Question

解微分方程 (y²-2xy)dx+x²dy=0

Answer 1

令：v=y/x,y=xv,dy=vdx+xdv

dy/dx = -(y^2-2xy)/x^2

(vdx+xdv)/dx = 2v - v^2

v+xdv/dx = 2v - v^2

xdv/dx = v - v^2

dv/[v(1 - v)] = dx/x

∫dv/[v(1 - v)] = ∫dx/x

∫dv/v + ∫dv/(1 - v)] = ∫dx/x

lnv - ln(1-v) = lnx

ln(v/(1-v))=lnx+lnc

v/(1-v)=cx

(y/x)/(1-(y/x))=cx

y/(x-y)=cx

y=cx(x-y)