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Equation data
Category:3. Linear Partial Differential Equations
Subcategory:3.5. Higher-Order Equations
Equation(s):$\displaystyle \frac{\partial ^2w}{\partial t^2}+ae^{bt}\frac{\partial^3w}{\partial x^3}=0$.
Solution(s),
Transformation(s),
Integral(s)
:
1. Partucular solutions:\hfill\break
$w_0(x,t)=1$,\\
$w_1(x,t)=x$,\\
$w_2(x,t)=x^2$,\\
$\displaystyle w_3(x,t)=x^3-\frac {3!a}{b^2}e^{bt}$,\\
$\displaystyle w_4(x,t)=x^4-\frac {4!a}{b^2}xe^{bt}$,\\
$\displaystyle w_5(x,t)=x^5-\frac {5!a}{2!b^2}x^2e^{bt}$,\\
$\displaystyle w_6(x,t)=x^6-\frac {6!a}{3!b^2}x^3e^{bt}+\frac {6!a^2}{4b^4}e^{2bt}$,\\
$\dots$,\\
$\displaystyle w_n(x,t)=x^n+\sum_{k=1}^{n-3k\geq0} \frac {(-1)^kn!a^ke^{kbt}x^{n-3k}}{(n-3k)!b^{2k}(k!)^2}$.
\medskip

2. Partucular solutions:\hfill\break
$w_0(x,t)=t$,\\
$w_1(x,t)=xt$,\\
$w_2(x,t)=x^2t$,\\
$\displaystyle w_3(x,t)=x^3t-\frac {3!ae^{bt}}{b^2}(t-\frac 2b)$,\\
$\displaystyle w_4(x,t)=x^4t-\frac {4!axe^{bt}}{b^2}(t-\frac 2b)$,\\
$\displaystyle w_5(x,t)=x^5t-\frac {5!ax^2e^{bt}}{2!b^2}(t-\frac 2b)$,\\
$\displaystyle w_6(x,t)=x^6t-\frac {6!ax^3e^{bt}}{3!b^2}(t-\frac 2b)+\frac {6!a^2e^{2bt}}{4b^4}(t-\frac 3b)$,\\
$\dots$,\\
$\displaystyle w_n(x,t)=x^nt+\sum_{k=1}^{n-3k\geq0} \frac {(-1)^kn!a^ke^{kbt}x^{n-3k}}{(n-3k)!b^{2k}(k!)^2}(t-\frac{c_k}{b})$,\\
where $c_1=2$, $c_2=3$, \dots, $\displaystyle c_k=c_{k-1}+\frac 2k$.
Novelty:New solution(s) / integral(s)
Author/Contributor's Details
Last name:Stepuchev
First name:Valeriy
Country:Latvija
City:Sigulda
Statistic information
Submission date:Mon 12 May 2008 18:57
Edits by author:1
Last edit by author:Tue 10 Nov 2009 19:15

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