大一常微分方程问题,求解dy/dx=-y/(a^2-y^2)^0.5;y(0)=a（a已知）

解微分方程的特dy/dx=-y/√(a²-y²)；y(0)=a
分离变量得(1/y)√(a²-y²)dy=-dx
积分之,得∫[(1/y)√(a²-y²)dy]dy=-∫dx=-x.(1)
∫[(1/y)√(a²-y²)dy]dy=∫{√[1-(y/a)²]/(y/a)}dy
令y/a=sinu,则y=asinu,dy=acosudu,故
∫{√[1-(y/a)²]/(y/a)}dy=a∫[(cos²u/sinu)]du=a∫[(1-sin²u)/sinu]du=a[∫du/sinu-∫sinudu]
=a{ln[tan(u/2)+cosu}=aln[sinu/(1+cosu)]+acosu
=aln{y/[a+√(a²-y²)]}+√(a²-y²)+C
代入(1)式得aln{y/[a+√(a²-y²)]}+√(a²-y²)+C=-x
代入初始条件：x=0时y=a,得C=0,
故得特解为aln{y/[a+√(a²-y²)]}+√(a²-y²)+x=0