Differential equations and initial conditions · Équations différentielles et conditions initiales
| English | Français |
|---|---|
| initial condition/ɪˈnɪʃl kənˈdɪʃn/ | condition initiale |
| characteristic root/ˌkærɪktəˈrɪstɪk ruːt/ | racine caractéristique |
A decision before an answer
- The same differential equation describes infinitely many motions until initial conditions select one.
- Your goal: Solve separable first-order equations.
Read the relationship
- For y′=ky, separation and integration give y=Ce^(kt); recover the zero solution if division by y was used. An initial value determines C.
- Solve constant-coefficient second-order equations.
For y′=2y and y(0)=3, the solution is:
The initial condition fixes C=3.
Use the defining rule
- A linear second-order constant-coefficient equation uses a characteristic polynomial. Distinct roots yield exponentials; a repeated root requires (C1+C2t)e^(rt).
- Use initial conditions and uniqueness conditions.
For a repeated root r=−1, the general solution includes:
The factor t supplies a second independent solution.
Check the conditions
- Complex roots α±iβ give e^(αt)(C1 cos βt+C2 sin βt). The real motion includes both amplitude and phase information.
- Use initial conditions and uniqueness conditions.
Solve y″+4y=0 with y(0)=3 and y′(0)=4. The characteristic roots are ±2i, so y=A cos2t+B sin2t. The first condition gives A=3; differentiating gives y′(0)=2B=4, so B=2.
For y′=3y and y(0)=2, C = ____.
At t=0 the exponential is 1.
Apply the task format
- Existence and uniqueness depend on hypotheses near the initial point. A singular coefficient or a failure of local Lipschitz behaviour can defeat the familiar uniqueness conclusion.
- Use initial conditions and uniqueness conditions.
A repeated characteristic root needs the factor t; two copies of the same exponential are not independent solutions.
Which answer fits this case? · Quelle réponse correspond à ce cas ?
Solve separable first-order equations · Résoudre les équations séparables du premier ordre
Initial conditions can select a unique solution when the relevant uniqueness hypotheses hold.
Check the hypotheses rather than assuming uniqueness at a singular point.
Keep the distinctions
- initial condition 初始条件 — A value of the solution or derivative specified at a starting point.
- characteristic root 特征根 — A root of the polynomial governing exponential solutions.
- Solve separable first-order equations.
- Solve constant-coefficient second-order equations.
- Use initial conditions and uniqueness conditions.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.