微分方程dy/dx=xy/y^2-x^2 ,当x=0,y=1的特解
问题描述：
微分方程dy/dx=xy/y^2-x^2 ,当x=0,y=1的特解
答
dy/dx=xy/(y^2-x^2)y^2dy-x^2dy=xydx(y^2-x^2)dy=xydxx=yu dx=ydu+udy(y^2-y^2u^2)dy=y^2u*(ydu+udy)(1-u^2)dy=uydu+u^2dy(1-2u^2)dy=uydudy/y=udu/(1-2u^2)lny=(-1/4)ln|1-2u^2|+lnCy=C/(1-2u^2)^(1/4)y=C/[1-2(x/y...谢谢 不过答案是y^4-2y^2x^2=1 我也不知道是那里不对……