第3个回答 2019-03-02
设 y' = dy/dx = p(y),
则 y'' = dp(y)/dx = [dp(y)/dy](dy/dx) = p(y)dp(y)/dy
微分方程 yy'' - (y')^2 = yy' 化为
ypdp/dy - p^2 = yp
p(ydp/dy-y-p) = 0
(1) ydp/dy-y-p = 0, 即 dp/dy - p/y = 1
p = e^(∫dy/y)[∫1e^(-∫dy/y)dy + C1] = y[∫dy/y + C1] = y(lny+C1)
即 dy/dx = y(lny+C1), dy/[y(lny+C1)] = dx,
d(lny+C1)/(lny+C1) = dx , ln(lny+C1) = x + lnC2
lny+C1 = C2 e^x, 通解是 y = e^(C2e^x - C1)。
(2) p = dy/dx = 0, 通解是 y = C本回答被网友采纳