dx/(x^2-y^2)=dy/(-2xy)如何积分?
问题描述：
dx/(x^2-y^2)=dy/(-2xy)如何积分?
答
dx/(x^2-y^2=dy/(-2xy)
两边积分
∫(1/(x^2-y^2))dx=∫dy/(-2xy)
-1/2y*ln((y+x)/(y-x))=-1/2y*lnx
(y+x)/(y-x)=x
y=(x^2+x)/(x-1)
答
2dx/dy = (y^2-x^2)/(xy) = y/x - x/y设x/y = p那么dx = pdy + ydp => dx/dy = p + y dp/dy所以2(p + y dp/dy) = 1/p - p2dp / (1/p-3p) = dy/y2pdp / (1-3p^2) = dy/y积分得 -1/3 ln(1-3p^2) = lnCy所以(Cy)^3 * (1...