Question

常微分方程问题

形如这种带有一阶导数平方项的方程怎么解啊？跪求大神指导OTL

Answer 1

y^2y'^2-2xyy'+2y^2-x^2=0
(yy'-x)^2=2x^2-2y^2
yy'-x=√2√(x^2-y^2) 或 yy'-x= -√2√(x^2-y^2)
dy/dx=x/y+√2[√((x/y)^2-1)]
y/x=u dy=udx+xdu
dy/dx=u+xdu/dx
u+xdu/dx=1/u+√2√(1/u)^2-1]
xdu/dx=(1-u^2)/u+√2√(1-u^2)/u
udu/[(1-u^2)+√[2(1-u^2)]]=dx/x
d(1-u^2)/[√(1-u^2)[√(1-u^2)+√2)]] =-2dx/x
t=1-u^2
dt/[√t(√t+√2)]=-2dx/x
ln(√t+√2)=-lnx+C0
[√(1-(y/x)^2)+√2]=C/x