1.y'=x(1-y^2)^(1/2).求y 2.(x^2-9)y'+xy=0,y(5)=1.求y

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解微分方程：1.y'=x√(1-y²)；求y 2.(x^2-9)y'+xy=0,y(5)=1.求y
1.dy/dx=x√(1-y²)
分离变量得 dy/√(1-y²)=xdx,两边分别积分之得 arcsiny=(1/2)x²+C；即y=sin[(1/2)x²+C]为
其通解.
2.(x²-9)y'+xy=0,y(5)=1；
(x²-9)(dy/dx)=-xy,分离变量得：(1/y)dy=-xdx/(x²-9)；即有(1/y)dy=-(1/2)d(x²-9)/(x²-9)
积分之,得lny=-(1/2)ln(x²-9)+lnC,故通解为 y=C/√(x²-9)；用y(5)=1代入得1=C/√(25-9)
故C=4,于是特解为 y=4/√(x²-9).

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