Second-order linear differential equations · Equações diferenciais lineares de segunda ordem
| English | Português |
|---|---|
| complementary function/ˌkɒmplɪˈmentəri ˈfʌŋkʃn/ | função complementar |
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? · Qual descrição define corretamente a função complementar?
The general solution of the associated homogeneous linear differential equation. · A solução geral da equação diferencial linear homogênea associada.
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 · Equações diferenciais lineares de segunda ordem
For y double prime+ay prime+by=f(x), solve the auxiliary quadratic for the homogeneous part · Para y''+ay'+by=f(x), resolva a equação auxiliar quadrática para a parte homogênea
Compare the model with the worked case and explain one change. · Compare o modelo com o caso resolvido e explique uma mudança.
For A+B=1 and A+2B=0, find A. · Para A+B=1 e A+2B=0, encontre A.
Subtract A+B=1 from A+2B=0: B=-1, hence A=2. · Subtraia A+B=1 de A+2B=0: B=-1, portanto 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. · Para essas condições, encontre B.
Subtract the first equation from the second to obtain B=-1. · Subtraia a primeira equação da segunda para obter B=-1.
One initial value is always sufficient to determine both constants in a second-order general solution. · Um único valor inicial é sempre suficiente para determinar ambas as constantes em uma solução geral de segunda ordem.
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. · Duas constantes arbitrárias exigem duas condições independentes. Se a integral particular de prova duplica um termo da função complementar, multiplique a prova por x conforme necessário. Não descarte uma solução oscilatória válida apenas porque as raízes auxiliares são complexas.
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. · Encontre a raiz maior de m²-3m+2=0.
Factor m²-3m+2=(m-1)(m-2); the larger root is 2. · Fatore m²-3m+2=(m-1)(m-2); a raiz maior é 2.
Match each part of a complete solution to its purpose. · Associe cada parte de uma solução completa ao seu propósito.
An assumption justifies the model; a check tests the result; interpretation connects it to the question. · Uma suposição justifica o modelo; uma verificação testa o resultado; a interpretação conecta-o à pergunta.
Use this in your course
- edexcel IAL pure 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.