Finding Particular Solutions Using Initial Conditions and Separation of Variables · 利用初始条件和分离变量法求特解
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| initial condition/ɪˈnɪʃl kənˈdɪʃn/ | 初始条件 | chū shǐ tiáo jiàn |
| particular solution/pəˈtɪkjʊlə səˈluːʃn/ | 特解 | tè jiě |
Pinning down the one true curve
- The general solution is a family with a constant $C$. Reality wants one curve.
- An initial condition 初始条件 — a known point $\big(x_0,y_0\big)$ — selects that single member.
- Solve the equation generally, then use the point to find $C$.
- The result is the particular solution 特解 for that situation.
锁定唯一那条真正的曲线
- 通解是带常数 $C$ 的一族。现实想要一条曲线。
- 初始条件——一个已知点 $\big(x_0,y_0\big)$——选出那单个成员。
- 先一般地解方程,再用那个点求 $C$。
- 结果就是那个情形的特解。
The single curve chosen by an initial condition is the ____ solution. · 由初始条件选出的唯一一条曲线称为 ____ 解。
One member of the general family. · 通解族中的某一个成员。
The full procedure
- 1. Separate the variables and integrate both sides (with $+C$) — the general solution.
- 2. Substitute the initial condition $\big(x_0,y_0\big)$ to solve for $C$.
- 3. Put that $C$ back in and simplify.
- 4. If asked, solve explicitly for $y$.
完整步骤
- 1. 分离变量并对两边积分(带 $+C$)——通解。
- 2. 代入初始条件 $\big(x_0,y_0\big)$ 求 $C$。
- 3. 把那个 $C$ 代回并化简。
- 4. 若被要求,显式解出 $y$。
The initial point picks one curve · 初始点确定其中一条曲线
The general solution is a whole family; the initial condition $y(x_0)=y_0$ selects the single curve through that point. · 通解是一个完整的解族；初始条件 $y(x_0)=y_0$ 从中选出经过该点的唯一一条曲线。
You apply the initial condition to find $C$... · 你应用初始条件来求解 $C$ ...
Get the general solution first, then use the point for $C$. · 先求出通解，然后用该点来确定 $C$ 。
Order the steps for a particular solution. · 排列求特解的步骤顺序。
Separate, integrate, apply the point, simplify. · 分离、积分、代入点、化简。
A worked particular solution
- Solve $\dfrac{dy}{dx}=xy$ with $y(0)=3$.
- General solution (from last lesson): $y=A\,e^{x^2/2}$.
- Apply the point: $3 = A\,e^{0}=A$, so $A=3$.
- Particular solution: $y=3\,e^{x^2/2}$.
一个特解范例
- 解 $\dfrac{dy}{dx}=xy$,带 $y(0)=3$。
- 通解(来自上一课):$y=A\,e^{x^2/2}$。
- 应用该点:$3 = A\,e^{0}=A$,所以 $A=3$。
- 特解: $y=3\,e^{x^2/2}$。
For · 支持 $y=x^2+C$ with $y(1)=5$, find $C$. · 对于 $y=x^2+C$ 且已知 $y(1)=5$ ，求 $C$ 。
$5=1+C\Rightarrow C=4$.
The particular solution is $y=x^2+4$. Find $y(3)$. · 特解为 $y=x^2+4$ 。求 $y(3)$ 。
$3^2+4=13$.
For · 支持 $\dfrac{dy}{dx}=xy$ with $y(0)=3$ (general $y=Ae^{x^2/2}$), the particular solution is... · 对于 $\dfrac{dy}{dx}=xy$ 且已知 $y(0)=3$ （通解为 $y=Ae^{x^2/2}$ ），特解是...
$y(0)=A=3$, so $y=3e^{x^2/2}$. · $y(0)=A=3$，所以 $y=3e^{x^2/2}$。
Watch the domain
- Solving for $y$ may involve $\ln$, roots, or reciprocals — mind where the solution is valid.
- The particular solution usually lives on an interval containing $x_0$, not necessarily all real $x$.
- A sign or an absolute value may need resolving using the given point (e.g. $y_0>0$ picks the $+$ branch).
- State the domain when the algebra restricts it.
注意定义域
- 解出 $y$ 可能涉及 $\ln$、根号或倒数——留意解在何处有效。
- 特解通常活在包含 $x_0$ 的一个区间上,不一定是所有实数 $x$。
- 符号或绝对值可能需要用给定点来确定(如 $y_0>0$ 选 $+$ 分支)。
- 当代数限制定义域时,说明定义域。
Apply the initial condition after integrating, to find $C$ — not before. And use the point to resolve any sign/branch ambiguity: if $\ln|y|=\dots$ and $y_0>0$, drop the absolute value as $y>0$. Substituting the point too early (like in separation of variables' warning) loses the general family.
在积分之后应用初始条件来求 $C$——不是之前。并用该点确定任何符号/分支:若 $\ln|y|=\dots$ 且 $y_0>0$,就按 $y>0$ 去掉绝对值。过早代入该点(如分离变量法警告所述)会丢失通解族。
Solve $\dfrac{dy}{dx}=2x$ with $y(1)=5$.
- General solution: $y=x^2+C$.
- Apply $y(1)=5$: $5 = 1^2 + C \Rightarrow C=4$.
- Particular solution: $y=x^2+4$.
解 $\dfrac{dy}{dx}=2x$,带 $y(1)=5$。
- 通解:$y=x^2+C$。
- 应用 $y(1)=5$:$5 = 1^2 + C \Rightarrow C=4$。
- 特解: $y=x^2+4$。
A particular solution is the one member of the general family fixed by an initial condition. Solve generally (integrate, keep $+C$), substitute the point $\big(x_0,y_0\big)$ to find $C$, and simplify. Use the point to settle any sign/branch and note the valid domain.
特解是由初始条件确定的通解族中的那一个成员。先一般地解(积分、保留 $+C$),代入点 $\big(x_0,y_0\big)$ 求 $C$,再化简。用该点确定任何符号/分支,并注明有效定义域。