Example 7.4.3: Solving a First-Order Linear Differential Equation | 例题7.4.3：求解一阶线性微分方程

📚 Example 7.4.3: Solving a First-Order Linear Differential Equation | 例题7.4.3：求解一阶线性微分方程

ample, we will solve the first-order linear differential equation given by dy/dx + 2y = e⁻ˣ expected to use an integrating factor to solve linear first-order differential equations.

在本例题中，我们将求解一阶线性微分方程 dy/dx + 2y = e⁻ˣ，并满足初始条件 y(0) = 1。此例题是 AQA A-Level 数学试卷2的典型内容，要求学生能够运用积分因子法求解一阶线性微分方程。

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1. Understanding the Problem | 理解题意

We are given a differential equation that involves a derivative of y with respect to x, a term 2y, and a non-homogeneous term e⁻ˣ. The equation is linear because y and its derivative appear only to the first power and are not multiplied together.

我们给出的微分方程包含 y 关于 x 的导数、一项 2y 以及非齐次项 e⁻ˣ。该方程是线性的，因为 y 及其导数只以一次幂出现，且不相互相乘。

ion will allow us to determine the particular solution from a family of general solutions.

初始条件 y(0) = 1 告诉我们当 x = 0 时 y 的值。该条件将使我们能够从一族通解中确定特解。

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2. Standard Form of a Linear Differential Equation | 线性微分方程的标准形式

and Q(x) are functions of x alone. Identifying P(x) and Q(x) correctly is essential before applying the integrating factor method.

一阶线性微分方程可以写成标准形式 dy/dx + P(x)y = Q(x)。其中 P(x) 和 Q(x) 仅是 x 的函数。在应用积分因子法之前，正确识别 P(x) 和 Q(x) 至关重要。

In our equation dy/dx + 2y = e⁻ˣ, we see that P(x) = 2 and Q(x) = e⁻ˣ. Both are simple functions, which makes the calculation straightforward.

在我们的方程 dy/dx + 2y = e⁻ˣ 中，可以看出 P(x) = 2，Q(x) = e⁻ˣ。两者都是简单函数，因此计算过程直接明了。

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3. The Integrating Factor Method | 积分因子法

). Multiplying both sides of the differential equation by I(x) transforms the left-hand side into the derivative of I(x)y.

对于标准形式 dy/dx + P(x)y = Q(x) 的方程，积分因子定义为 I(x) = e^(∫P(x) dx)。将微分方程两边乘以 I(x) 后，左边会变成 I(x)y 的导数。

This technique is powerful because it converts a differential equation into an exact differential, which can then be integrated directly.

这一技巧非常强大，因为它将微分方程转化为一个恰当微分，从而可以直接积分。

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4. Finding the Integrating Factor | 求积分因子

ting factor only needs a particular antiderivative, we ignore the constant of integration.

我们首先计算 ∫P(x) dx。这里 P(x) = 2，所以 ∫2 dx = 2x + C。由于积分因子只需要一个特定的原函数，我们忽略积分常数。

Therefore the integrating factor is:

I(x) = e^2x

This integrating factor will be used to multiply every term in the original differential equation.

这个积分因子将被用来乘以原微分方程的每一项。

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5. Multiplying Both Sides | 两边同乘积分因子

We multiply the original equation dy/dx + 2y = e⁻ˣ by I(x) = e^2x. This gives:

我们将原方程 dy/dx + 2y = e⁻ˣ 两边同乘 I(x) = e^2x，得到：

e^2x · dy/dx + 2e^2x · y = e^2x · e⁻ˣ

Simplifying the right-hand side, e^2x · e⁻ˣ = e^{2x − x} = eˣ. So the equation becomes:

化简右边，e^2x · e⁻ˣ = e^{2x − x} = eˣ。因此方程变为：

e^2x · dy/dx + 2e^2x · y = eˣ

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6. Recognising the Derivative of a Product | 识别乘积的导数

The left-hand side of the equation is exactly the derivative of e^2x · y with respect to x. By the product rule, d/dx [e^2x · y] = e^2x · dy/dx + 2e^2x · y.

方程的左边恰好是 e^2x · y 关于 x 的导数。根据乘积法则，d/dx [e^2x · y] = e^2x · dy/dx + 2e^2x · y。

Therefore, we can rewrite the entire equation as:

因此，我们可以将整个方程改写为：

d/dx [e^2x · y] = eˣ

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7. Integrating Both Sides | 两边同时积分

Now we integrate both sides with respect to x. The left side integrates to e^2x · y because integration is the inverse of differentiation. The right side integrates as ∫eˣ dx = eˣ + C, where C is the constant of integration.

现在我们对两边关于 x 进行积分。左边积分得到 e^2x · y，因为积分是微分的逆运算。右边积分为 ∫eˣ dx = eˣ + C，其中 C 是积分常数。

The result is:

结果为：

e^2x · y = eˣ + C

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8. Solving for y | 解出 y

To isolate y, we divide both sides of the equation by e^2x. This yields:

为了解出 y，我们将方程两边同时除以 e^2x，得到：

y = (eˣ + C) / e^2x

We can simplify this expression by separating the terms:

我们可以通过分离项来化简该表达式：

y = eˣ / e^2x + C / e^2x = e⁻ˣ + C·e⁻²ˣ

Thus the general solution is y = e⁻ˣ + C·e⁻²ˣ.

因此通解为 y = e⁻ˣ + C·e⁻²ˣ。

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9. Applying the Initial Condition | 应用初始条件

e are given y(0) = 1. Substitute x = 0 and y = 1 into the general solution. Recall that e⁻ˣ at x = 0 equals e⁰ = 1, and e⁻²ˣ at x = 0 also equals 1.

我们已知 y(0) = 1。将 x = 0 和 y = 1 代入通解。注意 e⁻ˣ 在 x = 0 时等于 e⁰ = 1，e⁻²ˣ 在 x = 0 时也等于 1。

So we have:

因此我们有：

1 = 1 + C·1

This simplifies to 1 = 1 + C, which gives C = 0.

化简得 1 = 1 + C，因此 C = 0。

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10. Particular Solution | 特解

Substituting C = 0 back into the general solution, we obtain the particular solution:

将 C = 0 代回通解，我们得到特解：

y = e⁻ˣ

This is the unique solution that satisfies both the differential equation and the initial condition.

这是同时满足微分方程和初始条件的唯一解。

tice that C = 0 makes the second term vanish, leaving a simple exponential decay function.

注意到 C = 0 使得第二项消失，留下一个简单的指数衰减函数。

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11. Verification | 验证解的正确性

We should always verify that the solution satisfies the original equation. Let y = e⁻ˣ. Then dy/dx = −e⁻ˣ. Substituting into dy/dx + 2y gives:

我们应当始终验证解是否满足原方程。令 y = e⁻ˣ，则 dy/dx = −e⁻ˣ。代入 dy/dx + 2y 得到：

−e⁻ˣ + 2e⁻ˣ = e⁻ˣ

t our solution is correct. The initial condition is also satisfied because y(0) = e⁻⁰ = 1.

这与原方程右边 e⁻ˣ 一致，证明我们的解是正确的。初始条件也满足，因为 y(0) = e⁻⁰ = 1。

Verification is a crucial step in examination answers, as it helps catch algebraic errors.

在考试作答中，验证步骤至关重要，有助于发现代数错误。

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12. Graphical Interpretation and Extensions | 图形解释与拓展

The particular solution y = e⁻ˣ → 0. As x → −∞, y → ∞; although the domain of this solution is typically all real numbers.

特解 y = e⁻ˣ 表示一条经过点 (0, 1) 的递减指数曲线。当 x → ∞ 时，y → 0；当 x → −∞ 时，y → ∞；通常该解的定义域为全体实数。

ates the general method for solving dy/dx + a·y = f(x). The integrating factor is always e^{∫a dx} = e^ax, and the same procedure follows. This technique extends to more complex functions P(x) and Q(x), as long as the integrals can be evaluated.

此例题说明了求解 dy/dx + a·y = f(x) 的一般方法。积分因子始终是 e^{∫a dx} = e^ax，后续步骤相同。只要积分能够计算，该技巧可以推广到更复杂的 P(x) 和 Q(x) 函数。

For AQA examinations, students must be comfortable with this method and also with cases where Q(x) is a polynomial, sine, cosine, or another exponential function.

对于 AQA 考试，学生必须熟练掌握这一方法，并能够处理 Q(x) 为多项式、正弦、余弦或其他指数函数的情形。

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