C.3 · Differential equations and initial conditions

GRE · GRE Subject Test · GRE 数学 · 知识点 8

Scope and prerequisites

Undergraduate GRE preparation. Local objectives within the reviewed ETS scope; this is original teaching, not an official test or score predictor.

Prerequisites: Differentiation, integration, exponentials and quadratic equations.

Solve separable first-order equations
Solve constant-coefficient second-order equations
Use initial conditions and uniqueness conditions

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.

词汇训练

English	中文	拼音
initial condition/ɪˈnɪʃl kənˈdɪʃn/	初始条件	chū shǐ tiáo jiàn
characteristic root/ˌkærɪktəˈrɪstɪk ruːt/	特征根	tè zhēng gēn

Choose and justify a method

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.

A linear second-order constant-coefficient equation uses a characteristic polynomial. Distinct roots yield exponentials; a repeated root requires (C1+C2t)e^(rt).

Complex roots α±iβ give e^(αt)(C1 cos βt+C2 sin βt). The real motion includes both amplitude and phase information.

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.

Worked reasoning

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.

Differential equations and initial conditions: course example — Original course illustration; its values belong to the worked example, not the later practice.

Conditions and counterexamples

A repeated characteristic root needs the factor t; two copies of the same exponential are not independent solutions.

Guided application

Solve $y''-4y'+4y=0$ with $y(0)=1$, $y'(0)=0$. Check both initial conditions and the equation.

Worked solution

The characteristic polynomial is $(r-2)^2$. Independent solutions are $e^{2t}$ and $te^{2t}$, hence $y=(C_1+C_2t)e^{2t}$. The initial value gives $C_1=1$; the derivative at zero gives $2C_1+C_2=0$, so $C_2=-2$.

$$y(t)=(1-2t)e^{2t},\quad y'(t)=-4te^{2t},\quad y''(t)=(-4-8t)e^{2t}.$$

Substitution gives $y''-4y'+4y=0$, and the two initial values hold. Two copies of $e^{2t}$ would not span the solution space.

Independent transfer

Show that the initial-value problem $y'=2\sqrt{|y|}$, $y(0)=0$ has more than one solution for $t\ge0$. Identify the missing standard uniqueness hypothesis.

Check after attempting

One solution is identically zero. For any $a\ge0$, set $y_a(t)=0$ for $0\le t\le a$ and $y_a(t)=(t-a)^2$ for $t>a$. The derivative is zero on the first interval and $2(t-a)$ on the second; both sides have derivative zero at a. Therefore $y_a$ is differentiable and satisfies the equation and initial value. The right side is continuous but is not locally Lipschitz in y at zero: $|2\sqrt h-0|/|h|=2/\sqrt h$ is unbounded. Existence does not imply uniqueness, and division by $\sqrt y$ would lose the waiting-time solutions.