方程ydx-xdy=(x^2+y^2)dx的通解
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方程ydx-xdy=(x^2+y^2)dx的通解
y dx - x dy = (x² + y²) dx
y - x • dy/dx = x² + y²
y' = y/x - y²/x - x
(令y = - xv,y' = - (xv' + v) = - xv' - v)
- xv' - v = - v - xv² - x
- v' = - v² - 1
dv/(v² + 1) = 1
arctan(v) = x + C
v = tan(x + C)
- y/x = tan(x + C)
y = - xtan(x + C)
y - x • dy/dx = x² + y²
y' = y/x - y²/x - x
(令y = - xv,y' = - (xv' + v) = - xv' - v)
- xv' - v = - v - xv² - x
- v' = - v² - 1
dv/(v² + 1) = 1
arctan(v) = x + C
v = tan(x + C)
- y/x = tan(x + C)
y = - xtan(x + C)
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