【1】最简便的方法,用matlab解.
>> y=dsolve('(2*t-y)+(y-t)*Dy=0')
y =
[ t+(-t^2+exp(C1)^2)^(1/2)]
[ t-(-t^2+exp(C1)^2)^(1/2)]
即：
y=
[ x+(-x^2+exp(C1)^2)^(1/2)]
[ x-(-x^2+exp(C1)^2)^(1/2)]
【2】用scrodinger 的方法
2xdx+ydy-ydx-xdy=0
d(2x^2+y^2-2xy)=0
2x^2+y^2-2xy-c＝0
判别式＝4x^2-4*(2x^2-c)=-4x^2-4c
y1=[2x+(-4x^2-4c)^(1/2)]/2=x+(-x^2-c)^(1/2)
y2=[2x-(-4x^2-4c)^(1/2)]/2=x-(-x^2-c)^(1/2)