle w_n(x,y)=y^n+\sum_{k=1}^{n-2k\geq0} \frac {n!(-1)^ky^{n-2k}x^{ka+2k}}{b^k(n-2k)!k!\left(\frac{a+1}{a+2}\right)_k(a+2)^{2k}}$.\\
\medskip

2. Particular solutions:\\
$w_0(x,y)=x$,\\
$w_1(x,y)=xy$,\\
$\displaystyle w_2(x,y)=xy^2-\frac {2x^{a+3}}{b(a+2)(a+3)}$,\\
$\displaystyle w_3(x,y)=xy^3-\frac {6yx^{a+3}}{b(a+2)(a+3)}$,\\
$\displaystyle w_4(x,y)=xy^4-\frac {12y^2x^{a+3}}{b(a+2)(a+3)}+\frac {24x^{2a+5}}{b^22!\left(\frac{a+3}{a+2}\right)_2(a+2)^4}$,\\
$\displaystyle w_5(x,y)=xy^5-\frac {20y^3x^{a+3}}{b(a+2)(a+3)}+\frac {120yx^{2a+5}}{b^22!\left(\frac{a+3}{a+2}\right)_2(a+2)^4}$,\\
$\dots$,\\
$\displaystyle w_n(x,y)=xy^n+\sum_{k=1}^{n-2k\geq0} \frac {n!(-1)^ky^{n-2k}x^{ka+2k+1}}{b^k(n-2k)!k!\left(\frac{a+3}{a+2}\right)_k(a+2)^{2k}}$.
Remarks:Here, $a\neq-2$; $\displaystyle a\neq-2\pm\frac 1k$, where $k=1,\,2,\,3,\,\dots$
Novelty:New solution(s) / integral(s)
Author/Contributor's Details
Last name:Stepuchev
First name:Valeriy
Country:Latviya
City:Sigulda
Statistic information
Submission date:Mon 28 Apr 2008 19:42
Edits by author:1
Last edit by author:Mon 05 Oct 2015 19:18

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