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

解答

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

y = ln (x + 1) (x + 1) + 1 + c_{1} (x + 1)

求解步骤

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

求解线性方程:

y = ln (x + 1) (x + 1) + 1 + c_{1} (x + 1)

y = ln (x + 1) (x + 1) + 1 + c_{1} (x + 1)

\int (2 x^{5}) d x

\frac{x d y}{d x} + y = - 8 x y^{2}, y (1) = 5

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

\int \frac{x + 1}{x ^{2} + 2 x - 3} d x

x \to - 1 - lim (x^{3} - 3)