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时间:2020-09-11
《微分方程练习题基础篇答案.pdf》由会员上传分享,免费在线阅读,更多相关内容在教育资源-天天文库。
1、.常微分方程基础练习题答案求下列方程的通解2xdydyxdx21.xy分离变量,yCe,C为任意常数dxy2dyx1x22.xydx1xdy0分离变量dx,yCe,C任意常数yx21dy1x3.xyylny0分离变量dx,yCeylnyx22ydyxdx224.(xyx)dx(xyy)dy0分离变量22,(1y)(1x)C1y1xdy2dudydu1u5.(2xy5)令u2xy5则2,2dx,arctanxC1dxdxdxu222y12dyxydyxydydu1u16.,原方程变为,令u,ux,代入得2dudxdxxydxyxdxdx1ux1xyyy2arctanuu
2、lnxCu回代得通解2arctanlnxC,xxx222dyyyydudx7.xyyxy0方程变形为10,令u,代入得2dxxxx1uxyyyarctanulnxCu回代得通解arctanlnxC,xxxdyydyyyydudxCx1Cx18.xyln,方程变形为ln,令u,,ue,yxedxxdxxxxu(lnu1)xdy2xdx2xdxx29.2xy4x,一阶线性公式法ye(4xedxC)Ce2dx11dyy2dx2dx3xx10.2x,一阶线性公式法ye(2xedxC)xCxdxx2222x4x14311.(x1)y2xy4x,方程变形为y2y2一阶线性公式法y
3、2(xC)x1x11x3'..2dydx3112312.(y6x)2y0,方程变形为xy一阶线性公式法yyCydxdyy2221dy11dz2dy13.y3xyxy,方程变形为23xx伯努利方程,令zy,y代入方程ydxydxdx得32dz1x123xzx一阶线性公式法再将z回代得Cedxy3dy1141dy11114.y(12x)y,方程变形为(12x)伯努利方程,令43dx33ydx3y33dz4dydzzy,3y代入方程得z2x1,一阶线性公式法再将z回代得dxdxdx1xCe2x13y215.y5y6y0,特征方程为r5r60,特征根为r12,r23,通解2x
4、3xyC1eC2e2316.16y24y9y0,特征方程为16r24r90,特征根为r1,2,通解43x4y(C1C2x)e2x17.yy0,特征方程为rr0,特征根为r10,r21,通解yC1C2e218.y4y5y0,特征方程为r4r50,特征根为r12i,r22i,通解2xye(C1cosxC2sinx)3322xx19.(xy)dxxdy0,全微分方程xdx(ydxxdy)0,dd(xy)0,通解xyC334233xy20.(xy)dx(xy)dy0,全微分方程xdx(ydxxdy)ydy0,dd(xy)d0,4242xy通解xyC42'..222221.(x
5、y)dx(2xyy)dy0全微分方程xdx(ydx2xydy)ydy0,3232x2yx2ydd(xy)d0,通解xyC323222.(xcosycosx)yysinxsiny0,全微分方程(xcosydysinydx)(cosxdyysinxdx)0,d(xsiny)d(ycosx)0,通解xsinyycosxC2222123.(3xy)dx(2xyx)dyC,3xdx2xydyydxxdy0,积分因子2,方程变xydxxdy2y2y为3dx2ydy20,d3xdyd0,通解3xyCxxx221xdxydy24.xdxydy(xy)dx,积分因子22,方程变为22d
6、x0,xyxy122122d[ln(xy)]dx0通解ln(xy)xC222222125.(xyy)dxxdy0,(xy)dxydxxdy0,积分因子22,方程变为xyydxxdyxxdx220,dxdarctan0,通解xarctanCxyyy3x(n)13x26.yesinx,可降阶yf(x)型,逐次积分得通解yesinxC1xC292227.y1y,可降阶令p(x)y,原方程化为p1p可分离变量型,得yptan(xC1),积分得通解ylncos(xC1)C228.yyx,可降阶yf(x,y)型,令p(x)y,原方程化为ppx,一阶线性非齐次xx12公式法得ypC
7、1ex1,积分得通解yC1exxC22'..3dpdp329.yyy,可降阶yf(y,y)型,令p(y)y,yp,原方程化为pppdydydp2dp2即p[(1p)]0,p0是方程的一个解,由(1p)0得dydyxCarctanpyC即yptan(yC),通解为yarcsine2C111xx230.y2yy4xe,二阶常系数非齐次f(x)ePm(x)型,1是特征方程210的x*2x重根,对应齐次方程的通解为Y(C1C2x)e,设特解为yx(axb)e,代入xx2*23x方程得(6ax2b)e4xe,得a,b0,故原方程的特解为yxe,原方33x23x
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