Second-Order Differential Equations with Initial Conditions | IB数学：二阶微分方程的初始条件

📚 Second-Order Differential Equations with Initial Conditions | IB数学：二阶微分方程的初始条件

In IB Mathematics (Analysis and Approaches HL), solving second-order differential equations with initial conditions is a key skill. This article guides you through the complete process, from standard form to particular solutions.

在IB数学（分析与方法HL）中，利用初始条件求解二阶微分方程是一项核心技能。本文将引导你贯穿完整过程，从标准形式到特解。

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1. Standard Form of a Second-Order Linear Differential Equation | 二阶线性微分方程的标准形式

A second-order linear differential equation with constant coefficients can be written as:

一个常系数二阶线性微分方程可以写成：

a y” + b y’ + c y = f(x)

Here, a, b, c are constants, and f(x) is a given function. When f(x) = 0, the equation is homogeneous; otherwise, it is non-homogeneous.

其中 a, b, c 是常数，f(x) 是已知函数。当 f(x) = 0 时方程为齐次的；否则为非齐次的。

• Homogeneous: a y” + b y’ + c y = 0
• 齐次：a y” + b y’ + c y = 0
• Non-homogeneous: a y” + b y’ + c y = f(x)
• 非齐次：a y” + b y’ + c y = f(x)

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2. The Characteristic Equation | 特征方程

For the homogeneous equation a y” + b y’ + c y = 0, we assume a solution of the form y = e^{rx}. Substituting into the equation gives the characteristic equation:

对于齐次方程 a y” + b y’ + c y = 0，我们假设解的形式为 y = e^{rx}。代入方程得到特征方程：

a r² + b r + c = 0

This quadratic equation determines the behaviour of the general solution. Its roots are found using the quadratic formula.

这个二次方程决定了通解的形式。其根通过求根公式得到。

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3. Real Distinct Roots | 两个相异实根

If the discriminant b² – 4ac > 0, the characteristic equation has two distinct real roots r₁ and r₂. The general solution is:

若判别式 b² – 4ac > 0，特征方程有两个相异实根 r₁ 和 r₂。通解为：

y = A e^{r₁x} + B e^{r₂x}

Here A and B are arbitrary constants determined by initial conditions.

其中 A 和 B 是由初始条件确定的任意常数。

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4. Repeated Root | 重根

If b² – 4ac = 0, there is one repeated root r. The general solution takes the form:

若 b² – 4ac = 0，有一个重根 r。通解形式为：

y = (A + Bx) e^{rx}

The extra factor x accounts for the linear independence of the two solutions.

额外的因子 x 用于保证两个解的线性无关性。

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5. Complex Roots | 复根

If b² – 4ac < 0, the roots are complex conjugates: r = α ± iβ. The general solution is:

若 b² – 4ac < 0，根为共轭复数：r = α ± iβ。通解为：

y = e^{αx} ( C cos(βx) + D sin(βx) )

This form is essential for describing oscillatory motion.

这一形式对于描述振荡运动至关重要。

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6. Applying Initial Conditions | 应用初始条件

Initial conditions for a second-order equation usually specify the value of y and y’ at a point, for example:

二阶方程的初始条件通常指定在某点 y 和 y’ 的值，例如：

y(x₀) = y₀,   y'(x₀) = y₁

Substitute x₀ into the general solution to get one equation. Then differentiate the general solution, substitute x₀ again, and solve the linear system for the constants.

将 x₀ 代入通解得到第一个方程；然后对通解求导，再代入 x₀，得到第二个方程；解这个线性方程组即可确定常数。

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7. Non-Homogeneous Equations | 非齐次方程

For a non-homogeneous equation a y” + b y’ + c y = f(x), the general solution is the sum of the complementary function y_c (solution of the homogeneous equation) and a particular integral y_p:

对于非齐次方程 a y” + b y’ + c y = f(x)，通解等于补函数 y_c（齐次方程的解）与一个特解 y_p 之和：

y = y_c + y_p

The particular integral is guessed using the method of undetermined coefficients, based on the form of f(x).

特解通过待定系数法，根据 f(x) 的形式进行假设。

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8. Example: Real Roots with Initial Conditions | 示例：实根与初始条件

Solve y” – 3y’ + 2y = 0, with y(0) = 1 and y'(0) = 0.

求解 y” – 3y’ + 2y = 0，初始条件为 y(0) = 1，y'(0) = 0。

The characteristic equation is r² – 3r + 2 = 0, giving r = 1 and r = 2. Thus the general solution is:

特征方程为 r² – 3r + 2 = 0，得到 r = 1 和 r = 2。因此通解为：

y = A e^{x} + B e^{2x}

Using y(0) = 1 gives A + B = 1. Differentiating: y’ = A e^{x} + 2B e^{2x}; y'(0) = 0 gives A + 2B = 0. Solving: A = 2, B = -1. Hence the particular solution is:

由 y(0) = 1 得 A + B = 1。求导：y’ = A e^{x} + 2B e^{2x}；由 y'(0) = 0 得 A + 2B = 0。解得 A = 2，B = -1。因此特解为：

y = 2e^{x} – e^{2x}

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9. Example: Complex Roots with Initial Conditions | 示例：复根与初始条件

Solve y” + 4y = 0, with y(0) = 3 and y'(0) = 8.

求解 y” + 4y = 0，初始条件为 y(0) = 3，y'(0) = 8。

Characteristic equation: r² + 4 = 0, so r = ±2i. Here α = 0, β = 2. General solution:

特征方程：r² + 4 = 0，所以 r = ±2i。这里 α = 0，β = 2。通解为：

y = C cos(2x) + D sin(2x)

y(0) = 3 gives C = 3. y’ = -2C sin(2x) + 2D cos(2x); y'(0) = 8 gives 2D = 8, so D = 4. Therefore:

y(0) = 3 得 C = 3。y’ = -2C sin(2x) + 2D cos(2x)；y'(0) = 8 得 2D = 8，所以 D = 4。因此：

y = 3 cos(2x) + 4 sin(2x)

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10. Common Mistakes and Pitfalls | 常见错误与陷阱

• Forgetting the extra x term when the root is repeated.
• 重根时忘记额外的 x 项。
• Using incorrect signs when differentiating the general solution.
• 对通解求导时符号出错。
• Mixing up the complementary function and particular integral.
• 混淆补函数和特解。
• Not checking that the particular integral is not already part of the complementary function.
• 未检查特解是否已经包含在补函数中。

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11. Applications in Physics | 物理应用

Second-order differential equations model simple harmonic motion, such as a mass on a spring: m y” + k y = 0. The solution describes oscillatory displacement with frequency √(k/m).

二阶微分方程可描述简谐运动，例如弹簧上的质量：m y” + k y = 0。其解描述了频率为 √(k/m) 的振荡位移。

With initial conditions specifying the starting position and velocity, you can determine the exact amplitude and phase of the motion.

通过指定初始位置和速度的初始条件，可以确定运动的精确振幅和相位。

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12. Exam Tips from TutorHao | TutorHao 考试提示

Always write the characteristic equation first. For non-homogeneous problems, find y_c before y_p. When applying initial conditions, use the full general solution, not just the complementary function.

务必先写出特征方程。对于非齐次问题，先求 y_c 再求 y_p。在应用初始条件时，使用完整的通解，而不仅仅是补函数。

Practice with past IB questions. Pay attention to whether the initial condition is given at x = 0 or elsewhere; the method remains the same, but substitute the correct value.

练习IB历年真题。注意初始条件是在 x = 0 还是其他位置；方法相同，但要代入正确的值。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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