求微分方程dy/dx=2x√1-y^2满足初始条件y(0)=1的特解

Question

求微分方程dy/dx=2x√1-y^2满足初始条件y(0)=1的特解最后我算得arcsiny=x^2，后面怎么写？

Answer 1

dy/dx = 2x√(1-y^2)
dy/√(1-y^2) = 2xdx
arcsiny = x^2 + C
y(0) = 1 代入， 得 C = π/2
特解 arcsiny = x^2 + π/2本回答被提问者采纳