y+6y^'+13y=10,e^{2t}y=-13

解答

y + 6 y^{^{'}} + 13 y = 10, e^{2 t} y = - 13

y = - \frac{e ^{\frac{- 7 t - c _{1}}{3}}}{7} + \frac{5}{7}

求解步骤

y + 6 y^{'} (t) + 13 y = 10

求解可分离方程:

y = - \frac{e ^{\frac{- 7 t - c _{1}}{3}}}{7} + \frac{5}{7}

y = - \frac{e ^{\frac{- 7 t - c _{1}}{3}}}{7} + \frac{5}{7}

\frac{d y}{d t} = 5 - \frac{7}{400} y

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

x^{2} sin (y) d x - cot (y) d y = 0

\frac{d y}{d x} + (x^{2} - 2) y = x^{2} - 2

y^{^{'}} + 9 y^{3} = 0, y (0) = y^{0}