10.9.4 Exercises

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62 求下列微分方程的通解:

(a) $\frac{d^2x}{dt^2} - 2\frac{dx}{dt} - 3x = t$

(b) $\frac{d^2x}{dt^2} - 2\frac{dx}{dt} - 5x = t^2 - 2t$

(c) $\frac{d^2x}{dt^2} - \frac{dx}{dt} - x = 5e^t$

63 求下列微分方程的通解:

(a) $\frac{d^2x}{dt^2} - 3\frac{dx}{dt} + 4x = \cos 4t - 2 \sin 4t$

(b) $9\frac{d^2x}{dt^2} - 12\frac{dx}{dt} + 4x = e^{-3t}$

(c) $2\frac{d^2x}{dt^2} + 4\frac{dx}{dt} - 7x = 7 \cos 2t$

(d) $\frac{d^2x}{dt^2} + \frac{dx}{dt} + 4x = 5t - 7$

(e) $16\frac{d^2x}{dt^2} + 8\frac{dx}{dt} + x = t + 6$

(f) $\frac{d^2x}{dt^2} - 8\frac{dx}{dt} + 16x = -3 \sin 3t$

(g) $\frac{d^2x}{dt^2} - 4\frac{dx}{dt} + 7x = e^{-5t}$

(h) $3\frac{d^2x}{dt^2} + 3\frac{dx}{dt} - x = t^2 + e^{-2t}$

(i) $\frac{d^2x}{dt^2} + 2\frac{dx}{dt} - 3x = 5e^{-3t} + \sin 2t$

(j) $\frac{d^2x}{dt^2} + 16x = 1 + 2 \sin 4t$

(k) $\frac{d^2x}{dt^2} - 4\frac{dx}{dt} = 7 - 3e^{4t}$

64 证明……微分方程的特征方程

$$\frac{d^4x}{dt^4} - 3\frac{d^3x}{dt^3} - 5\frac{d^2x}{dt^2} + 9\frac{dx}{dt} - 2x = 0$$

is

$$(m^2 + m - 2)(m^2 - 4m + 1) = 0$$

从而求出这些方程的通解

(a) $\frac{d^4x}{dt^4} - 3\frac{d^3x}{dt^3} - 5\frac{d^2x}{dt^2} + 9\frac{dx}{dt} - 2x = \cos 2t$

(b) $\frac{d^4x}{dt^4} - 3\frac{d^3x}{dt^3} - 5\frac{d^2x}{dt^2} + 9\frac{dx}{dt} - 2x = e^{2t} + e^{-2t}$

(c) $\frac{d^4x}{dt^4} - 3\frac{d^3x}{dt^3} - 5\frac{d^2x}{dt^2} + 9\frac{dx}{dt} - 2x = t^2 - 1 + e^{-t}$

65 证明该微分方程的特征方程

$$\frac{d^3x}{dt^3} - 9\frac{d^2x}{dt^2} + 27\frac{dx}{dt} - 27x = 0$$

is

$$(m - 3)^3 = 0$$

从而求出这些方程的通解

(a) $\frac{d^3x}{dt^3} - 9\frac{d^2x}{dt^2} + 27\frac{dx}{dt} - 27x = \cos t - \sin t + t$

(b) $\frac{d^3x}{dt^3} - 9\frac{d^2x}{dt^2} + 27\frac{dx}{dt} - 27x = e^t$

(c) $\frac{d^3x}{dt^3} - 9\frac{d^2x}{dt^2} + 27\frac{dx}{dt} - 27x = e^{3t} + t$