Second-order linear differential equations
導入済み| English | 意味 |
|---|---|
| complementary function/ˌkɒmplɪˈmentəri ˈfʌŋkʃn/ | 補関数 |
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.
Why does a second-order equation need TWO conditions to fix a solution?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The complementary function y=Ae^x+Be^(2x) carries two arbitrary constants. Each condition gives one linear equation in A and B; only two independent conditions fix both.
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
For y double prime+ay prime+by=f(x), solve the auxiliary quadratic for the homogeneous part
Compare the model with the worked case and explain one change.
For A+B=1 and A+2B=0, find A.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Subtract A+B=1 from A+2B=0: B=-1, hence 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.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Subtract the first equation from the second to obtain B=-1.
One initial value is always sufficient to determine both constants in a second-order general solution.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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.
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.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Factor m²-3m+2=(m-1)(m-2); the larger root is 2.
Match each part of y″-3y′+2y=f(x) to its role.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
The general solution is complementary function plus particular integral; the auxiliary roots decide the form, with special cases for repeated or complex roots.
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.