微分方程dx/dy=(2xy-y^2)/(x^2-2xy)满足y(1)=-2的特解是?
人气:414 ℃ 时间:2019-08-18 05:25:45
解答
dx/dy=(2xy-y²)/(x²-2xy)dy/dx=(x²-2xy)/(2xy-y²) 分子分母同时除以x²dy/dx=(1-1y/x)/[2y/x-(y/x)²]设y/x=uu+xu'=(1-2u)/(2u-u²)xdu/dx=(u³-2u²-2u+1)/(2u-u²)分...
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