Question

求解：微分方程dy/dx=xy/y^2-x^2 ,当x=0，y=1的特解

求详细过程……谢谢……

Answer 1

dy/dx=xy/(y^2-x^2)
y^2dy-x^2dy=xydx
(y^2-x^2)dy=xydx
x=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=uydu
dy/y=udu/(1-2u^2)
lny=(-1/4)ln|1-2u^2|+lnC
y=C/(1-2u^2)^(1/4)
y=C/[1-2(x/y)^2]^(1/4)
x=0,y=1,C=1
特解y=1/(1-2(x/y)^2) ^(1/4)

追问

谢谢 不过答案是y^4-2y^2x^2=1 我也不知道是那里不对……