(xdy-ydx)/x^2=(1+(y/x)^2)xdx
d(y/x)=(1+(y/x)^2)xdx
d(y/x)/(1+(y/x)^2)=xdx
两边积分：arctan(y/x)=x^2/2+C
y/x=tan(x^2/2+C)
y=xtan(x^2/2+C)
常微分方程 xdy-ydx=(x^2+y^2)xdx的通解 希望有过程 谢谢
(xdy-ydx)/x^2=(1+(y/x)^2)xdx
d(y/x)=(1+(y/x)^2)xdx
d(y/x)/(1+(y/x)^2)=xdx
两边积分：arctan(y/x)=x^2/2+C
y/x=tan(x^2/2+C)
y=xtan(x^2/2+C)