(u^2+u)dx+u^2(xdu+udx)=0

解答

(u^{2} + u) d x + u^{2} (x d u + u d x) = 0

x = \frac{e ^{\frac{a r c t a n ( \frac{2 u + 1}{3} ) + c _{1}}{3}}}{2 u ^{2} + u + 1}

求解步骤

(u^{2} + u) d x + u^{2} (x d u + u d x) = 0

求解可分离方程:

x = \frac{e ^{\frac{a r c t a n ( \frac{2 u + 1}{3} ) + c _{1}}{3}}}{2 u ^{2} + u + 1}

x = \frac{e ^{\frac{a r c t a n ( \frac{2 u + 1}{3} ) + c _{1}}{3}}}{2 u ^{2} + u + 1}

x \to 4 lim (e^{x} + 4 x)

\int 3 2 x d x

\frac{\partial}{\partial x} (7 x sin (5 x^{2} y))

a re a x^{2} = 4 y, 3 x - 4 y + 4 = 0

y^{^{'}} + y = e^{- t} sin (t), y (0) = 1