y^'=2xy^2,y(2)=1

解答

y^{^{'}} = 2 x y^{2}, y (2) = 1

y = - \frac{1}{x ^{2} - 5}

求解步骤

y^{'} (x) = 2 x y^{2}

求解可分离方程:

y = - \frac{1}{x ^{2} - 5}

y = - \frac{1}{x ^{2} - 5}

\frac{d y}{d x} x y = x^{2} - x

\frac{d y}{d x} = \frac{5 x y}{( ln ( y ) ) ^{6}}

2 q^{^{'}} + 6 q = 2

\frac{d y}{d t} = 3 - \frac{3}{10} y

\frac{1}{2 y} = y^{^{'}} \frac{x ^{2}}{y ^{2} - 8}, y (1) = 9