一道微积分 (y^2(1-x^2))^(1/2)dy=arcsinxdx,y(0)=1

显示全部楼层 · 2011-9-8 19:14:34

求解微分方程：y2[√(1-x2)]dy=(arcsinx)dx,，y(0)=1解：分离变量：y2dy=[(arcinx)/√(1-x2)]dx...........(1)积分之，得y3/3=∫[(arcinx)/√(1-x2)]dx令arcsinx=u，则x=sinu，dx=cosudu，故∫[(arcinx)/√(1-x2)]dx=∫ucosudu/√(1-sin2u)=∫udu=u2/2=(arcsinx)2/2，代入(1)式得y3/3=(arcsinx)2/2+C/3即得通解：y3=(3/2)(a...

千问 · 2011-9-8 19:14:34

(y^2(1-x^2))^(1/2)dy=arcsinxdx
分离变量=>ydy = arcsinx *(1-x^2)^(-1/2) dx=> ∫ ydy = ∫ arcsinx *(1-x^2)^(-1/2) dx
积分=> y^2 /2 = (1/2) (arcsinx)^2 + C1=> y^2 =(a...