yln(x)y^'=((y+1)/x)^2

解答

y ln (x) y^{^{'}} = (\frac{y + 1}{x})^{2}

y = - 1 - \frac{1}{W ( - \frac{1}{e ^{\frac{c _{1} x - 1}{x}}} )}

求解步骤

y ln (x) y^{'} (x) = (\frac{y + 1}{x})^{2}

求解贝努力方程:

y = - 1 - \frac{1}{W ( - \frac{1}{e ^{\frac{c _{1} x - 1}{x}}} )}

y = - 1 - \frac{1}{W ( - \frac{1}{e ^{\frac{c _{1} x - 1}{x}}} )}

d er i v a t i v e \frac{e ^{2 x}}{x}

d er i v a t i v e f (x) = \frac{x ^{2} - 2}{2 x + 1}

\frac{d y}{d x} = x^{8} y^{- 5}

(3 x y^{2} - 3) d x + (3 x^{2} y + 4) d y = 0

\int \frac{( x + 7 ) ^{3}}{2 x} d x