Second-order linear differential equations · 二阶线性微分方程
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| complementary function/ˌkɒmplɪˈmentəri ˈfʌŋkʃn/ | 互补函数 | hù bǔ hán shù |
Why do we need two initial conditions?
- A displacement model involves both velocity and acceleration. Solving only a first-order rate equation cannot capture both initial conditions.
- This lesson studies complementary function 互补函数: The general solution of the associated homogeneous linear differential equation.
Choose the mathematical structure
- For y double prime+ay prime+by=f(x), solve the auxiliary quadratic for the homogeneous part. Add a suitable particular integral for the forcing term. Repeated and complex roots require their correct forms.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines complementary function? · 下列哪项描述正确定义了齐次解（补函数）？
The general solution of the associated homogeneous linear differential equation. · 相关齐次线性微分方程的通解。
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
For y double prime-3y prime+2y=0, the auxiliary equation is m²-3m+2=0 with roots 1,2. Hence y=Ae^x+Be^(2x). With y(0)=1 and y prime(0)=0, A+B=1 and A+2B=0, so A=2,B=-1.
Second-order linear differential equations · 二阶线性微分方程
For y double prime+ay prime+by=f(x), solve the auxiliary quadratic for the homogeneous part · 对于 y''+ay'+by=f(x)，求解对应齐次部分的辅助二次方程
Compare the model with the worked case and explain one change. · 对比模型与已解案例，并说明其中一处变化。
For A+B=1 and A+2B=0, find A. · 已知 A+B=1 且 A+2B=0，求 A。
Subtract A+B=1 from A+2B=0: B=-1, hence A=2. · 从 A+B=1 中减去 A+2B=0：得 B=-1，因此 A=2。
Test a tempting shortcut
- Two arbitrary constants need two independent conditions. If the trial particular integral duplicates a complementary-function term, multiply the trial by x as required. Do not discard a valid oscillatory solution because the auxiliary roots are complex.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
One initial value is always sufficient to determine both constants in a second-order general solution. This claim is false. Explain which definition or assumption it violates.
For those conditions, find B. · 在这些条件下，求 B 的值。
Subtract the first equation from the second to obtain B=-1. · 用第二个方程减去第一个方程，得到 B=-1。
One initial value is always sufficient to determine both constants in a second-order general solution. · 在二阶通解中，通常只需一个初值即可确定两个常数。
Two arbitrary constants need two independent conditions. If the trial particular integral duplicates a complementary-function term, multiply the trial by x as required. Do not discard a valid oscillatory solution because the auxiliary roots are complex. · 两个任意常数需要两个独立条件。若特解试探形式与齐次解项重复，需乘以 x 进行调整。不要因辅助根为复数而丢弃有效的振荡解。
Interpret a new situation
- Differentiate the final expression and substitute into the original differential equation. Check both initial conditions separately, and interpret the permitted solution interval.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find the larger root of m²-3m+2=0. · 求 m²-3m+2=0 的较大根。
Factor m²-3m+2=(m-1)(m-2); the larger root is 2. · 因式分解 m²-3m+2=(m-1)(m-2)；较大根为 2。
Match each part of a complete solution to its purpose. · 将完整解答的每个部分与其目的相匹配。
An assumption justifies the model; a check tests the result; interpretation connects it to the question. · 假设用于论证模型的合理性；检查用于验证结果；解释用于将其与问题建立联系。
Use this in your course
- edexcel IAL further mathematics; official unit FP2. Other-unit enrichment is identified in the scope review; it is not an extra cash-in requirement.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The general solution of the associated homogeneous linear differential equation. Choose the relationship, show the method, check its assumptions and interpret the result.