欧拉方程解法--6-11-6-1 ，2-3-1

欧拉方程

形如 $f(x)=x^{n}y^{(n)}+p_{1}x^{n-1}y^{(n-1)}+cdotcdotcdot+p_{n-1}xy'+p_{n}y$ 的方程（其中 $p_{1},p_{2},cdotcdotcdot,p_{n}$ 为常数），叫做欧拉方程。

$做变换x=e^t或t=lnx,将自变量x换成t,我们有$

$frac{dy}{dx}=frac{dy}{dt}cdotfrac{dt}{dx}=frac{1}{x}frac{dy}{dt}\frac{d^2y}{dx^2}=frac{1}{x^2}(frac{d^2y}{dt^2}-frac{dy}{dt})\ frac{d^3y}{dx^3}=frac{1}{x^3}(frac{d^3y}{dt^3}-3frac{d^2y}{dt^2}+2frac{dy}{dt})\frac{d^4y}{dx^4}=frac{1}{x^4}(frac{d^4y}{dt^4}-6frac{d^3y}{dt^3}+11frac{d^2y}{dt^2}-6frac{dy}{dt})$

如果采用记号D表示对t求导的运算 $frac{d}{dt}$ ,那么上述计算结果可以写成

$xy'=Dy$

$x^{2}y''=frac{d^2y}{dt^2}-frac{dy}{dt}=(frac{d^2}{dt^2}-frac{d}{dt})y=(D^2-D)y=D(D-1)y$
$x^3y'''=frac{d^3y}{dt^3}-3frac{d^2y}{dt^2}+2frac{dy}{dt}=(D^3-3D^2+2D)y=D(D-1)(D-2)y$

一般地，有 $x^ky^{(k)}=D(D-1)cdotcdotcdot(D-k+1)y$
把它代入欧拉方程，便得到一个以t为自变量的c常系数线性微分方程。在求出这个方程的解后，把 $t$ 换成 $lnx$ ,即得原方程的解。

以下以一个在弹性力学中常见的四阶变系数线性微分方程的求解为例

$frac{d^4f(p)}{dp^4}+frac{2}{p}frac{d^3f(p)}{dp^3}-frac{9}{p^2}frac{d^2f(p)}{dp^2}+frac{9}{p^3}frac{df(p)}{dp}=0$
$令p=x,f(p)=y.$
$则原式为，frac{d^4y}{dx^4}+frac{2}{p}frac{d^3y}{dx^3}-frac{9}{p^2}frac{d^2y}{dx^2}+frac{9}{p^3}frac{dy}{dx}=0$

$令x=e^t作自变量代换，有t=lnx,从而frac{dt}{dx}=frac{1}{x}$

$frac{dy}{dx}=frac{dy}{dt}frac{dt}{dx}=frac{1}{x}frac{dy}{dt},则xfrac{dy}{dx}=frac{dy}{dt}$

$frac{d^2y}{dx^2}=frac{d}{dx}(frac{1}{x}frac{dy}{dt})=-frac{1}{x^2}frac{dy}{dt}+frac{1}{x}frac{d}{dx}(frac{dy}{dt})=-frac{1}{x^2}frac{dy}{dt}+frac{1}{x^2}frac{d^2y}{dt^2}$

$frac{d^3y}{dx^3}=frac{d}{dx}(-frac{1}{x^2}frac{dy}{dt}+frac{1}{x^2}frac{d^2y}{dt^2})$

$=(frac{2}{x^3}frac{dy}{dt}-frac{1}{x^2}frac{d}{dx}(frac{dy}{dt})-frac{2}{x^3}frac{d^2y}{dt^2}+frac{1}{x^2}frac{d}{dx}(frac{d^2y}{dt^2}))$
$=frac{2}{x^3}frac{dy}{dt}-frac{1}{x^3} frac{d^2y}{dt^2}-frac{2}{x^3}frac{d^2y}{dt^2}+frac{1}{x^3} frac{d^3y}{dt^3}$
$=frac{2}{x^3}frac{dy}{dt}-frac{3}{x^3} frac{d^2y}{dt^2}+frac{1}{x^3} frac{d^3y}{dt^3}$
$frac{d^4y}{dx^4}=frac{d}{dx}(frac{2}{x^3}frac{dy}{dt}-frac{3}{x^3} frac{d^2y}{dt^2}+frac{1}{x^3} frac{d^3y}{dt^3})$
$=-frac{6}{x^4}frac{dy}{dt}+frac{2}{x^3}frac{d}{dx}(frac{dy}{dt})+frac{9}{x^4}frac{d^2y}{dt^2}-frac{3}{x^3}frac{d}{dx}(frac{d^2y}{dt^2})-frac{3}{x^4}frac{d^3y}{dt^3}+frac{1}{x^3}frac{d}{dx}(frac{d^3y}{dt^3})$

$=-frac{6}{x^4}frac{dy}{dt}+frac{11}{x^4}frac{d^2y}{dt^2}-frac{6}{x^4}frac{d^3y}{dt^3}+frac{1}{x^4}frac{d^4y}{dt^4}$

$将所求得的frac{d^ny}{dx^n}的一二三四阶形式替换原式中的frac{d^ny}{dx^n}的一二三四阶形式$ $转化为特征方程得,lambda^4-4lambda^3-4lambda^2+16lambda=0Rightarrowlambda(lambda^2-4)(lambda-4)=0$
$lambda_{1}=0,lambda_{2}=2,lambda_{3}=-2,lambda_{4}=4.$
$则其对应的通解为ar{y}=C_{1}+C_{2}e^{2t}+C_{3}e^{-2t}+C_{4}e^{4t}$
$由上文中x=e^t做代换得，ar{y}=C_{1}+C_{2}x^{2}+C_{3}x^{-2}+C_{4}x^{4}$
$又因为p=x,f(p)=y.所以,f(p)=C_{1}+C_{2}p^{2}+C_{3}p^{-2}+C_{4}p^{4}$