y^'=x+xy+1+y

解答

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

y = e^{\frac{x ^{2}}{2} + x + c_{1}} - 1

求解步骤

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

求解可分离方程:

y = e^{\frac{x ^{2}}{2} + x + c_{1}} - 1

y = e^{\frac{x ^{2}}{2} + x + c_{1}} - 1

\frac{d y}{d x} = \frac{9 + 17 x}{x y ^{2}}, x > 0, y (1) = 2

so l v e f or t, e^{- 4 H (t)} d H (t) = e^{6 t} d t

(1 + y^{2}) d x - y (x + x^{3}) d y = 0

\frac{d y}{d x} = \frac{4}{y ^{2}} x

x d x = x^{2} e^{H (x)} d H (x)