x^2+y^2=e^(arctan(y⼀x)),求dy⼀dx

2025-03-01 02:57:04
推荐回答(1个)
回答1:

x^2+y^2-e^(arctan(y/x))=0
2x+2y*y'-(arctan(y/x))'e^(arctan(y/x))=0
2x+2y*y'-1/(1+y^2/x^2)*(y'x-y)/x^2 *e^(arctan(y/x))=0
2x+2y*y'-(y'x-y)/(x^2+y^2)*e^(arctan(y/x))=0
2x+2y*y'-y'x/(x^2+y^2)*e^(arctan(y/x))+y/(x^2+y^2)*e^(arctan(y/x))=0
y'=(2x++y/(x^2+y^2)*e^(arctan(y/x)))/(x/(x^2+y^2)*e^(arctan(y/x))-2y)