📚 Formation of Second Order Differential Equations | 二阶微分方程的建立
A second order differential equation contains a second derivative, such as d²y/dx², and it is often formed by eliminating two arbitrary constants from a family of curves or by applying physical laws involving acceleration.
二阶微分方程包含二阶导数，例如 d²y/dx²，它通常通过消去曲线族中的两个任意常数或应用涉及加速度的物理定律来建立。
1. What Is a Second Order Differential Equation? | 什么是二阶微分方程
case is often written as a d²y/dx² + b dy/dx + c y = f(x), where a, b and c are constants.
二阶微分方程是包含未知函数、它的一阶导数和二阶导数的方程。在 AQA 记法中，线性情形常写作 a d²y/dx² + b dy/dx + c y = f(x)，其中 a、b 和 c 为常数。
The highest derivative appearing in the equation is the second derivative, and this is what makes the equation 'second order'. The equation may also contain x and y, but it must include d²y/dx² in some form.
方程中出现的最高阶导数是二阶导数，这正是方程被称为“二阶”的原因。方程中也可以含有 x 和 y，但必须以某种形式包含 d²y/dx²。
2. Why Form a Differential Equation? | 为什么要建立微分方程
Forming a differential equation reverses the process of solving one. Given a family of curves with two independent parameters, we can produce a single equation that every member of that family satisfies, and this equation often reveals the underlying model or physical law.
建立微分方程是求解过程的逆运算。给定含有两个独立参数的曲线族，我们可以构造一个该族中每一条曲线都满足的单一方程，这个方程通常能揭示背后的模型或物理规律。
B sin ωt, forming the differential equation shows that the acceleration is proportional to the displacement, which is the key idea in simple harmonic motion.
例如，如果我们知道一个质点按照 y = A cos ωt + B sin ωt 振动，那么建立微分方程就能表明加速度与位移成正比，这是简谐运动的核心思想。
3. Arbitrary Constants in a General Solution | 通解中的任意常数
differential equation normally contains exactly two arbitrary constants, such as y = A e²ˣ + B e⁻²ˣ constants are independent, so the equation cannot be formed by differentiating only once.
二阶微分方程的通解通常恰好包含两个任意常数，例如 y = A e²ˣ + B e⁻²ˣ。这些常数相互独立，因此仅求一次导是无法建立方程的。
In general, the number of independent arbitrary constants in the solution tells you the order of the differential equation: one constant leads to a first order equation, and two constants lead to a second order equation.
一般来说，解中独立任意常数的个数决定了微分方程的阶数：一个常数对应一阶方程，两个常数对应二阶方程。
4. The Standard Formation Strategy | 标准建立策略
To form a second order differential equation from a given relation involving two arbitrary constants, start by differentiating the relation twice with respect to the independent variable, usually x.
要从含有两个任意常数的已知关系建立二阶微分方程，首先对自变量（通常是 x）将该关系求导两次。
/dx², use the original equation and the two differentiated equations to eliminate the two arbitrary constants by substitution or comparison.
在得到 y′ = dy/dx 和 y″ = d²y/dx² 之后，利用原方程和两个求导后的方程，通过代入或比较来消去两个任意常数。
The final equation should contain only x, y, y′ and y″, with no arbitrary constants remaining. If a constant is still present, you have not differentiated enough times.
最终方程应只含有 x、y、y′ 和 y″，不能留下任意常数。如果仍有常数，说明求导次数不够。
5. Example 1: Exponential Solutions | 示例 1：指数型解
Consider the general solution y = A e²ˣ + B e⁻²ˣ, where A and B are arbitrary constants. Differentiate once to obtain y′ = 2A e²ˣ − 2B e⁻²ˣ, and differentiate again to obtain y″ = 4A e²ˣ + 4B e⁻²ˣ.
考虑通解 y = A e²ˣ + B e⁻²ˣ，其中 A 和 B 是任意常数。求一次导得到 y′ = 2A e²ˣ − 2B e⁻²ˣ，再求一次导得到 y″ = 4A e²ˣ + 4B e⁻²ˣ。
Notice that y″ = 4(A e²ˣ + B e⁻²ˣ), so y″ = 4y. Rearranging gives the second order differential equation y″ − 4y = 0.
注意 y″ = 4(A e²ˣ + B e⁻²ˣ)，因此 y″ = 4y。移项得到二阶微分方程 y″ − 4y = 0。
y = A e²ˣ + B e⁻²ˣ ⇒ y″ − 4y = 0
This shows that every function of the form y = A e²ˣ + B e⁻²ˣ atisfies the same second order differential equation, regardless of the values of A and B.
这表明，无论 A 和 B 取何值，所有形如 y = A e²ˣ + B e⁻²ˣ 的函数都满足同一个二阶微分方程。
6. Example 2: Trigonometric Solutions | 示例 2：三角函数型解
= −3A sin 3x + 3B cos 3x, and a second differentiation gives y″ = −9A cos 3x − 9B sin 3x.
取通解 y = A cos 3x + B sin 3x。求导得 y′ = −3A sin 3x + 3B cos 3x，再求导得 y″ = −9A cos 3x − 9B sin 3x。
B sin 3x), we can write y″ = −9y. Rearranging gives the differential equation y″ + 9y = 0.
因为 y″ = −9(A cos 3x + B sin 3x)，所以可以写作 y″ = −9y。移项得到微分方程 y″ + 9y = 0。
y = A cos 3x + B sin 3x ⇒ y″ + 9y = 0
This is a standard equation for simple harmonic motion with angular frequency 3. The two arbitrary constants A and B determine the amplitude and phase of the motion.
这是角频率为 3 的简谐运动的标准方程。两个任意常数 A 和 B 决定了运动的振幅和相位。
7. Example 3: Polynomial Solutions | 示例 3：多项式型解
ary constants. The first derivative is y′ = 2Ax + B, and the second derivative is y″ = 2A.
考虑函数族 y = Ax² + Bx，其中 A 和 B 是任意常数。一阶导数为 y′ = 2Ax + B，二阶导数为 y″ = 2A。
ute A and B into the original equation y = Ax² + Bx to obtain y = (y″/2)x² + (y′ − x y″)x.
由 y″ = 2A 可得 A = y″/2。将其代入 y′ = 2Ax + B，得到 B = y′ − x y″。再把 A 和 B 代入原方程 y = Ax² + Bx，得到 y = (y″/2)x² + (y′ − x y″)x。
)y″. Multiplying through by 2 and rearranging produces the required differential equation.
化简得 y = x y′ − (x²/2)y″。两边乘以 2 并移项，得到所求的微分方程。
x² y″ − 2x y′ + 2y = 0
This example shows that the resulting differential equation is not always homogeneous with constant coefficients; the coefficients can involve x as well.
这个例子说明，得到的微分方程并不总是常系数齐次方程，系数也可以含有 x。
8. Forming Equations from Physical Contexts | 从物理情境建立方程
Second order differential equations frequently arise from Newton's second law of motion, which links force to acceleration. If a particle of mass m is attached to a spring and displaced by x from its equilibrium position, the restoring force is often modeled as −kx.
二阶微分方程经常来自牛顿第二运动定律，该定律将力与加速度联系起来。如果一个质量为 m 的质点连接在弹簧上，偏离平衡位置的位移为 x，则回复力通常建模为 −kx。
Newton' roduces d²x/dt² + (k/m)x = 0. Writing ω² = k/m gives the standard form d²x/dt² + ω² x = 0.
牛顿第二定律给出 m d²x/dt² = −kx，两边除以 m 得到 d²x/dt² + (k/m)x = 0。令 ω² = k/m，就得到标准形式 d²x/dt² + ω² x = 0。
m d²x/dt² + kx = 0 ⇒ d²x/dt² + ω² x = 0
m a physical law, and its general solution can then be written as x = A cos ωt + B sin ωt.
这里的二阶方程直接由物理定律建立，其通解随后可以写成 x = A cos ωt + B sin ωt。
9. Verifying a Derived Equation | 验证导出的方程
To check that a formed differential equation is correct, substitute the original general solution back into the equation and confirm that both sides simplify identically.
要检查建立的微分方程是否正确，可以将原通解代回方程，确认两边化简后恒等。
in ωt = −ω² y. Therefore y″ + ω² y = 0, so the solution definitely satisfies the equation.
例如，对于 y = A cos ωt + B sin ωt，求导两次得到 y″ = −Aω² cos ωt − Bω² sin ωt = −ω² y。因此 y″ + ω² y = 0，所以该解确实满足方程。
Verification is useful in an exam because it can reveal sign errors, missing factors or incorrect elimination of constants before you finalise your answer.
验证在考试中非常有用，因为它可以在你确定最终答案之前发现符号错误、漏乘因子或常数消去不正确等问题。
10. Common Mistakes and How to Avoid Them | 常见错误与避免方法
A common mistake is differentiating only once when two arbitrary constants need to be removed. One differentiation can only eliminate one constant, so you must differentiate twice for a second order equation.
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