C.3 · Phương trình vi phân và điều kiện ban đầu
GRE Subject Test · GRE Mathematics Chủ đề 8 2:38 Giọng đọc tiếng Anh · phụ đề tiếng Anh + 中文 được ghi trực tiếp
Chương
Biên bản ghi âm
This unit solves separable first-order equations and constant-coefficient second-order equations, and uses initial conditions and uniqueness conditions.
本单元求解可分离的一阶方程和常系数二阶方程，并使用初始条件和唯一性条件。
For y prime equals k y, separate and integrate to get y equals C e to the k t.
对 y′ = ky，分离变量并积分得 y = Ce^{kt}。
Dividing by y hid the zero solution, so recover it: C equals zero covers it.
除以 y 时隐藏了零解，所以要把它找回来：C = 0 包含它。
The initial value then fixes C.
然后由初值确定 C。
A constant-coefficient equation uses its characteristic polynomial.
常系数方程使用特征多项式。
Distinct roots give two exponentials.
不同的根给出两个指数函数。
A repeated root needs C1 plus C2 t, times the exponential.
重根需要 (C₁ + C₂t) 乘以指数函数。
Complex roots alpha plus or minus i beta give a decaying or growing oscillation.
复根 α ± iβ 给出衰减或增长的振荡。
Existence and uniqueness depend on hypotheses near the initial point.
存在性和唯一性取决于初始点附近的条件。
A singular coefficient, or a failure of local Lipschitz behaviour, can defeat the familiar uniqueness conclusion.
奇异的系数，或不满足局部利普希茨条件，都可能使熟悉的唯一性结论失效。
The characteristic roots are plus or minus 2i, so y equals A cos 2t plus B sin 2t.
特征根是 ±2i，所以 y = A cos 2t + B sin 2t。
The first condition gives A equals 3; the derivative at zero is 2B, which is 4, so B is 2.
第一个条件给出 A = 3；零点处导数为 2B = 4，所以 B = 2。
So y equals 3 cos 2t plus 2 sin 2t: an oscillation of amplitude root 13.
于是 y = 3cos 2t + 2sin 2t：振幅为 √13 的振荡。
A repeated characteristic root needs the factor t; two copies of the same exponential are not independent solutions.
重特征根需要因子 t；同一个指数函数的两份拷贝不是两个独立的解。
The characteristic polynomial is r minus 2, squared, so y equals C1 plus C2 t, times e to the 2t.
特征多项式是 (r − 2)²，所以 y = (C₁ + C₂t)e^{2t}。
The initial value gives C1 equals 1; the derivative gives 2 C1 plus C2 equals 0, so C2 is minus 2.
初值给出 C₁ = 1；导数条件给出 2C₁ + C₂ = 0，所以 C₂ = −2。
Substituting back checks the equation and both conditions.
代回去检验方程和两个条件。
One solution is identically zero.
一个解是恒等于零。
For any a, another waits at zero until a, then rises as t minus a squared; it is differentiable at a and satisfies the equation.
对任意 a，另一个解在零处停留到 a，然后按 (t − a)² 上升；它在 a 处可导并满足方程。
The right side is continuous but not Lipschitz at zero, so existence holds and uniqueness does not.
右端连续，但在零处不满足利普希茨条件，所以存在性成立，唯一性不成立。
Check yourself.
自我检测。
y prime equals k y gives C e to the k t.
y′ = ky 的解是 Ce^{kt}。
A repeated root needs the factor t.
重根需要因子 t。
Roots plus or minus 2i give cos 2t and sin 2t.
根 ±2i 给出 cos 2t 和 sin 2t。
The square-root equation has many solutions from zero.
平方根方程从零出发有许多解。
And continuity alone gives existence, not uniqueness.
仅有连续性只保证存在，不保证唯一。