Finding Particular Solutions Using Initial Conditions and Separation of Variables
Introduced| English | Meaning |
|---|---|
| initial condition/ɪˈnɪʃl kənˈdɪʃn/ | A specified starting value used to determine a particular solution of an equation or model. |
| particular solution/pəˈtɪkjʊlə səˈluːʃn/ | A complete solution whose arbitrary constants have been fixed by conditions. |
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.
The single curve chosen by an initial condition is the ____ solution.
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One member of the general family.
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 $y$.
The constant gives a family of curves; an initial condition picks out one
The initial point picks one curve
The general solution is a whole family; the initial condition $y(x_0)=y_0$ selects the single curve through that point.
You apply the initial condition to find $C$...
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Get the general solution first, then use the point for $C$.
Order the steps for a particular solution.
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Separate, integrate, apply the point, simplify.
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: $y=3\,e^{x^2/2}$.
For $y=x^2+C$ with $y(1)=5$, find $C$.
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$5=1+C\Rightarrow C=4$.
The particular solution is $y=x^2+4$. Find $y(3)$.
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$3^2+4=13$.
For $\dfrac{dy}{dx}=xy$ with $y(0)=3$ (general $y=Ae^{x^2/2}$), the particular solution is...
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$y(0)=A=3$, so $y=3e^{x^2/2}$.
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.
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.
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: $y=x^2+4$.
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.
Write from an initial condition
- The general solution is y = x squared + C.
- The curve passes through (1, 5).
What is the particular solution?
5 - 1 fixes the constant.
y = x squared + C passes through (1, 5). The particular solution is what?
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C = 5 - 1 = 4.