微分方程(y)^3 2xy^(4)=0的阶数

y^4dx=dy-2xy^3dyy^4dx/dy+2xy^3=1y^2dx/dy+2xy=1/y^2d(xy^2)/dy=1/y^2d(xy^2)=dy/y^2两边积分：xy^2=-1/y+Cx=-1

x^2*dy/dx=xy-y^2dy/dx=y/x-y^2/x^2u=y/xy=xuy'=u+xu'代入：u+xu'=u+u^2xu'=u^2du/u^2=dx/x-1/u=lnx+lnCCx=e^(

1.(dy/dx)-2xy=y+4x-2dy/dx=(2x-1)(y+2),dy/(y+2)=(2x-1)dx,ln(y+2)=x^2-x+C,y=e^(x^2-x+C)-2.2.y''+2y(y')

dy/dx=(xy+3x-y-3)/(xy-2x+4y-8)=(x-1)(y+3)/(x+4)(y-2)再问：然后呢？再答：(y-2)dy/(y+3)=(x-1)dx/(x+4)已经是变量分离方程，两

dy/dx=（1+y^2)/(xy)[y/（1+y^2)]dy=dx/x两边积分得1/2[ln（1+y^2)]+c1=ln|x|+c2,c1,c2为任意常数两边都以e为底数得1+y^2=cx^2,c为

dy/dx=xy+x+y+1dy/dx=(x+1)(y+1)分离变量dy/(y+1)=dx*(x+1)两边积分ln(y+1)=(x²/2)+x+lnC两边取以e为底的幂y+1=Ce^[(x&

dhy2603,这题太容易了,xy'-ylny=0①,两边再对x求一次导得到y'+xy''-y'lny-yy'/y=0,即有xy''-y'lny=0②,联立两式得,ylny*y''/y'-y'lny=

令z=1/x,则dx=-x²dz代入原方程得(x²y³+xy)dy=-x²dz==>dz/dy+y/x=-y³==>dz/dy+yz=-y³

∵xy"+y'=0==>xdy'/dx+y'=0==>dy'/y'=-dx/x==>ln│y'│=-ln│x│+ln│C1│(C1是积分常数)==>y'=C1/x∴y=∫C1/xdx=C1ln│x│+

左右除以x^2,y'/x+y(1/x)'=e^(x-1/x).左边就是(y/x)',两边关于x积分就能得到y=x(右边的不定积分+C).不过e^(x-1/x)不定积分没有初等函数表示啊……是不是抄错了

设x=e^t则d^2y/dt^2-5dy/dt+6y=e^ty=C1*e^(3t)+C2*e^(2t)+1/2e^t=C1*x^3+C2*x^2+x/2再问：设x=e^t则d^2y/dt^2-5dy/

伯努利方程xy'+y=2y^3->x/y^3*y'+1/y^2=2令1/y^2=t-x/2*dt/dx+t=2解这个一阶方程得(2x^(-2)+c)*x^2

令z=1/x,则dx=-x²dz代入原方程得(x²y³+xy)dy=-x²dz==>dz/dy+y/x=-y³==>dz/dy+yz=-y³

xdy+ydx-(x^2+3x+2)dx=0设dz(x,y)=xdy+ydx-(x^2+3x+2)dx∂z/∂y=x,z=xy+g(x),∂z/∂x=y

该微分方程只能用级数解法

令y'/x=t原方程化为：y"=(y'/x)(lny'/x)=tlnty'=txy"=t+t'x=tlntdx/x=dt/[tln(t/e)]即lnx=lnln(t/e)+lnlnC1x=ln(t/e

左端是x*y'还是(xy)'再问：x*y'再答：令t=lnx,x=e^tdy/dx=(dy/dt)/(dx/dt)=(dy/dt)/e^tx*dy/dx=e^t*(dy/dt)/e^t=dy/dtdy

答案,X=1Y=0或者X=0Y=1再问：是求微分。不是微分方程。答案是dxdy-3^（xy）•ln3（dx•ydy•x）=0再问：求过程

方程没有出现y,令y'=p,则y''=p'则方程化为xp'+p=xlnx即是p'+(1/x)p=lnx用一次微分方程求解公式：p=(1/2)xlnx-(1/4)x+cy是p再积分一次y=(1/4)x^

xy'+y＝xy^3(xy)'=xy*y^2令xy=u,y=u/x原式化为u'=u*(u/x)^2即du/u^3=dx/x^2两边对x积分得-1/2*1/u^2=-1/x+C1即1/(xy)^2=2/