求微分方程(1+y)dx+(x-1)dy=0的通解
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求微分方程(1+y)dx+(x-1)dy=0的通解 (1+y)dx+(x-1)dy=0 ===> dx+ydx+xdy-dy=0 ===> ydx+xdy=dy-dx ===> d(xy)=d(y-x) ===> xy=y-x+C ===> y=[x/(1-x)]+C
答:若按x>0求解: dy/dx=[y-√(x^2+y^2)]/x =y/x-√[1+(y/x)^2] 令u=y/x,则y=xu,y'=u+xu' 代入原方程,得 ...详情>>
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