IS Math-2023-02

题目来源：Problem 2 日期：2024-08-13 题目主题：Math-常微分方程-二阶线性微分方程

解题思路

这组问题涉及求解二阶线性常微分方程，重点在于找到满足特定边界条件的解。主要步骤包括：

1. 齐次方程的求解：首先解决齐次部分，找出通解。
2. 非齐次方程的求解：使用特解法（常数变易法、待定系数法等）求解非齐次方程的特解。
3. 边界条件的应用：利用边界条件来确定通解中的常数。

对于第二个问题，我们需要同时求解两个联立微分方程，并考虑它们之间的耦合关系。对于第三个问题，我们通过变量代换将非线性方程转化为线性方程。

Solution

Problem 1

We begin by solving the homogeneous part of the equation:

$$\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}+2\frac{\mathrm{d}x}{\mathrm{d}t}+x=0.$$

The characteristic equation associated with this differential equation is:

$$r^2+2r+1=0.$$

This is a perfect square, so it has a repeated root $r=-1$. Therefore, the general solution to the homogeneous equation is:

$$x_h(t)=(C_1+C_{2}t)e^{-t},$$

where $C_1$ and $C_2$ are arbitrary constants.

Next, we seek a particular solution to the non-homogeneous equation. The non-homogeneous term is $\cos (t)$, so we propose a particular solution of the form:

$$x_p(t)=A\cos (t)+B\sin (t).$$

We differentiate to find the first and second derivatives:

$$\frac{\mathrm{d}{x}_p}{\mathrm{d}t}=-A\sin (t)+B\cos (t),$$ $$\frac{{\mathrm{d}}^2{x}_p}{\mathrm{d}{t}^2}=-A\cos (t)-B\sin (t).$$

Substituting $x_p(t)$ and its derivatives into the original differential equation gives:

$$(-A\cos (t)-B\sin (t))+2(-A\sin (t)+B\cos (t))+(A\cos (t)+B\sin (t))=\cos (t).$$

Simplifying this expression, we collect terms involving $\cos (t)$ and $\sin (t)$:

$$(2B)\cos (t)+(-2A)\sin (t)=\cos (t).$$

By comparing coefficients, we obtain:

$$2B=1,-2A=0.$$

Solving these gives $A=0$ and $B=\frac{1}{2}$. Therefore, the particular solution is:

$$x_p(t)=\frac{1}{2}\sin (t).$$

The general solution to the differential equation is then:

$$x(t)=(C_1+C_{2}t)e^{-t}+\frac{1}{2}\sin (t).$$

Since we require that the solution is bounded as $t\to -\infty$, we must have $C_2=0$ because the term $t{e}^{-t}$ becomes unbounded. Therefore, the final solution is:

$$x(t)=C_1{e}^{-t}+\frac{1}{2}\sin (t).$$

Problem 2

We are asked to solve the following system of differential equations:

$$\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}+2\frac{\mathrm{d}x}{\mathrm{d}t}+x-y=\cos (t),$$ $$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}+2\frac{\mathrm{d}y}{\mathrm{d}t}+y-x=0.$$

Step 1: Solve the homogeneous system

First, consider the homogeneous part of the system:

$$\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}+2\frac{\mathrm{d}x}{\mathrm{d}t}+x-y=0,$$ $$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}+2\frac{\mathrm{d}y}{\mathrm{d}t}+y-x=0.$$

Adding and subtracting these equations, we can uncouple the system.

Let’s define:

$$u(t)=x(t)+y(t),v(t)=x(t)-y(t).$$

The equations for $u(t)$ and $v(t)$ are:

$$\frac{{\mathrm{d}}^{2}u}{\mathrm{d}{t}^2}+2\frac{\mathrm{d}u}{\mathrm{d}t}+2u=0,$$ $$\frac{{\mathrm{d}}^{2}v}{\mathrm{d}{t}^2}+2\frac{\mathrm{d}v}{\mathrm{d}t}=0.$$

These can be solved separately.

For $v(t)$: The characteristic equation is:

$$r^2+2r=0,$$

which gives roots $r=0$ and $r=-2$. Thus, the general solution for $v(t)$ is:

$$v(t)=C_1+C_2{e}^{-2t}.$$

For $u(t)$:

The characteristic equation is:

$$r^2+2r+2=0,$$

which has complex roots $r=-1\pm i$. Therefore, the general solution for $u(t)$ is:

$$u(t)=e^{-t}(A\cos (t)+B\sin (t)).$$

Using the definitions of $u(t)$ and $v(t)$:

$$x(t)=\frac{u(t)+v(t)}{2},y(t)=\frac{u(t)-v(t)}{2}.$$

Thus, the homogeneous solutions for $x(t)$ and $y(t)$ are:

$$x_h(t)=\frac{1}{2}{e}^{-t}(A\cos (t)+B\sin (t))+\frac{1}{2}(C_1+C_2{e}^{-2t}),$$ $$y_h(t)=\frac{1}{2}{e}^{-t}(A\cos (t)+B\sin (t))-\frac{1}{2}(C_1+C_2{e}^{-2t}).$$

Step 2: Find a particular solution

Now, consider the non-homogeneous system with the forcing term $\cos (t)$.

We look for a particular solution $x_p(t)$ and $y_p(t)$. Since the right-hand side of the first equation is $\cos (t)$, we try a particular solution of the form:

$$x_p(t)=P\cos (t)+Q\sin (t),y_p(t)=R\cos (t)+S\sin (t).$$

Substituting into the original equations:

1. Substituting into the first equation:

$$(-P\cos (t)-Q\sin (t))+2(-P\sin (t)+Q\cos (t))+P\cos (t)+Q\sin (t)-R\cos (t)-S\sin (t)=\cos (t).$$

Simplifying, we get:

$$(2Q-R)\cos (t)+(-2P-S)\sin (t)=\cos (t).$$

Thus, comparing coefficients:

$$2Q-R=1,-2P-S=0.$$

2. Substituting into the second equation:

$$(-R\cos (t)-S\sin (t))+2(-R\sin (t)+S\cos (t))+R\cos (t)+S\sin (t)-P\cos (t)-Q\sin (t)=0.$$

Simplifying, we get:

$$(2S-P)\cos (t)+(-2R-Q)\sin (t)=0.$$

Thus, comparing coefficients:

$$2S-P=0,-2R-Q=0.$$

From the first set of equations:

$$P=2S,Q=-2R.$$

Substituting into the second set of equations:

$$4R-R=1\Rightarrow R=\frac{1}{3},Q=-\frac{2}{3},$$ $$S=\frac{1}{6},P=\frac{1}{3}.$$

Thus, the particular solution is:

$$x_p(t)=\frac{1}{3}\cos (t)-\frac{2}{3}\sin (t),$$ $$y_p(t)=\frac{1}{3}\cos (t)+\frac{1}{6}\sin (t).$$

Step 3: General Solution

The general solution is the sum of the homogeneous and particular solutions:

$$x(t)=\frac{1}{2}{e}^{-t}(A\cos (t)+B\sin (t))+\frac{1}{2}(C_1+C_2{e}^{-2t})+\frac{1}{3}\cos (t)-\frac{2}{3}\sin (t),$$ $$y(t)=\frac{1}{2}{e}^{-t}(A\cos (t)+B\sin (t))-\frac{1}{2}(C_1+C_2{e}^{-2t})+\frac{1}{3}\cos (t)+\frac{1}{6}\sin (t).$$

Step 4: Bounded Solution

For the solution to be bounded as $t\to -\infty$, the terms involving $t{e}^{-t}$ must vanish. Therefore, $C_2=0$. So, the bounded solution is:

$$x(t)=\frac{1}{2}{e}^{-t}(A\cos (t)+B\sin (t))+\frac{1}{2}{C}_1+\frac{1}{3}\cos (t)-\frac{2}{3}\sin (t),$$ $$y(t)=\frac{1}{2}{e}^{-t}(A\cos (t)+B\sin (t))-\frac{1}{2}{C}_1+\frac{1}{3}\cos (t)+\frac{1}{6}\sin (t).$$

Problem 3

We are given the nonlinear differential equation:

$$e^{-t}{x}^2-2\frac{\mathrm{d}x}{\mathrm{d}t}+x=0$$

and we are asked to solve for $x(t)$ with the initial condition $x(0)=\frac{1}{2}$.

Step 1: Simplify the equation using substitution

To simplify the equation, let’s propose a substitution:

$$x(t)=e^{\frac{t}{2}}u(t),$$

where $u(t)$ is a new function to be determined. Let’s substitute this into the original equation.

First, compute the derivatives:

$$\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left(e^{\frac{t}{2}}u(t)\right)=\frac{1}{2}{e}^{\frac{t}{2}}u(t)+e^{\frac{t}{2}}\frac{\mathrm{d}u}{\mathrm{d}t}.$$

Substituting $x(t)=e^{\frac{t}{2}}u(t)$ and its derivative into the original differential equation:

$$e^{-t}\left(e^t{u}^2(t)\right)-2\left(\frac{1}{2}{e}^{\frac{t}{2}}u(t)+e^{\frac{t}{2}}\frac{\mathrm{d}u}{\mathrm{d}t}\right)+e^{\frac{t}{2}}u(t)=0.$$

Simplifying the equation:

$$u^2(t)-e^{\frac{t}{2}}\left(e^{\frac{t}{2}}u(t)+2\frac{\mathrm{d}u}{\mathrm{d}t}\right)+e^{\frac{t}{2}}u(t)=0.$$

This simplifies further to:

$$u^2(t)=2e^{\frac{t}{2}}\frac{\mathrm{d}u}{\mathrm{d}t}.$$

This is a separable differential equation.

Step 2: Separate variables and integrate

We separate the variables:

$$\frac{\mathrm{d}u}{u^2}=\frac{1}{2}{e}^{-\frac{t}{2}}\mathrm{d}t.$$

Integrating both sides:

$$\int u^{-2}\mathrm{d}u=\frac{1}{2}\int e^{-\frac{t}{2}}\mathrm{d}t.$$

The left side integrates to:

$$-\frac{1}{u}=-e^{-\frac{t}{2}}+C_1.$$

Taking the reciprocal:

$$u(t)=\frac{1}{C_1-e^{-\frac{t}{2}}}.$$

Substituting back for $x(t)$:

$$x(t)=\frac{e^{\frac{t}{2}}}{C_1-e^{-\frac{t}{2}}}.$$

Step 3: Apply initial condition

We apply the initial condition $x(0)=\frac{1}{2}$:

$$\frac{e^0}{C_1-1}=\frac{1}{2}.$$

This gives:

$$C_1=3.$$

Thus, the solution is:

$$x(t)=\frac{e^{\frac{t}{2}}}{3-e^{-\frac{t}{2}}}.$$

This is the final solution for $x(t)$ that satisfies the initial condition $x(0)=\frac{1}{2}$.

知识点

二阶常微分方程常微分方程特征方程特解法未完成

解题技巧和信息

1. 常微分方程的齐次解：求解二阶常微分方程时，首先要求齐次方程的解，即求解其对应的特征方程。
2. 非齐次方程的特解：根据右边的非齐次项选择特解形式。
3. 边界条件的应用：边界条件对于确定通解中的常数至关重要，特别是当要求解在无穷远处有界时。

重点词汇

• Differential Equation: 微分方程
• Homogeneous Solution: 齐次解
• Particular Solution: 特解
• Characteristic Equation: 特征方程

参考资料

1. Boyce, W. E., & DiPrima, R. C. (2017). Elementary Differential Equations and Boundary Value Problems (10th Edition). Wiley.