齊次微分方程今天說(shuō)一下微分方程的幾種特別的思路1.齊次微分何為齊次微分如果一階微分方程可化成\frac{dy}{dx}=\varphi\left(\frac{y}{x}\right)\\的形式,那么就稱這方程為齊次方程,例如\left(xy-y^{2}\right)dx-\left(x^{2}-2xy\right)dy=0\\是齊次方程,因?yàn)樗苫蒤frac{dy}{dx}=\frac{xy-y^{2}}{x^{2}-2xy}\\即\frac{dy}{dx}=\frac{\frac{y}{x}-\left(\frac{y}{x}\right)^{2}}{1-2\left(\frac{y}{x}\right)}\\解題步驟在齊次方程\frac{dy}{dx}=\varphi\left(\frac{y}{x}\right)\\中,引進(jìn)新的未知函數(shù)u=\frac{y}{x}\text{(1)}\\就可把它化為可分離變量的方程.因?yàn)橛?1)有y=ux,\frac{dy}{dx}=u+x\frac{du}{dx}\\便得方程u+x\frac{\mathrmqyssyuqu}{\mathrm{~d}x}=\varphi(u)\Rightarrowx\frac{\mathrmuekmoyau}{\mathrm{~d}x}=\varphi(u)-u\\。分離變量,得\frac{du}{\varphi(u)-u}=\frac{dx}{x}\Rightarrow\int\frac{du}{\varphi(u)-u}=\int\frac{dx}{x}\\。求出積分后,再以\frac{y}{x}代替u,使得所給齊次方程的通解.例題解方程y^{2}+x^{2}\frac{dy}{dx}=xy\frac{dy}{dx}\\解：原方程可寫(xiě)成\frac{\mathrmwyaceicy}{\mathrm{~d}x}=\frac{y^{2}}{xy-x^{2}}=\frac{\left(\frac{y}{x}\right)^{2}}{\frac{y}{x}-1}\\。因此是齊次方程.令\frac{y}{x}=u，則y=xu,\frac{dy}{dx}=u+x\frac{du}{dx}\\于是原方程變?yōu)閡+x\frac{du}{dx}=\frac{u^{2}}{u-1}\Rightarrowx\frac{du}{dx}=\frac{u}{u-1}\\分離變量,得\left(1-\frac{1}{u}\right)du=\frac{dx}{x}\\兩端積分,得u-\ln|u|+C_{1}=\ln|x|\Rightarrow\ln|xu|=u+C_{1}\\以代上式中的u,使得所給方程的通解為\ln|y|=\frac{y}{x}+C_{1}\text{或}y=Ce^{\frac{y}{x}}\left(C=\pme^{c_{1}}\right)\\2.一階線性方程定義方程\frac{dy}{dx}+P(x)y=Q(x)(1)\\叫做一階線性微分方程,因?yàn)樗鼘?duì)于未知函數(shù)y及其導(dǎo)數(shù)是一次方程.如果Q(x)\equiv0那么方程(1)稱為齊次的。如果Q(x)\neq0,那么方程(1)稱為非齊次的。齊次方程的求解為了求出非齊次線性方程(1)的解,我們先把Q(x)換成零而寫(xiě)出方程\frac{dy}{dx}+P(x)y=0(2)\\(2)叫做對(duì)應(yīng)于非齊次線性方程(1)的齊次線性方程.方程(2)是可分離變量的,分離變量后得方程\frac{dy}{y}=-P(x)dx\\兩端積分,得\ln|y|=-\intP(x)dx+C_{1}\\即y=Ce^{-\intP(x)dx}\left(C=\pme^{c_{1}}\right)\\這是對(duì)應(yīng)的齊次線性方程(2)的通解。非齊次方程求解:常數(shù)變易法求非齊次線性方程\frac{dy}{dx}+P(x)y=Q(x)\\的通解.這方法是把(2)的通解中的C換成的未知函數(shù)u(x),即作變換y=u\mathrm{e}^{-\intp(x)dx}(3)\\于是\frac{dy}{dx}=u^{\prime}e^{-\intp(x)dx}-uP(x)e^{-\intP(x)dx}(4)\\將(3)(4)代入(1)可得:u^{\prime}\mathrm{e}^{-\intP(x)dx}-uP(x)\mathrm{e}^{-\intP(x)\mathrm6w02k0ox}\\+P(x)u\mathrm{e}^{-\intP(x)\mathrmq22sum2x}=Q(x)\\。即u^{\prime}e^{-\intp(x)dx}=Q(x),u^{\prime}=Q(x)e^{\intp(x)dx}\\兩端積分,得u=\intQ(x)e^{\intp(x)dx}dx+C\\把上式代人(3)可得：y=e^{-\intP(x)dx}\left(\intQ(x)e^{\intP(x)dx}dx+C\right)\\即y=Ce^{-\intP(x)dx}+e^{-\intP(x)dx}\intQ(x)e^{\intP(x)dx}dx\\例題求方程：\frac{dy}{dx}-\frac{2y}{x+1}=(x+1)^{\frac{5}{2}}\\通解。解：這是一個(gè)非齊次線性方程,先求對(duì)應(yīng)的齊次方程的通解\frac{dy}{dx}-\frac{2}{x+1}y=0;\frac{dy}{y}=\frac{2dx}{x+1}\\\ln|y|=2\ln|x+1|+C_{1}\\\Rightarrowy=C(x+1)^{2}\left(C=\pm\mathrm{e}^{c_{1}}\right)\\。用常數(shù)變易法,把C換成u,即令y=u(x+1)^{2}(1)\\那么\frac{\mathrmgkcwucky}{\mathrm{~d}x}=u^{\prime}(x+1)^{2}+2u(x+1)\\代入所給非齊次方程,得u^{\prime}=(x+1)^{\frac{1}{2}}\\