2y^'-yy^'=x^2+1

解答

2 y^{^{'}} - y y^{^{'}} = x^{2} + 1

y = - \frac{- 12 + 144 + 12 ( - c _{1} - 6 x - 2 x ^{3} )}{6}, y = - \frac{- 12 - 144 + 12 ( - c _{1} - 6 x - 2 x ^{3} )}{6}

求解步骤

2 y^{'} (x) - y y^{'} (x) = x^{2} + 1

求解可分离方程:

y = - \frac{- 12 + 144 + 12 ( - c _{1} - 6 x - 2 x ^{3} )}{6}, y = - \frac{- 12 - 144 + 12 ( - c _{1} - 6 x - 2 x ^{3} )}{6}

y = - \frac{- 12 + 144 + 12 ( - c _{1} - 6 x - 2 x ^{3} )}{6}, y = - \frac{- 12 - 144 + 12 ( - c _{1} - 6 x - 2 x ^{3} )}{6}

\frac{2 d y}{d t} + 4 y = 6, y (0) = 1 (\frac{1}{2})

\frac{d y}{d x} = (2 - y) sin (x), x = 2, c^{1} = 3

x^{^{'}} + 2^{2} \cdot 3^{2} \cdot x = 0

y^{^{'}} + \frac{y}{x} = x^{3} y

\frac{d P}{d t} = a \cdot P