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当前位置:首页 > 商业/管理/HR > 经营企划 > 四川大学常微分方程-(张伟年-著)-高等教育出版社-课后答案
12.11.(2.1.1)g(y)6=0,.:(2.1.1).g(y0)=0,y=y0(2.1.1).g(y0)6=0.y='(x)(a;b)'(x0)=y0(2.1.1)d'(x)dx=h(x)g('(x));8axb:(a;b)g('(x))6=0.˜x02(a;b)g('(˜x0))=0,y=Ã(x)´'(˜x0)()(2.1.1).',Ã(˜x0;'(˜x0))(2.1.1).,'´Ã´'(˜x0).g(y0)=g('(x0))=g('(˜x0))=0:g(y0)6=0.g('(x))6=01g('(x))¢d'(x)dx=h(x):x2(a;b)Zxx0'0(t)g('(t))dt=Zxx0h(t)dt:¿='(t),Z'(x)'(x0)d¿g(¿)d¿=Zxx0h(t)dt;y='(x)Zyy0d¿g(¿)=Zxx0h(t)dt;'(x0)=y0;.,y=Ã(x)y=Ã(x)(x0;y0)(2.1.1).,axb,ZÃ(x)y0d¿g(¿)=Zxx0h(t)dt:x1g(Ã(x))¢dÃ(x)dx=h(x):dÃ(x)dx=h(x)g(Ã(x)):y=Ã(x)(2.1.1).2Zyy0d¿g(¿)=Zxx0h(t)dt+C;y=Ã(x;C)(2.1.1),C.,g(y)=0y,g(y)(2.1.1)Zdyg(y)=Zh(x)dx;.2..(1)dydx=¡xy:(3)dydx=2xy:(4)xy(1+x2)dy=(1+y2)dx:(9)dydx=p1¡y2p1¡x2:(12)dydx=cosx3y2+ey::(1)ydy=¡xdx,,y22=¡x22+C1;x2+y2=C;C=2C1.(3)y6=0,1ydy=2xdx;lnjyj=x2+C1;y=0.y=Cex2;C.(4)ydy1+y2=dxx(1+x2);12ln(1+y2)=lnjxj¡12ln(1+x2)+C1;ln(1+x2)(1+y2)=lnx2+2C1;(1+x2)(1+y2)=Cx2;C=e2C1.3(9)y6=§1,dyp1¡y2=dxp1¡x2;arcsiny=arcsinx+C;C,y=sin(arcsinx+C);C.y=1y=¡1.(12)(3y2+ey)dy=cosxdx;y3+ey=sinx+C;C.12.13.,.(2)dydx=x¡y+1x+y¡3:(8)dydx=x2+y2xy:(14)dydx=sin(x+y+1)::(2)x=»+1,y=´+2,d´d»=»¡´»+´;u=´»,»dud»+u=1¡u1+u;u2+2u¡1=C1»¡2,C1.u=´»=y¡2x¡1;»=x¡1;y2+2xy¡x2¡6y¡2x=C;C.(8)dydx=xy+yx;u=yx,xdudx+u=1u+u;u2=lnx2+C,uyxy2=x2(lnx2+C),C.(14)u=x+y+1,dudx=1+sinu;tanu¡secu=x+C,ux+y+1tan(x+y+1)¡sec(x+y+1)=x+C;C.4..(1)dydx¡2yx+1=(x+1)52:(6)dydx¡2xy=x::2(1)a(x)=2x+1,f(x)=(x+1)52.y=exp(Z2x+1dx)(C+Z(x+1)52exp(¡Z2x+1dx)dx)=C(x+1)2+23(x+1)72;y=C(x+1)2+23(x+1)72,C.(6)a(x)=2x,f(x)=x.y=exp(2Zxdx)(C+Zxexp(¡2Zxdx)dx)=¡12+Cex2;y=¡12+Cex2,C.6..(2)y(1+x2)dy=x(1+y2)dx;y(0)=1.(5)eydydx¡x¡x3=0;y(1)=1.:(2),y1+y2dy=x1+x2dx;1+y2=C(1+x2),C.C=2.y=p1+2x2.(5)u=ey,dudx=x+x3;u=12x2+14x4+C,ey=12x2+14x4+C;C.e=34+C,C=e¡34.ey=12x2+14x4+e¡34:7.Bernoulli(1)dydx=6yx¡xy2:(3)xdydx¡4y=2x2py(x6=0;y0)::(1)y6=0,z=y¡1,dzdx=¡6xz+x;,z=1x6(C+18x8);3x6y¡x88=C;C.,y=0.(3)z=py,dzdx=2xz+x;,z=x2(lnjxj+C),y=x4(lnjxj+C)2,C.11.y1(x),y2(x)dydx+p(x)y=q(x);.y(x),y(x)¡y1(x)y2(x)¡y1(x)=C;C.:Ã(x)=y(x)¡y1(x),Á(x)=y2(x)¡y1(x),dÃdx+p(x)Ã(x)=0;dÁdx+p(x)Á(x)=0:k1,k26=0Ã(x)=k1exp(¡Zxx0p(t)dt);Á(x)=k2exp(¡Zxx0p(t)dt);x02R.y(x)¡y1(x)y2(x)¡y1(x)=Ã(x)Á(x)=k1k2=C;C=k1k2.12.21.,:(2)(cosx+1y)dx+(1y¡xy2)dy=0:(3)(5x4+3xy2¡y3)dx+(3x2y¡3xy2+y2)dy=0:(5)dydx=¡6x+y+2x+8y¡3:(9)3y+ex+(3x+cosy)dydx=0::(2)M(x;y)=cosx+1y,N(x;y)=1y¡xy2,@M@y=¡1y2=@N@x;.x0=0,y0=1,U(x;y)=Zx0(cosx+1y)dx+Zy11ydy=sinx+xy+lnjyj:sinx+xy+lnjyj=C,C.(3)M(x;y)=5x4+3xy2¡y3,N(x;y)=3x2y¡3xy2+y2,@M@y=6xy¡3y2=@N@x;.x0=0,y0=0,U(x;y)=Zx0(5x4+3xy2¡y3)dx+Zy0y2dy=x5+32x2y2¡xy3+y33:x5+32x2y2¡xy3+y33=C,C.(5)(6x+y+2)dx+(x+8y¡3)dy=0;M(x;y)=6x+y+2,N(x;y)=x+8y¡3,@M@y=1=@N@x;.x0=0,y0=0,U(x;y)=Zx0(6x+y+2)dx+Zy0(8y¡3)dy=3x2+xy+2x+4y2¡3y:3x2+xy+2x+4y2¡3y=C,C.2(9)(3y+ex)dx+(3x+cosy)dy=0;M(x;y)=3y+ex,N(x;y)=3x+cosy,@M@y=3=@N@x;.x0=0,y0=0,U(x;y)=Zx0(3y+ex)dx+Zy0cosydy=ex+3xy+siny:ex+3xy+siny=C,C.3.:(1)ydx+(y¡x)dy=0:(3)(x2+y2+y)dx¡xdy=0:(5)2xylnydx+(x2+y2p1+y2)dy=0::(1)M(x;y)=y,N(x;y)=y¡x,E=@M@y¡@N@x=2;.¡EM=¡2yx,y¹(x)=e¡R2ydy=¡1y2:1ydx+y¡xy2dy=0,x0=0,y0=1,U(x;y)=Zx01ydx+Zy11ydy=lnjyj+xy:lnjyj+xy=C,C.,y=0.(3)M(x;y)=x2+y2+y,N(x;y)=¡x,E=@M@y¡@N@x=2(y+1);.(x2+y2+y)dx¡xdy=(x2+y2)(dx+ydx¡xdyx2+y2)=(x2+y2)(dx+d(arctan(xy)))=(x2+y2)d(x+arctan(xy))1x2+y2,x+arctan(xy)=C;C.3(5)M(x;y)=2xylny,N(x;y)=x2+y2p1+y2,E=@M@y¡@N@x=2xlny;.¡EM=¡1yx,y¹(y)=1y.2xlnydx+(x2y+yp1+y2)dy=0,x0=0,y0=1,U(x;y)=Zx02xlnydx+Zy1yp1+y2dy=x2lny+13(1+y2)32¡23p2:x2lny+13(1+y2)32=C,C.5.Bernoulli.:Bernoulli(a(x)y+f(x)y®)dx¡dy=0;z=y1¡®,[(1¡®)a(x)z+(1¡®)f(x)]dx¡dz=0:z,1.2¹0(x)=e¡(1¡®)Ra(x)dx:¹0(x)[(1¡®)a(x)z+(1¡®)f(x)]dx¡¹0(x)dz=0,dz=(1¡®)y¡®dy,¹0(x)y¡®[a(x)y+f(x)y®]dx¡¹0(x)y¡®dy=0,Bernoulli¹=y¡®e(1¡®)RP(x)dx:8.(x2+y)dx+f(x)dy=0¹=x,f(x).:M(x;y)=x2+y,N(x;y)=f(x),¹=x@(xM)@y=@(xN)@x;x=xf0(x)+f(x),f(x)f0(x)=¡f(x)x+1;¹=xf(x)f(x)=Cx+x2;C.12.41..(2)y=(dydx)2¡xdydx+x22:(3)y2(1¡dydx)=(2¡dydx)2:(4)(dydx)3¡4xydydx+8y2=0::(2)p=dydx,:y=p2¡xp+x22:x,p=2pdpdx¡p¡xdpdx+x;(2p¡x)dpdx=(2p¡x);2p¡x6=0,dpdx=1,p=x+C,C,y=x22+Cx+C2.2p¡x=0,y=x24.(3)dydx=p,2¡p=yt,p=1¡t2;y=t+1t:p6=0,:dx=dyp=¡1t2dt:x=¡Z1t2dt=1t+C:,8:x=1t+C;y=t+1t;C.:y=x+1x¡C¡C:,p=0,y=§2.(4)p=dydx,x=p24y+2yp:x,1=(p2y¡2yp2)pdpdy¡p24y2+2;2p3¡4y22yp2dpdy=p3¡4y24y2p;dpdy=p2yp3¡4y2=0:dpdy=p2yp=C1y12,C1.x=C214+2y12C1;y=C(x¡C)2,C=C214.p3¡4y2=0p=(4y2)13,x3=274y;y=427x3.2.,().(1)(dydx)2+y2¡1=0:(2)x(dydx)2¡ydydx+1=0:(6)dydx=¡x+px2+2y::(1)dydx=p,p=cost;y=sint:p6=0,:dx=dyp=dt:x=t¡C.,x=t¡C;y=sint;C.y=sin(x+C).,p=0,y=§1.y=sin(x+C)C-:y¡sin(x+C)=0;¡cos(x+C)=0:Cy=§1,.(2)p=dydx,p6=0.y=xp+1p:x,p=p+xdpdx¡1p2dpdx;3(x¡1p2)dpdx=0;x¡p¡26=0,dpdx=0,p=C,C,y=Cx+1C.x¡p¡2=0,y2=4x.y=Cx+1CC-:y¡Cx¡1C=0;¡x+1C2=0:Cy2=4x,.(6)z=x2+2y,dzdx=2x+2dydx=2pz:z6=0,z=(x+C)2,y=Cx+C22,C.z=0,y=¡x22.y=Cx+C22C-:y¡Cx¡C22=0;¡x¡C=0:Cy=¡x22,.3..(1)d5ydx5¡1xd4ydx4=0:(10)4d4ydx4=d2ydx2::(1)p=d4ydx4,dpdx=px:,p=A1x,A1.d4ydx4=A1x:y=C1x5+C2x3+C3x2+C4x+C5,C1=A15!,C2,C3,C4,C5.(10)p=d2ydx2,4d2pdx2=p;p0dx2d(p0)2=pdp,4(p0)2=p2+A1,A1.2dpdx=§pp2+A1:2dpdx=pp2+A1;,p+pp2+A1=A2ex2;4A2.p+pp2+A1=A2ex2;p=A22ex2¡A12A2e¡x2;d2ydx2=A22ex2¡A12A2e¡x2;y=C1ex2+C2e¡x2+C3x+C4;C1=2A2,C2=¡2A1A2,C3,C4.2dpdx=¡pp2+A1;.7.y='(x)dydx=p(x)siny'(0)=0,'(x)´0,p(x)¡1x1.:.'(x)6´0,x0'(x0)6=0,x00.¯x0=supf
本文标题:四川大学常微分方程-(张伟年-著)-高等教育出版社-课后答案
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