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View Equation

The database contains 327 equations (8 equations are awaiting activation).

Equation data
Category:3. Linear Partial Differential Equations
Subcategory:3.5. Higher-Order Equations
Equation(s):$\displaystyle \frac{\partial w}{\partial t}+at^b\frac{\partial^4w}{\partial x^4}=0$,
\qquad $b\neq-1$.
Solution(s),
Transformation(s),
Integral(s)
:
Particular solutions:\hfill\break
$w_0(x,t)=1$,\\
$w_1(x,t)=x$,\\
$w_2(x,t)=x^2$,\\
$w_3(x,t)=x^3$,\\
$\displaystyle w_4(x,t)=x^4-\frac {4!a}{b+1}t^{b+1}$,\\
$\displaystyle w_5(x,t)=x^5-\frac {5!a}{b+1}xt^{b+1}$,\\
$\displaystyle w_6(x,t)=x^6-\frac {6!a}{2!(b+1)}x^2t^{b+1}$,\\
$\displaystyle w_7(x,t)=x^7-\frac {7!a}{3!(b+1)}x^3t^{b+1}$,\\
$\displaystyle w_8(x,t)=x^8-\frac {8!a}{4!(b+1)}x^4t^{b+1}+\frac {8!a^2}{(b+1)(2b+2)}t^{2b+2}$,\\
$\displaystyle w_9(x,t)=x^9-\frac {9!a}{5!(b+1)}x^5t^{b+1}+\frac {9!a^2}{(b+1)(2b+2)}xt^{2b+2}$,\\
$\displaystyle w_{10}(x,t)=x^10-\frac {10!a}{6!(b+1)}x^6t^{b+1}+\frac {10!a^2}{2!(b+1)(2b+2)}x^2t^{2b+2}$,\\
$\dots$,\\
$\displaystyle w_n(x,t)=\sum_{k=0}^{n-4k\geq0} \frac {(-1)^kn!a^kt^{kb+k}x^{n-4k}}{(n-4k)!k!(b+1)^k}$.
Novelty:New solution(s) / integral(s)
Author/Contributor's Details
Last name:Stepuchev
First name:Valeriy
Country:Latvija
City:Sigulda
Statistic information
Submission date:Wed 07 May 2008 20:58
Edits by author:1
Last edit by author:Wed 11 Jun 2008 14:16

Edit (Only for author/contributor)


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