Phương trình vi phân và điều kiện ban đầu
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| initial condition/ɪˈnɪʃl kənˈdɪʃn/ | Điều kiện ban đầu |
| characteristic equation/ˌkærɪktəˈrɪstɪk ɪˈkweɪʒn/ | The polynomial equation in r obtained by trying solutions of the form e to the rt. |
One equation, many motions
- $y'' + 4y = 0$ describes every oscillation of a certain spring. Two initial conditions 初始条件 — position and velocity — pick out one.
- GRE differential equations test separable and linear first-order equations, constant-coefficient second-order equations, and uniqueness.
Separable equations
- $y' = ky$: separate, $\int \frac{dy}{y} = \int k\,dt$, so $y = Ce^{kt}$. Dividing by $y$ hid $y = 0$; it is included as $C = 0$.
- $y' = xy^2$, $y(0) = 1$: $-\frac1y = \frac{x^2}{2} - 1$, so $y = \dfrac{2}{2 - x^2}$, valid only for $|x| < \sqrt2$.
$y' = 2y$, $y(0) = 5$. Find $y(\ln 2)$.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
$y = 5e^{2t}$, $e^{2\ln2} = 4$.
Characteristic equations 特征方程
- $ay'' + by' + cy = 0$ → $ar^2 + br + c = 0$.
- Distinct real roots: $C_1e^{r_1t} + C_2e^{r_2t}$. Repeated root: $(C_1 + C_2t)e^{rt}$. Complex $\alpha \pm i\beta$: $e^{\alpha t}(C_1\cos\beta t + C_2\sin\beta t)$.
$y'' + 4y = 0$, $y(0) = 3$, $y'(0) = 4$. Roots $\pm 2i$: $y = A\cos 2t + B\sin 2t$. $y(0) = A = 3$; $y'(0) = 2B = 4$, so $B = 2$. Then $y(\pi/4) = 2$.
Which solution form?
Read the general solution from the characteristic roots.
The general solution of $y'' - 6y' + 9y = 0$ is
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The characteristic equation r² − 6r + 9 = (r − 3)² = 0 has the repeated root 3, so the second solution is t e^{3t}: y = (C₁ + C₂t)e^{3t}. Writing C₁e^{3t} + C₂e^{3t} gives only one independent solution.
$y'' + 9y = 0$, $y(0) = 0$, $y'(0) = 6$. Find $y(\pi/6)$.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
$y = 2\sin 3t$; $\sin(\pi/2) = 1$.
The characteristic roots are $3 \pm 5i$. The solutions oscillate as $\cos \beta t$ and $\sin \beta t$. Find $\beta$.
Độ khó 2/5 · 140 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
Roots $\alpha \pm i\beta$ give $e^{\alpha t}\cos\beta t$ and $e^{\alpha t}\sin\beta t$.
Existence and uniqueness
- $y' = f(t, y)$ has a unique local solution when $f$ and $\partial f/\partial y$ are continuous near the initial point.
- $y' = 3y^{2/3}$, $y(0) = 0$ has both $y = 0$ and $y = t^3$: $\partial f/\partial y$ blows up at $y = 0$.
$y' = 3y^{2/3}$, $y(0) = 0$ has exactly one solution.
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
False. y = 0 and y = t³ both satisfy y′ = 3y^{2/3} with y(0) = 0. Uniqueness fails because 3y^{2/3} is not Lipschitz in y near y = 0.
Common traps
- A repeated root needs the factor $t$; $C_1e^{rt} + C_2e^{rt}$ is just one solution.
- Losing $y = 0$ (or another constant solution) when dividing during separation.
Recap
- Separate and integrate; check constant solutions and the interval of validity.
- Read the solution form from the characteristic roots; use both initial conditions.
- Uniqueness needs $f$ and $f_y$ continuous near the initial point.
- characteristic equation: The polynomial equation in r obtained by trying solutions of the form e to the rt.