(f) $\frac{d^2x}{dt^2} + 3\frac{dx}{dt} + \frac{13}{4}x = \sin t$

中文 · PDF 920

(g) $\frac{d^3x}{dt^3} - 5\frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 8x = t^2 - t$

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

(i) $\frac{d^3x}{dt^3} - 5\frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 8x = e^{2t} + e^t$

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

21 求解下列初值问题：

(a) $\frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 5x = 1, \quad x(0) = 0, \quad \frac{dx}{dt}(0) = 0$

(b) $3\frac{d^2x}{dt^2} - 2\frac{dx}{dt} - x = 2t - 1,$

$x(0) = 7, \quad \frac{dx}{dt}(0) = 2$

$x(0) = 0, \quad \frac{dx}{dt}(0) = 2$

(d) $\frac{d^2x}{dt^2} - \frac{dx}{dt} = -2e^{2t}, \quad x(0) = 0, \quad \frac{dx}{dt}(0) = 1$

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

$x(0) = 0, \quad \frac{dx}{dt}(0) = 1$

(f) $\frac{d^3x}{dt^3} + 5\frac{d^2x}{dt^2} + 17\frac{dx}{dt} + 13x = 1,$

$x(0) = 1, \quad \frac{dx}{dt}(0) = 1, \quad \frac{d^2x}{dt^2}(0) = 0$

22 求由下列二阶线性常系数微分方程所描述的系统的阻尼参数和固有频率：

(a) $\frac{d^2x}{dt^2} + 7\frac{dx}{dt} + 2x = 0$

(b) $\frac{d^2x}{dt^2} + p\frac{dx}{dt} + p^{1/2}x = 0$

(d) $\frac{d^2x}{dt^2} + 14\frac{dx}{dt} + 2\alpha x = 0$

23 确定所需适当参数的值，使由下列二阶线性常系数微分方程所描述的系统具有所给出的阻尼参数和固有频率：

(a) $\frac{d^2x}{dt^2} + \frac{a}{2}\frac{dx}{dt} + bx = 0, \quad \zeta = 0.25, \quad \omega = 2$

(b) $\frac{d^2x}{dt^2} + a\frac{dx}{dt} + bx = 0, \quad \zeta = 2, \quad \omega = \pi$

(d) $p\frac{d^2x}{dt^2} + q^2\frac{dx}{dt} + 6x = 0, \quad \zeta = 1.2, \quad \omega = 0.2$

24 证明：通过作代换

$$v = \frac{dx}{dt}$$

方程

$$\frac{d^2x}{dt^2} + \frac{dx}{dt} = 1$$

可以表示为

$$\frac{dv}{dt} + v = 1$$

证明该方程的解为 $v = 1 + Ce^{-t}$ 并由此求出 $x(t)$ 。

这一技术是求解因变量本身不显式出现的二阶微分方程的标准方法。用同样的方法求下列微分方程的解

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

(b) $\frac{d^2x}{dt^2} - \left(\frac{dx}{dt}\right)^2 = 1$