y^'=-(1+y)/(1+x+y)

解答

y^{^{'}} = - \frac{1 + y}{1 + x + y}

y = - 1 - x + x^{2} + c_{1} + 1, y = - x - x^{2} + c_{1} + 1 - 1

求解步骤

y^{'} (x) = - \frac{1 + y}{1 + x + y}

求解恰当方程:

y = - 1 - x + x^{2} + c_{1} + 1, y = - x - x^{2} + c_{1} + 1 - 1

y = - 1 - x + x^{2} + c_{1} + 1, y = - x - x^{2} + c_{1} + 1 - 1

\frac{d y}{d x} = 7 y (7 - y)

d er i v a t i v e \frac{5 x ^{3} + 9 x ^{2}}{x}

\int \frac{1}{5 cos ( x ) - 4} d x

\int x^{2} \cdot lo g_{e} (x) d x

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