y^'=y(4-ty)

解答

y^{^{'}} = y (4 - t y)

y = \frac{16 e ^{4 t}}{4 e ^{4 t} t - e ^{4 t} + c _{1}}

求解步骤

y^{'} (t) = y (4 - t y)

求解贝努力方程:

y = \frac{16 e ^{4 t}}{4 e ^{4 t} t - e ^{4 t} + c _{1}}

y = \frac{16 e ^{4 t}}{4 e ^{4 t} t - e ^{4 t} + c _{1}}

\frac{d}{d x} (cos^{2} (x) + sin^{2} (x))

d er i v a t i v e f (x) = \frac{56}{x ^{9}}

x \to 0 lim (x^{2} - 3)

x \to 2 lim (\frac{x + 5}{x - 2})

\int 2 sin (x) - 1 d x