求齐次方程xy'-y-√(y^2-x^2)=0的通解
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求齐次方程xy'-y-√(y^2-x^2)=0的通解
求齐次方程xy'-y-√(y^2-x^2)=0的通解
求齐次方程xy'-y-√(y^2-x^2)=0的通解
∵xy'-y-√(y-x)=0 ==>y'-y/x-√(y/x-1)=0
∴设y=xt,则y'=xt'+t
代入方程得xt'-√(t-1)=0 ==>dt/√(t-1)=dx/x
==>ln(t+√(t-1))=ln│x│+ln│C│ (C是积分常数)
==>t+√(t-1)=Cx
==>y/x+√(y/x-1)=Cx
==>y+√(y-x)=Cx
故原方程的通解是y+√(y-x)=Cx (C是积分常数).
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