(t^2+x^2)dt+(2tx-x)dx=0

解答

(t^{2} + x^{2}) d t + (2 t x - x) d x = 0

x = - \frac{2 t ^{3}}{3 ( 2 t - 1 )} + \frac{c _{1}}{2 t - 1}, x = - - \frac{2 t ^{3}}{3 ( 2 t - 1 )} + \frac{c _{1}}{2 t - 1}

求解步骤

(t^{2} + x^{2}) d t + (2 t x - x) d x = 0

求解贝努力方程:

x = - \frac{2 t ^{3}}{3 ( 2 t - 1 )} + \frac{c _{1}}{2 t - 1}, x = - - \frac{2 t ^{3}}{3 ( 2 t - 1 )} + \frac{c _{1}}{2 t - 1}

x = - \frac{2 t ^{3}}{3 ( 2 t - 1 )} + \frac{c _{1}}{2 t - 1}, x = - - \frac{2 t ^{3}}{3 ( 2 t - 1 )} + \frac{c _{1}}{2 t - 1}

x \to 0 - lim (- 7 x^{2} + 5 x)

d er i v a t i v e y = \frac{6 x - 1}{8 x + 3}

t a y l or tan^{- 1} (x)

\frac{d}{d x} ((2 - 6 x)^{0.5})

\frac{d}{d x} (0.1 x^{2})