y^'=(2x)/(-8y-10),y(0)=-8

解答

y^{^{'}} = \frac{2 x}{- 8 y - 10}, y (0) = - 8

y = - \frac{- 4 x ^{2} + 729 + 5}{4}

求解步骤

y^{'} (x) = \frac{2 x}{- 8 y - 10}

求解可分离方程:

y = - \frac{- 4 x ^{2} + 729 + 5}{4}

y = - \frac{- 4 x ^{2} + 729 + 5}{4}

\frac{d y}{d x} + 4 x y = 4 \cdot e^{6} x

3 y^{2} y^{^{'}} = 1 + y^{3}

x^{^{'}} - \frac{2}{t ^{2}} x = \frac{1}{t ^{2}}

\frac{d y}{d x} = 10 + 3 y

2 y^{^{'}} - 0.8 y = 100