x^{''}+2x^'-4x=6e^{3t}-4t+3

解答

x^{^{''}} + 2 x^{^{'}} - 4 x = 6 e^{3 t} - 4 t + 3

x = c_{1} e^{(- 1 + 5) t} + c_{2} e^{(- 1 - 5) t} + \frac{6}{11} e^{3 t} + t - \frac{1}{4}

求解步骤

x^{''} (t) + 2 x^{'} (t) - 4 x = 6 e^{3 t} - 4 t + 3

求解线性方程:

x = c_{1} e^{(- 1 + 5) t} + c_{2} e^{(- 1 - 5) t} + \frac{6}{11} e^{3 t} + t - \frac{1}{4}

x = c_{1} e^{(- 1 + 5) t} + c_{2} e^{(- 1 - 5) t} + \frac{6}{11} e^{3 t} + t - \frac{1}{4}

y^{^{''}} - 25 y = 3 x^{2} - 2 + e^{x}, y (0) = 1, y^{^{'}} (0) = 0

x^{2} y^{^{''}} - 8 x^{2} y^{^{'}} + 16 x^{2} y = 3 e^{4 x}

y^{^{''}} + 3 y^{^{'}} + 2 y = 12 e^{t}, y (0) = 2, y^{^{'}} (0) = 3

y^{^{''}} + 4 y^{^{'}} + 3 y = 4 x^{2} + 5

y^{^{''}} - y^{^{'}} - 6 y = 73 cos (3 x)