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求解：(x^2+y^2)dy/dx=2xy
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求解：(x^2+y^2)dy/dx=2xy
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答案和解析
2xy-(x^2+y^2)y'=0
设y=xt,则dy=xdt+tdx
于是,代入原方程得2xydx-(x^2+y^2)dy=0
==>2x²tdx-(x²+x²t²)(xdt+tdx)=0
==>2tdx-(1+t²)(xdt+tdx)=0
==>t(1-t²)dx=x(1+t²)dt
==>dx/x=(1+t²)dt/(t(1-t²))
==>dx/x=[1/t+1/(1-t)-1/(1+t)]dt
==>ln│x│=ln│t│-ln│1-t│-ln│1+t│+ln│C│ (C是积分常数)
==>x=Ct/(1-t²)
==>x(1-t²)=Ct
==>x(1-y²/x²)=Cy/x
==>x²-y²=Cy
故原微分方程的通解是x²-y²=Cy (C是积分常数).
设y=xt,则dy=xdt+tdx
于是,代入原方程得2xydx-(x^2+y^2)dy=0
==>2x²tdx-(x²+x²t²)(xdt+tdx)=0
==>2tdx-(1+t²)(xdt+tdx)=0
==>t(1-t²)dx=x(1+t²)dt
==>dx/x=(1+t²)dt/(t(1-t²))
==>dx/x=[1/t+1/(1-t)-1/(1+t)]dt
==>ln│x│=ln│t│-ln│1-t│-ln│1+t│+ln│C│ (C是积分常数)
==>x=Ct/(1-t²)
==>x(1-t²)=Ct
==>x(1-y²/x²)=Cy/x
==>x²-y²=Cy
故原微分方程的通解是x²-y²=Cy (C是积分常数).
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