求下列微分方程的解(1) (x+y)dy+(x-y)dx=0 (2)ylnydx+(x-lny)dy=0 (3) y'=(2-x+y)^2 (4)y''=3*y^(1/2)

求下列微分方程的解(1) .(x+y)dy+(x-y)dx=0(x+y)dy=(y-x)dx,故dy/dx=(y-x)/(y+x)=(y/x-1)/(y/x+1).(1)；令y/x=u,即y=ux；因为dy/dx=u+xdu/dx；于是方程(1)变为：u+xdu/dx=(u-1)/(u+1)；也就是xdu/dx=(u-1)/(u+1)-u=-...回答有点小问题，不过思路很对，但重要的是后2个(4)y''=3y^(1/2)设y′=dy/dx=p，则y′′=dp/dx=(dp/dy)(dy/dx)=p(dp/dy)代入原式得p(dp/dy)=3y^(1/2)分离变量得pdp=3[y^(1/2)]dy积分之得(1/2)p²=2y^(3/2)+(1/2)C₁p²=4y^(3/2)+C₁故p=√[4y^(3/2)+C₁]于是得dy/dx=√[4y^(3/2)+C₁]dy/√[4y^(3/2)+C₁]=dx下面的问题就是求解这个积分。(3) y'=(2-x+y)² 令2-x+y=z，两边对x取导数得-1+dy/dx=dz/dx，即有dy/dx=1+dz/dx，代入原式得：1+dz/dx=z²，即有dz/dx=z²-1；故dz/(z²-1)=dx，积分之得∫dz/(z²-1)=(1/2)∫[1/(z-1)-1/(z+1)]dz=(1/2)[ln(z-1)-ln(z+1)]=x+(1/2)C即有ln[(z-1)/(z+1)=2x+C，将z=2-x+y代入即得原方程的通解为：ln[(1-x+y)/(3-x+y)]=2x+C.（3）如何积分(4)y''=3y^(1/2)设y′=dy/dx=p，则y′′=dp/dx=(dp/dy)(dy/dx)=p(dp/dy)代入原式得p(dp/dy)=3y^(1/2)；分离变量得pdp=3[y^(1/2)]dy；积分之得(1/2)p²=2y^(3/2)+2C₁p²=4y^(3/2)+4C₁；故p=2√[y^(3/2)+C₁]于是得dy/dx=2√[y^(3/2)+C₁]；dy/√[y^(3/2)+C₁]=2dx；令y=u²，则dy=2udu，代入上式得：2∫udu/√(u³+C₁)=2X+C₂；(有事离开，待续，请勿中断答题程序)