求伯努利方程的通解?
伯努利方程 y' + P(x)y = Q(x)y^a (a ≠ 1) 令 y^(1-a) = z, 则 y = z^[1/(1-a)], y' = [1/(1-a)]z^[a/(1-a)]z' 可将伯努利方程化为一阶线性微分方程,求其通解后, 将 z = y^(1-a) 回代即可。例伯努利方程: dy/dx -y/x = y^3令 1/y^2 = z, 则 y = z^(-1/2), dy/dx = (-1/2) z^(-3/2) dz/dx得 (-1/2) z^(-3/2) dz/dx - z^(-1/2)/x = z^(-3/2)将伯努利方程化为了一阶线性微分方程 z' +2z/x = -2通解为 z = e^(-∫2dx/x) [ ∫-2e^(∫2dx/x)dx + C ]= (1/x^2) [ ∫-2x^2dx + C ] = (1/x^2) [ (-2/3)x^3 + C ]= (1/x^2) [ (-2/3)x^3 + C ] = (-2/3)x + C/x^2即 y^2[(-2/3)x + C/x^2] = 1