y^'=5y+y^5

解答

y^{^{'}} = 5 y + y^{5}

y = 4 \frac{5}{e ^{- 20 t - c_{1}} - 1}, y = - 4 \frac{5}{e ^{- 20 t - c_{1}} - 1}

求解步骤

y^{'} (t) = 5 y + y^{5}

求解贝努力方程:

y = 4 \frac{5}{e ^{- 20 t - c_{1}} - 1}, y = - 4 \frac{5}{e ^{- 20 t - c_{1}} - 1}

y = 4 \frac{5}{e ^{- 20 t - c_{1}} - 1}, y = - 4 \frac{5}{e ^{- 20 t - c_{1}} - 1}

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

d er i v a t i v e x^{\frac{1}{3}} - 1

\frac{d}{d t} (\frac{1}{( t + 1 ) ^{2}})

\int 3 x y^{2} d y

\frac{\partial}{\partial x} (e^{e^{y} (1 + e^{x}) - 2})