y^'=1+2y

解答

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

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

求解步骤

y^{'} (t) = 1 + 2 y

求解可分离方程:

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

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

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

y x = - y^{^{'}}

e^{x} cos (y) d x - e^{x} sin (y) d y = 0

2 sin (y^{2}) + t y cos (y^{2}) y^{^{'}} = 0

x^{2} sin (3 y) + cos (3 y) \frac{d y}{d x} = 4 x^{2}