東京大学 新領域創成科学研究科 人間環境学専攻 2023年8月実施 微分方程

标签：

• Tokyo-University
• Laplace-Transform
• Initial-Value-Problem
• Systems-of-ODEs

Author

思齐塾, 祭音Myyura

Description

以下の連立微分方程式の解 $x(t)$ , $y(t)$ を求めよ。 ただし、初期値は $x(0)=0$ , $y(0) = 1$ とする。

$$\begin{cases} \frac{dx}{dt} = 4y - 6e^{-3t}\sin{2t} \\ \frac{dy}{dt} = -x - 3e^{-3t}\cos{2t} \end{cases}$$

题目描述

求下列线性微分方程组的解 $x(t)$、$y(t)$：

$$\begin{cases} \dfrac{dx}{dt}=4y-6e^{-3t}\sin 2t,\\ \dfrac{dy}{dt}=-x-3e^{-3t}\cos 2t, \end{cases}$$

并满足初值

$$x(0)=0,\qquad y(0)=1.$$

Kai

Let $X(s) = L[x(t)]$ and $Y(s) = L[y(t)]$ . Taking the Laplace transform of the given system of differential equations, we get:

$sX(s) - x(0) = 4Y(s) - 6L[e^{-3t}\sin{2t}]$ $sY(s) - y(0) = -X(s) - 3L[e^{-3t}\cos{2t}]$

Using the initial conditions $x(0) = 0$ and $y(0) = 1$ , and the Laplace transforms $L[e^{-3t}\sin{2t}] = \frac{2}{(s+3)^2 + 4}$ and $L[e^{-3t}\cos{2t}] = \frac{s+3}{(s+3)^2 + 4}$ , we have:

$sX(s) = 4Y(s) - \frac{12}{(s+3)^2 + 4}$ $sY(s) - 1 = -X(s) - \frac{3(s+3)}{(s+3)^2 + 4}$

Multiplying the second equation by 4 gives: $4sY(s) - 4 = -4X(s) - \frac{12(s+3)}{(s+3)^2 + 4}$

From the first equation, $Y(s) = \frac{sX(s)}{4} + \frac{3}{(s+3)^2 + 4}$ . Substituting this into the above equation:

$4s(\frac{sX(s)}{4} + \frac{3}{(s+3)^2 + 4}) - 4 = -4X(s) - \frac{12(s+3)}{(s+3)^2 + 4}$ $s^2X(s) + \frac{12s}{(s+3)^2 + 4} - 4 = -4X(s) - \frac{12(s+3)}{(s+3)^2 + 4}$ $(s^2 + 4)X(s) = 4 - \frac{12s}{(s+3)^2 + 4} - \frac{12(s+3)}{(s+3)^2 + 4}$ $(s^2 + 4)X(s) = 4 - \frac{12(2s+3)}{(s^2 + 6s + 13)}$ $X(s) = \frac{4(s^2 + 6s + 13) - 12(2s+3)}{(s^2 + 4)(s^2 + 6s + 13)} = \frac{4s^2 + 24s + 52 - 24s - 36}{(s^2 + 4)(s^2 + 6s + 13)} = \frac{4s^2 + 16}{(s^2 + 4)(s^2 + 6s + 13)} = \frac{4(s^2+4)}{(s^2 + 4)(s^2 + 6s + 13)} = \frac{4}{s^2 + 6s + 13} = \frac{4}{(s+3)^2 + 4}$

Therefore, $X(s) = \frac{4}{(s+3)^2 + 4} = 2\frac{2}{(s+3)^2 + 4}$ , so $x(t) = 2e^{-3t}\sin{2t}$ .

$Y(s) = \frac{sX(s)}{4} + \frac{3}{(s+3)^2 + 4} = \frac{s(\frac{4}{(s+3)^2 + 4})}{4} + \frac{3}{(s+3)^2 + 4} = \frac{s + 3}{(s+3)^2 + 4}$ So $y(t) = e^{-3t}\cos{2t}$ .

Therefore, $x(t) = 2e^{-3t}\sin{2t}$ and $y(t) = e^{-3t}\cos{2t}$ .

Verification: $\frac{dx}{dt} = 2(-3e^{-3t}\sin{2t} + 2e^{-3t}\cos{2t}) = -6e^{-3t}\sin{2t} + 4e^{-3t}\cos{2t}$ $4y - 6e^{-3t}\sin{2t} = 4e^{-3t}\cos{2t} - 6e^{-3t}\sin{2t}$

$$\frac{dy}{dt} = -3e^{-3t}\cos{2t} - 2e^{-3t}\sin{2t}$$

$-x - 3e^{-3t}\cos{2t} = -2e^{-3t}\sin{2t} - 3e^{-3t}\cos{2t}$ $x(0) = 2e^0\sin{0} = 0$ $y(0) = e^0\cos{0} = 1$