(2x y-4)dx (x y-1)dy=0 求通解

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

解析2xdx+ydx+xdy+3y²dy=0(2x+y)dx+(x+3y²)dy=0(2x+y)dx=-(x+3y²)dydy/dx=(2x+y)/-(x+3y²

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

1.2(Xy+Xy)-3(Xy-xy)-4Xy=2*2xy-0-4xy=4xy-4xy=02.1/2ab-5aC-(3acb)+(3aC-4aC)=1/2ab-5ac-3acb-ac=1/2ab-6a

虽说结果与路径无关,但是怎么知道起点与终点的位置如何?如果透过格林公式的结果是0,用参数方程的结果又是0,那又如何解释呢?那只有起点和终点的位置都一样,重合了.起点无论从曲线哪处开始也好,都绕曲线正向

答：dy/dx=1+x+y^2+xy^2y'=(1+x)(1+y^2)y'/(1+y^2)=1+x(arctany)'=1+x积分得：arctany=x+x²/2+Cy=tan(x+x

dx/dy=(2xy-y²)/(x²-2xy)dy/dx=(x²-2xy)/(2xy-y²)分子分母同时除以x²dy/dx=(1-1y/x)/[2y/

令y/x=u,dy=u+xdu,原方程化为:u+xdu/dx=x/(u^2)+u,即du/dx=1/(u^2)通解为:y=x*[(3x+3c)^(1/3)]

别人一般问一道题,你一下子5道?我给你个提示：1.所有5道题全部可以化成y'=f(y/x)的形式.比如5：:y’=√(1-y^2/x^2)+y/x2.设y/x=uy=xuy'=u+xu',代入：u+x

xx+2xy-yy=-3两式相加即可

是xy-[1/(x^2y)]dx-[1/(xy^2)]dy=0还是[(xy-1)/(x^2y)]dx-[1/(xy^2)]dy=0请表达清楚,无歧义!再问：[(xy-1)/(x^2y)]dx-[1/(

你好!两边对x求导：e^(xy)*(y+xy')-y^2=y'cosy解得y'=(y^2-ye^(xy))/(xe^(xy)-cosy)

dx/dy-3xy=xy^2dx/x=(y^2+3y)dy两边积分得：lnx=y^3/3+3y^2/2+c==>x=exp(y^3/3+3y^2/2+c)=Cexp(y^3/3+3y^2/2)C常数

分离,有dy/(2-y)=2xdx,d(2-y)=-dy,所以-d(2-y)/(2-y)=2xdx,两边积分,有-ln|2-y|=x^2+C>=0,所以ln|2-y|=0,y=1或3,x=0,C=0

∵(1-x^2)dy/dx+xy=1==>(1-x^2)dy+xydx=dx==>dy/(1-x^2)^(1/2)+xydx/(1-x^2)^(3/2)=dx/(1-x^2)^(3/2)(等式两端同除

=(x²y²-4-x²y²-xy+4)/(-2xy)=-xy/(-2xy)=1/2

y=-1/4-(1/4)x+(3/8)tan(6x-c)再问：求过程

[(xy-2)(-xy-2)-4(xy-1)^2]除以(-xy)=[-x²y²+4-4(x²y²-2xy+1)]÷(-xy)=(-x²y²+

dx/dy+xy=-1积分因子：exp(∫ydy)=exp(y²/2)=e^(y²/2)dx/dy•e^(y²/2)+xy•e^(y²/

dy/dx=xy²+3xydy/dx=x(y²+3y)∫1/[y(y+3)]dy=∫xdx(1/3)∫(3+y-y)/[y(y+3)]dy=∫xdx∫[1/y-1/(y+3)]dy