高数 08.06 微分方程习题课 01

微分方程习题课01
一、考试内容
1、常微分方程的基本概念
2、变量可分离方程
3、齐次微分方程
4、一阶线性微分方程
二、考试要求
1、了解常微分方程及其阶、解、通解、初始条件和特解的概念
2、掌握变量可分离的微分方程、一阶线性微分方程的求解方法
3、会解齐次微分方程
三、基本知识
1、微分方程的概念
2、变量可分离方程
dydx=f(x)⋅g(y)→∫dyg(y)=∫f(x)dx\dfrac{dy}{dx} = f(x) \cdot g(y) \rightarrow \int \dfrac{dy}{g(y)} = \int f(x) dx
3、齐次微分方程
dydx=f(yx)令u=yx,y′=u+xu′\dfrac{dy}{dx} = f(\dfrac{y}{x}) \quad 令 u = \dfrac{y}{x}, y^{\prime} = u + xu^{\prime}
4、一阶线性微分方程
y′+P(x)y=Q(x)通解公式:y=e−∫P(x)dx[C+∫Q(x)e∫P(x)dxdx]y^{\prime} + P(x)y = Q(x) \\ 通解公式:y = e^{-\int P(x) dx}[C + \int Q(x) e^{\int P(x) dx} dx]

或者以x为因变量：x′+P(y)x=Q(y)通解公式:x=e−∫P(y)dy[C+∫Q(y)e∫P(y)dydy]或者以x为因变量：\\ x^{\prime} + P(y)x = Q(y) \\ 通解公式:x = e^{-\int P(y) dy}[C + \int Q(y) e^{\int P(y) dy} dy]

(一)单选题
例1.下列微分方程中,是变量可分离的微分方程式( C )例1.下列微分方程中,是变量可分离的微分方程式(\ \ {\color{blue}{C} }\ \ )
A.xsin(xy)dy+ydy=0B.y′=ln(x+y)C.dydx=xsinyD.y′+1xy=ex⋅y2A. x\sin(xy)dy + ydy = 0 \quad B.y^{\prime} = \ln(x + y) \\ C.\dfrac{dy}{dx} = x\sin{y} \quad D.y^{\prime} + \dfrac{1}{x} y = e^x \cdot y^2

例2.方程(y−x3)dx+xdy=2xydx+x2dy是( C )例2.方程(y - x^3)dx + xdy = 2xydx + x^2dy是(\ \ {\color{blue}{C} }\ \ )
A.变量可分离方程B.齐次方程C.一阶线性方程D.不属于以上三类方程A. 变量可分离方程 \quad B.齐次方程 \\ C.一阶线性方程 \quad D. 不属于以上三类方程
(y−x3)dx+xdy=2xydx+x2dydydx=y−x3−2xyx2−xdydx+2x−1x2−xy=−x3x2−x一阶线性方程(y - x^3)dx + xdy = 2xydx + x^2dy \\ \dfrac{dy}{dx} = \dfrac{y - x^3 - 2xy}{x^2 - x} \\ \dfrac{dy}{dx} + \dfrac{2x - 1}{x^2 - x}y= \dfrac{- x^3}{x^2 - x} \\ 一阶线性方程

例3.微分方程y′=e−x2的通解是( C )例3.微分方程y^{\prime} = e^{-\frac{x}{2}}的通解是(\ \ {\color{blue}{C} }\ \ )
A.y=e−12x+CB.y=e12x+CC.y=−2e−12x+CD.y=Ce−12xA.y = e^{-\frac{1}{2}x} + C \quad B.y = e^{\frac{1}{2}x} + C \\ C.y = -2e^{-\frac{1}{2}x} + C \quad D.y = Ce^{-\frac{1}{2}x}
解:y=∫e−x2dx=−2∫e−x2d(−x2)=−2e−x2+C解:\\ y = \int e^{-\frac{x}{2}} dx = -2\int e^{-\frac{x}{2}} d(-\frac{x}{2}) = -2e^{-\frac{x}{2}} + C

例4.微分方程y′−y=1的通解是( C )例4.微分方程y^{\prime} - y = 1的通解是(\ \ {\color{blue}{C} }\ \ )
A.y=CexB.y=Cex+1C.y=Cex−1D.y=(C+1)exA. y = Ce^x \quad B. y = Ce^x + 1 \\ C. y = Ce^x - 1 \quad D. y = (C + 1)e^x
P(x)=−1,Q(x)=1y=e−∫P(x)dx[C+∫Q(x)e∫P(x)dxdx]=e−∫−1dx[C+∫1e∫−1dxdx]=ex[C−e−x]=Cex−1 P(x) = -1, Q(x) = 1\\ y = e^{-\int P(x) dx} [C + \int Q(x) e^{\int P(x) dx} dx] \\ = e^{-\int -1 dx} [C + \int 1 e^{\int -1 dx} dx] \\ = e^{x} [C -e^{-x}] \\ = Ce^{x} - 1 \\

例5.微分方程{xy′+y=3y|x=1=0的特解是( A )例5.微分方程 \left \{ \begin{array}{l} xy^{\prime} + y = 3 \\ y|_{x = 1} = 0 \end{array} \right. 的特解是(\ \ {\color{blue}{A} }\ \ )
A.y=3(1−1x)B.y=3(1−x)C.y=1−1xD.y=1−xA. y = 3(1 - \dfrac{1}{x}) \quad B. y = 3(1 - x) \\ C. y = 1 - \dfrac{1}{x} \quad D.y = 1 -x
解：xy′+6=3→y′+1xy=3xP(x)=1x,Q(x)=3x,y=e−∫P(x)dx[C+∫Q(x)e∫P(x)dxdx]=e−∫1xdx[C+∫3xe∫1xdxdx]=1x[C+∫3xxdx]=1x[C+3x]∵y|x=1=0,11[C+3⋅1]=0,C=−3y==1x[−3+3x]y=3(1−1x)解：\\ xy^{\prime} + 6 = 3 \rightarrow y^{\prime} + \dfrac{1}{x}y = \dfrac{3}{x} \\ P(x) = \dfrac{1}{x}, Q(x) = \dfrac{3}{x}, \\ y = e^{-\int P(x) dx} [C + \int Q(x) e^{\int P(x) dx} dx] \\ = e^{- \int \frac{1}{x} dx} [C + \int \frac{3}{x} e^{\int \frac{1}{x} dx} dx] \\ = \dfrac{1}{x} [C + \int \frac{3}{x} x dx] \\ = \dfrac{1}{x} [C + 3x] \\ \because y|_{x = 1} = 0, \dfrac{1}{1}[C + 3 \cdot 1] = 0, C = -3 \\ y = = \dfrac{1}{x} [-3 + 3x] \\ y = 3(1 - \dfrac{1}{x})

(二)填空题
例6.微分方程d2xdy2+xy=0,的自变量为 y −−−,未知函数为 x −−−,方程阶数为 2 −−−例6.微分方程 \dfrac{d^2 x }{d y^2} + xy = 0, 的自变量为\underline{\ \ y\ \ }, \\ 未知函数为\underline{\ \ x \ \ }, 方程阶数为\underline{\ \ 2 \ \ }

例7.微分方程t(x′)2−2tx′+t=0的自变量为 t −−−,未知函数为 x −−−,方程的阶数为 1 −−−例7.微分方程t(x^{\prime})^2 - 2tx^{\prime} + t = 0的自变量为\underline{\ \ t \ \ }, \\ 未知函数为\underline{\ \ x \ \ },方程的阶数为\underline{\ \ 1 \ \ }

例8.微分方程xyy′′+x(y′)3−y4y′=0的阶数为 2 −−−例8.微分方程xyy^{\prime \prime} + x(y^{\prime})^3 - y^4y^{\prime} = 0的阶数为\underline{\ \ 2 \ \ }

例9.微分方程dydx=e2x−y为一阶变量可分离 −−−−−−−−−方程例9.微分方程\dfrac{dy}{dx} = e^{2x - y}为一阶\underline{\ \ 变量可分离 \ \ }方程
原式可变为:eydy=e2xdx原式可变为: e^ydy = e^{2x} dx

例10.微分方程(y2−6x)y′+2y=0为一阶一阶线性方程 −−−−−−−−−−−方程例10.微分方程(y^2 - 6x)y^{\prime} + 2y = 0为一阶\underline{\ \ 一阶线性方程 \ \ }方程
解:(y2−6x)y′+2y=0dydx=2y6x−y2dxdy=6x−y22ydxdy−3yx=−12yP(y)=−3y,Q(x)=−1yy是以y为自变量的一阶线性方程解:\\ (y^2 - 6x)y^{\prime} + 2y = 0 \\ \dfrac{dy}{dx} = \dfrac{2y}{6x - y^2} \\ \dfrac{dx}{dy} = \dfrac{6x - y^2}{2y} \\ \dfrac{dx}{dy} - \dfrac{3}{y} x = -\dfrac{1}{2}y \\ P(y) = -\dfrac{3}{y}, Q(x) = -\dfrac{1}{y}y \\ 是以y为自变量的一阶线性方程

例11.微分方程xy′−y−y2−x2−−−−−−√=0为一阶齐次 −−−−−方程例11.微分方程 xy^{\prime} - y - \sqrt{y^2 - x^2} = 0为一阶\underline{\ \ 齐次 \ \ }方程
解:xy′−y−y2−x2−−−−−−√=0y′=yx+(yx)2−1−−−−−−−√(x>0)y′=yx−(yx)2−1−−−−−−−√(x<0)为一阶齐次方程解:\\ xy^{\prime} - y - \sqrt{y^2 - x^2} = 0 \\ y^{\prime} = \dfrac{y}{x} + \sqrt{(\dfrac{y}{x})^2 - 1} \quad (x > 0) \\ y^{\prime} = \dfrac{y}{x} - \sqrt{(\dfrac{y}{x})^2 - 1} \quad (x

例12.微分方程y′−2y=0的通解是 y=Ce2x −−−−−−−−−例12.微分方程y^{\prime} - 2y = 0的通解是\underline{\ \ y = Ce^{2x} \ \ }
解:dyy=2dx∫dyy=∫2dxlny=2x+C1y=e2x+C1=eC1⋅e2x=Ce2x解:\\ \dfrac{dy}{y} = 2dx \\ \int \dfrac{dy}{y} = \int 2dx \\ \ln y = 2x + C_1 \\ y = e^{2x + C_1} \\ = e^{C_1} \cdot e^{2x} \\ = Ce^{2x}

例13.微分方程ylnxdx=xlnydy满足y|x=1=1的特解为 lny=±lnx −−−−−−−−−−−−例13.微分方程y \ln {x} dx = x \ln {y} dy 满足 \\ y|_{x = 1} = 1的特解为\underline{\ \ \ln {y} = \pm \ln {x} \ \ }
解:ylnxdx=xlnydylnydyy=lnxdxx∫lnydyy=∫lnxdxx∫lnydlny=∫lnxdlnx12(lny)2=12(lnx)2+C∵y|x=1=1C=0lny=±lnx解:\\ y \ln {x} dx = x \ln {y} dy \\ \dfrac{\ln{y} dy}{y} = \dfrac{\ln{x} dx }{x} \\ \int \dfrac{\ln{y} dy}{y} = \int \dfrac{\ln{x} dx }{x} \\ \int \ln{y} d\ln{y} = \int \ln{x} d\ln{x} \\ \dfrac{1}{2}(\ln{y})^2 = \dfrac{1}{2}(\ln{x})^2 + C \\ \because y|_{x = 1} = 1 \\ C = 0 \\ \ln {y} = \pm \ln {x}

例14.微分方程y′sinx=ylny,满足y|x=π2=e的特解为 y=etanx2 −−−−−−−−−−例14.微分方程y^{\prime}\sin{x} = y \ln y,\\ 满足y|_{x = \frac{\pi}{2}} = e的特解为\underline{\ \ y = e^{\tan{\dfrac{x}{2}}} \ \ }
解:y′sinx=ylnydyylny=dxsinx∫dyylny=∫dxsinx∫dlnylny=∫dxsinxln(lny)=∫cscxdxln(lny)=ln(tanx2)+C又由y|x=π2=e，有0=0+C,C=0ln(lny)=ln(tanx2)lny=tanx2y=etanx2解:\\ y^{\prime}\sin{x} = y \ln y \\ \dfrac{dy}{y \ln y} = \dfrac{dx}{\sin{x}} \\ \int \dfrac{dy}{y \ln y} = \int \dfrac{dx}{\sin{x}} \\ \int \dfrac{d \ln y}{\ln y} = \int \dfrac{dx}{\sin{x}} \\ \ln (\ln y) = \int csc{x} dx \\ \ln (\ln y) = \ln (\tan{\dfrac{x}{2}}) + C \\ 又由y|_{x = \frac{\pi}{2}} = e，有 0 = 0 + C, C = 0 \\ \ln (\ln y) = \ln (\tan{\dfrac{x}{2}}) \\ \ln y = \tan{\dfrac{x}{2}} \\ y = e^{\tan{\dfrac{x}{2}}}

(三)解答题
例15.求微分方程dydx=−yx的通解.例15. 求微分方程\dfrac{dy}{dx} = -\dfrac{y}{x}的通解.
解:dyy=−dxx∫dyy=∫−dxxlny=−lnx+C1y=e−lnx+C1)=eC1⋅e−lnx=C1x(C为任意常数)解:\\ \dfrac{dy}{y} = -\dfrac{dx}{x} \\ \int \dfrac{dy}{y} = \int -\dfrac{dx}{x} \\ \ln y = - \ln x + C_1 \\ y = e^{-\ln x} + C_1) \\ = e^ {C_1} \cdot e^ {-\ln x} \\ = C\dfrac{1}{x} (C为任意常数)

例16.求微分方程dydx=e2x−y的通解,并求满足初始条件y(0)=0的特解例16.求微分方程\dfrac{dy}{dx} = e^{2x - y}的通解,\\ 并求满足初始条件y(0) = 0的特解
解:dydx=e2x−y变形eydy=e2xdx∫eydy=∫e2xdx∫dey=12∫de2xey=12e2x+C通解:y=ln(12e2x+C)把y(0)=0代入,1=12+C,C=12特解:y=lne2x+12解:\\ \dfrac{dy}{dx} = e^{2x - y} 变形\\ e^y dy = e^{2x} dx \\ \int e^y dy = \int e^{2x} dx \\ \int de^y = \dfrac{1}{2} \int de^{2x} \\ e^y = \dfrac{1}{2}e^{2x} + C \\ 通解:y = \ln (\dfrac{1}{2}e^{2x} + C) 把y(0) = 0代入, 1 = \dfrac{1}{2} + C, C = \dfrac{1}{2} \\ 特解:y = \ln \dfrac{e^{2x} + 1}{2}

例17.求微分方程(1+y2)xdx+(1+x2)ydy=0的通解例17.求微分方程(1 + y^2)xdx + (1 + x^2)ydy = 0的通解
解:(1+y2)xdx+(1+x2)ydy=0dydx=−(1+y2)x(1+x2)ydy2dx2=−1+y21+x2d(y2+1)d(x2+1)=−1+y21+x2d(y2+1)y2+1=−d(x2+1)x2+1∫d(y2+1)y2+1=∫−d(x2+1)x2+1ln(y2+1)=−ln(x2+1)+C1(x2+1)(y2+1)=eC1(x2+1)(y2+1)=C解:\\ (1 + y^2)xdx + (1 + x^2)ydy = 0 \\ \dfrac{dy}{dx} = -\dfrac{(1 + y^2)x}{(1 + x^2)y} \\ \dfrac{dy^2}{dx^2} = -\dfrac{1 + y^2}{1 + x^2} \\ \dfrac{d(y^2 + 1)}{d(x^2 + 1)} = -\dfrac{1 + y^2}{1 + x^2} \\ \dfrac{d(y^2 + 1)}{y^2 + 1} = -\dfrac{d(x^2 + 1)}{x^2 + 1} \\ \int \dfrac{d(y^2 + 1)}{y^2 + 1} = \int -\dfrac{d(x^2 + 1)}{x^2 + 1} \\ \ln (y^2 + 1) = -\ln (x^2 + 1) + C_1 \\ (x^2 + 1)(y^2 + 1) = e^C_1 \\ (x^2 + 1)(y^2 + 1) = C

例18.求微分方程(xy2+x)dx+(y−x2y)dy=0的通解例18.求微分方程(xy^2 + x)dx + (y - x^2y)dy = 0的通解
解:(xy2+x)dx+(y−x2y)dy=0变形ydyy2+1=xdxx2−1∫ydyy2+1=∫xdxx2−112∫dy2y2+1=12∫dx2x2−1∫d(y2+1)y2+1=∫d(x2−1)x2−1ln(y2+1)=ln(x2−1)+C1lny2+1x2−1=C1y2+1x2−1=C(C>0)注意:x≠±1解:\\ (xy^2 + x)dx + (y - x^2y)dy = 0 变形\\ \dfrac{ydy}{y^2 + 1} = \dfrac{xdx}{x^2 - 1} \\ \int \dfrac{ydy}{y^2 + 1} = \int \dfrac{xdx}{x^2 - 1} \\ \dfrac{1}{2}\int \dfrac{dy^2}{y^2 + 1} = \dfrac{1}{2}\int \dfrac{dx^2}{x^2 - 1} \\ \int \dfrac{d(y^2 + 1)}{y^2 + 1} = \int \dfrac{d(x^2 - 1)}{x^2 - 1} \\ \ln (y^2 + 1) = \ln (x^2 - 1) + C_1 \\ \ln {\dfrac{y^2 + 1}{x^2 - 1}} = C_1 \\ \dfrac{y^2 + 1}{x^2 - 1} = C ( C > 0) \\ 注意:x \neq \pm 1

例19.求微分方程(1+ex)yy′=ex满足y(1)=1的特解例19.求微分方程(1 + e^x)yy^{\prime} = e^x满足y(1) = 1的特解
解:方程变形得:ydy=ex1+exdx∫ydy=∫ex1+exdx∫d(12y2)=∫11+exd(1+ex)12y2=ln(1+ex)+C1y2=2ln(1+ex)+C代入y(1)=1,12=2ln(1+e1)+C,C=1−2ln(1+e)特解:y2=2ln(1+ex)+1−2ln(1+e)解:\\ 方程变形得:\\ ydy = \dfrac{e^x}{1 + e^x} dx \\ \int ydy = \int \dfrac{e^x}{1 + e^x} dx \\ \int d(\dfrac{1}{2}y^2) = \int \dfrac{1}{1 + e^x} d(1 + e^x) \\ \dfrac{1}{2}y^2 = \ln (1 + e^x) + C_1 \\ y^2 = 2\ln (1 + e^x) + C \\ 代入y(1) = 1, 1^2 = 2 \ln (1 + e^1) + C, C = 1 - 2 \ln (1 + e) \\ 特解: y^2 = 2\ln(1 + e^x) + 1 - 2\ln (1 + e)

例20.求微分方程dydx=y2xy−x2的通解例20.求微分方程\dfrac{dy}{dx} = \dfrac{y^2}{xy - x^2}的通解
解:方程变形:dxdy=xy−x2y2dxdy=xy−(xy)2令u=xy,u+u′y=x′=u−u2dyy=−duu2∫dyy=∫−duu2lny=1u−lnC1lnC1y=yxy=Ceyx解:\\ 方程变形:\\ \dfrac{dx}{dy} = \dfrac{xy - x^2}{y^2} \\ \dfrac{dx}{dy} = \dfrac{x}{y} - (\dfrac{x}{y})^2 \\ 令u = \dfrac{x}{y}, \\ u + u^{\prime} y = x^{\prime} = u - u^2 \\ \dfrac{dy}{y} = -\dfrac{du}{u^2} \\ \int \dfrac{dy}{y} = \int -\dfrac{du}{u^2} \\ \ln y = \dfrac{1}{u} - \ln C_1 \\ \ln C_1y = \dfrac{y}{x} \\ y = Ce^{\frac{y}{x}}

例21.求微分方程y′=yx+tanyx的通解,并求满足初始条件y(1)=πy的特解例21.求微分方程y^{\prime} = \dfrac{y}{x} + \tan{y}{x}的通解,并求满足初始条件y(1) = \dfrac{\pi}{y}的特解
解:设u=yx,u+u′x=y′=u+tanudutanu=dxx∫dutanu=∫dxxlnsinu=lnx+lnCsinu=Cx通解：sinyx=Cx把y(1)=π6代入,C=sinπ6=12特解:x=2sinyx解:\\ 设 u = \dfrac{y}{x}, u + u^{\prime}x = y^{\prime} = u + \tan{u} \\ \dfrac{du}{\tan{u}} = \dfrac{dx}{x} \\ \int \dfrac{du}{\tan{u}} = \int \dfrac{dx}{x} \\ \ln \sin{u} = \ln {x} + \ln C \\ \sin u = Cx \\ 通解：\sin{\dfrac{y}{x}} = Cx \\ 把y(1) = \dfrac{\pi}{6} 代入, C = \sin{\dfrac{\pi}{6}} = \dfrac{1}{2}\\ 特解:x = 2\sin{\dfrac{y}{x}}

例22.求微分方程(1+e−xy)ydx+(y−x)dy=0的通解例22.求微分方程(1 + e^{-\frac{x}{y}})ydx + (y - x)dy = 0的通解
解：原方程变形得:dxdy=xy−11+e−xy令u=xy,u+u′y=x′=u−11+e−uu′y=u−1−u−ue−u1+e−u=−1−ue−u1+e−u=−eu−ueu+1dueu+ueu+1=−dyy(eu+1)dueu+u=−dyy∫(eu+1)dueu+u=∫−dyyln(eu+u)=−lny+lnClny(eu+u)=lnCy(eu+u)=Cyexy+x=C解：原方程变形得:\\ \dfrac{dx}{dy} = \dfrac{\frac{x}{y} - 1}{1 + e^{-\frac{x}{y}}} \\ 令 u = \dfrac{x}{y}, \\ u + u^{\prime}y = x^{\prime} = \dfrac{u - 1}{1 + e^{-u}} \\ u^{\prime}y = \dfrac{u - 1 -u - ue^{-u}}{1 + e^{-u}} \\ = \dfrac{-1 - ue^{-u}}{1 + e^{-u}} \\ = \dfrac{-e^u - u}{e^u + 1} \\ \dfrac{du}{\dfrac{e^u + u}{e^u + 1}} = -\dfrac{dy}{y} \\ \dfrac{(e^u + 1)du}{e^u + u} = -\dfrac{dy}{y} \\ \int \dfrac{(e^u + 1)du}{e^u + u} = \int -\dfrac{dy}{y} \\ \ln (e^u + u) = -\ln {y} + \ln C \\ \ln y(e^u + u) = \ln C \\ y(e^u + u) = C \\ ye^{\frac{x}{y}} + x= C

例23.求微分方程dydx+2y=4x的通解例23.求微分方程\dfrac{dy}{dx} + 2y = 4x的通解
解：使用一阶线性方程通解公式:P(x)=2,Q(x)=4xy=e−∫P(x)dx[C+∫Q(x)e∫P(x)dxdx]=e−∫2dx[C+∫4xe∫2dxdx]=e−2x[C+∫4xe2xdx]=e−2x[C+∫(2x)e2xd(2x)]=e−2x[C+∫(2x)de2x]=e−2x[C+2xe2x−∫e2xd(2x)]=e−2x[C+2xe2x−e2x]=Ce−2x+2x−1(C为任意常数)解：使用一阶线性方程通解公式:\\ P(x) = 2, Q(x) = 4x\\ y = e^{-\int P(x) dx} [C + \int Q(x) e^{ \int P(x) dx} dx ] \\ = e^{-\int 2 dx} [C + \int 4x e^{ \int 2 dx} dx ] \\ = e^{-2x} [C + \int 4x e^{2x} dx ] \\ = e^{-2x} [C + \int (2x) e^{2x} d(2x) ] \\ = e^{-2x} [C + \int (2x) de^{2x} ] \\ = e^{-2x} [C + 2x e^{2x} - \int e^{2x} d(2x)] \\ = e^{-2x} [C + 2x e^{2x} - e^{2x}] \\ = Ce^{-2x} + 2x - 1 (C为任意常数)

例24.求微分方程dydx+y=e−x的通解例24.求微分方程\dfrac{dy}{dx} + y = e^{-x}的通解
解:使用通解公式，P(x)=1,Q(x)=e−x,y=e−∫P(x)dx[C+∫Q(x)e∫P(x)dxdx]=e−∫1dx[C+∫e−xe∫1dxdx]=e−x[C+∫e−xexdx]=e−x[C+∫1dx]=e−x(C+x)(C为任意常数)解:\\ 使用通解公式，P(x) = 1, Q(x) = e^{-x},\\ y = e^{-\int P(x) dx} [C + \int Q(x) e^{\int P(x) dx} dx] \\ = e^{-\int 1 dx} [C + \int e^{-x} e^{\int 1 dx} dx] \\ = e^{-x} [C + \int e^{-x} e^{x} dx] \\ = e^{-x} [C + \int 1 dx] \\ = e^{-x} (C + x) (C为任意常数)

例25.求微分方程x2dy+(2xy−x+1)dx=0满足y(1)=0的特解例25.求微分方程x^2 dy + (2xy - x + 1)dx = 0满足y(1) = 0的特解
解:x2dy=−(2xy−x+1)dxdydx+2xy=x−1x2P(x)=2x,Q(x)=x−1x2,y=e−∫P(x)dx[C+∫Q(x)e∫P(x)dxdx]=e−∫2xdx[C+∫x−1x2e∫2xdxdx]=1x2[C+∫x−1x2elnx2dx]=1x2[C+∫x−1x2x2dx]=1x2[C+∫(x−1)dx]=1x2[C+12x2−x]=Cx2−1x+12将y(1)=0代入，得C=12y=12x2−1x+12解:\\ x^2 dy = - (2xy - x + 1)dx \\ \dfrac{dy}{dx} +\dfrac{2}{x}y = \dfrac{x - 1}{x^2} \\ P(x) = \dfrac{2}{x}, Q(x) = \dfrac{x - 1}{x^2}, \\ y = e^{-\int P(x) dx} [C + \int Q(x) e^{\int P(x) dx} dx] \\ = e^{-\int \dfrac{2}{x} dx} [C + \int \dfrac{x - 1}{x^2} e^{\int \dfrac{2}{x} dx} dx] \\ = \dfrac{1}{x^2} [C + \int \dfrac{x - 1}{x^2} e^{\ln x^2} dx] \\ = \dfrac{1}{x^2} [C + \int \dfrac{x - 1}{x^2} x^2 dx] \\ = \dfrac{1}{x^2} [C + \int (x - 1) dx] \\ = \dfrac{1}{x^2} [C + \dfrac{1}{2}x^2 - x] \\ = \dfrac{C}{x^2} - \dfrac{1}{x} + \dfrac{1}{2} \\ 将y(1) = 0代入，得C = \dfrac{1}{2} \\ y = \dfrac{1}{2x^2} - \dfrac{1}{x} + \dfrac{1}{2}

例26.求微分方程(y2−6x)y′+2y=0的通解例26.求微分方程(y^2 - 6x)y^{\prime} + 2y = 0的通解
解:1y′=6x−y22ydxdy=6x−y22ydxdy−3yx=−y2原方程变形为关于y的一阶线性方程:P(y)=−3y,Q(y)=−y2;x=e−∫P(y)dy[C+∫Q(y)e∫P(y)dydy]=e−∫−3ydy[C+∫−y2e∫−3ydydy]=e−∫−3ydy[C+∫−y2e∫−3ydydy]=elny3[C+∫−y2e−lny3dy]=y3[C+∫−y21y3dy]=y3[C+∫−12y2dy]=y3[C+1y]=Cy3+y2解:\\ \dfrac{1}{y^{\prime}} = \dfrac{6x - y^2}{2y} \\ \dfrac{dx}{dy} = \dfrac{6x - y^2}{2y} \\ \dfrac{dx}{dy} -\dfrac{3}{y} x = -\dfrac{y}{2} \\ 原方程变形为关于y的一阶线性方程:\\ P(y) = -\dfrac{3}{y}, Q(y) = -\dfrac{y}{2};\\ x = e^{-\int P(y) dy} [C + \int Q(y) e^{\int P(y) dy} dy] \\ = e^{-\int -\dfrac{3}{y} dy} [C + \int -\dfrac{y}{2} e^{\int -\dfrac{3}{y} dy} dy] \\ = e^{-\int -\dfrac{3}{y} dy} [C + \int -\dfrac{y}{2} e^{\int -\dfrac{3}{y} dy} dy] \\ = e^{\ln y^3} [C + \int -\dfrac{y}{2} e^{-\ln y^3} dy] \\ = y^3 [C + \int -\dfrac{y}{2} \dfrac{1}{y^3} dy] \\ = y^3 [C + \int -\dfrac{1}{2y^2} dy] \\ = y^3 [C + \dfrac{1}{y}] \\ = Cy^3 + y^2

例27.求微分方程y′=yy−x的通解例27.求微分方程y^{\prime} = \dfrac{y}{y -x}的通解
解：原方程变型得:dxdy=y−xydxdy+1yx=1P(y)=1y,Q(y)=1,x=e−∫P(y)dy[C+∫Q(y)e∫P(y)dydy]=e−∫1ydy[C+∫1e∫1ydydy]=eln1y[C+∫1elnydy]=1y[C+∫ydy]=1y[C+12y2]=Cy+y22xy−y2−C=0(C为任意常数)解：原方程变型得:\\ \dfrac{dx}{dy} = \dfrac{y - x}{y} \\ \dfrac{dx}{dy} + \dfrac{1}{y} x= 1 \\ P(y) = \dfrac{1}{y}, Q(y) = 1, \\ x = e^{-\int P(y) dy} [C + \int Q(y) e^{\int P(y) dy} dy] \\ = e^{-\int \dfrac{1}{y} dy} [C + \int 1 e^{\int \dfrac{1}{y} dy} dy] \\ = e^{\ln {\frac{1}{y}}} [C + \int 1 e^{\ln y} dy] \\ = \dfrac{1}{y} [C + \int y dy] \\ = \dfrac{1}{y} [C + \dfrac{1}{2}y^2] \\ = \dfrac{C}{y} + \dfrac{y}{2} \\ 2xy - y^2 - C = 0 (C为任意常数)

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