積分因子と非斉次微分方程式
導入済み| English | 意味 |
|---|---|
| integrating factor/ˈɪntɪɡreɪtɪŋ ˈfæktə/ | 積分因子 |
| particular solution/pəˈtɪkjʊlə səˈluːʃn/ | 特解 |
A decision before an answer
- A forcing input changes a system’s response. Solving only the unforced equation leaves out the sustained contribution; dividing by y can also lose a valid solution.
- Your goal: Solve first-order linear equations using an integrating factor.
Read the relationship
- A first-order linear equation has the form y′+p(x)y=q(x) on an interval where its coefficients are continuous. It need not be separable. Set μ(x)=exp(∫p(x)dx). Since μ′=pμ, multiplying gives (μy)′=μq, hence y=μ⁻¹(∫μq dx+C). Choose any convenient antiderivative for p; its integration constant only rescales μ and cancels from the solution. Apply initial data after integration, and keep any singular coefficient points outside the chosen interval.
- Construct the homogeneous and particular parts of a constant-coefficient solution.
Which integrating factor works for y′+3y=x?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
p(x)=3, so μ=exp(∫3 dx)=e^(3x). Its derivative is 3μ.
Use the defining rule
- For y′+2xy=x with y(0)=1, μ=e^(x²). Then (e^(x²)y)′=x e^(x²), whose integral is e^(x²)/2+C. Thus y=1/2+C e^(−x²), and the initial condition gives C=1/2. Direct substitution checks y′+2xy=x. The integrating factor makes the left side a product derivative; it is not an extra factor that remains multiplying the original right side in the final answer.
- Handle resonance and verify the result in the original equation.
For y″−3y′+2y=4, which constant is a particular solution?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Both derivatives vanish for a constant. The equation reduces to 2y=4, so y=2.
Check the conditions
- For ay″+by′+cy=r(x) with a nonzero, solve the characteristic equation aλ²+bλ+c=0 for the homogeneous part. Distinct real roots give two exponentials; a repeated root λ gives (C₁+C₂x)e^(λx); roots α±iβ give e^(αx)(C₁ cos βx+C₂ sin βx). The complete solution is y_h+y_p. A polynomial forcing suggests a polynomial particular trial; an exponential or sine/cosine forcing suggests the corresponding family, with enough coefficients to account for differentiation.
- Handle resonance and verify the result in the original equation.
For y″−3y′+2y=4, the characteristic roots are 1 and 2. A constant particular solution y_p=2 gives 2y_p=4, so y=C₁e^x+C₂e^(2x)+2. For y′+2xy=x and y(0)=1, the separate first-order solution is y=(1+e^(−x²))/2. These illustrate distinct methods; neither equation becomes homogeneous just because its left side is linear.
For y′+2xy=x with y(0)=1, y=1/2+C e^(−x²). Give C as a decimal: ____.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
At x=0 the exponential is 1, so 1=1/2+C gives C=1/2.
Apply the task format
- If a particular trial duplicates a homogeneous solution, it cannot produce the forcing: multiply by x once for a simple root, twice for a repeated root. For y″−3y′+2y=e^x, the naive Ke^x is annihilated. Trying Kxe^x gives −Ke^x, so y_p=−xe^x. Initial conditions determine the homogeneous constants only after adding y_p. Substitute the final function into the original equation to check signs and forcing; an initial-value check alone cannot verify the differential equation.
- Handle resonance and verify the result in the original equation.
Keep the particular solution. A resonant trial needs an x factor. The integrating-factor derivation applies on a valid coefficient interval and should not divide by y or silently discard zero solutions.
このケースに当てはまる答えはどれか？
積分因子を用いて一次線形方程式を解く
Ke^x can be a particular solution of y″−3y′+2y=e^x for a suitable constant K.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
e^x is homogeneous, so the differential operator sends every Ke^x to zero. A resonant trial requires an extra x factor.
Keep the distinctions
- integrating factor 积分因子 — A multiplier that turns a first-order linear equation into a product derivative.
- particular solution 特解 — One solution supplying the specified nonhomogeneous forcing.
- Solve first-order linear equations using an integrating factor.
- Construct the homogeneous and particular parts of a constant-coefficient solution.
- Handle resonance and verify the result in the original equation.
Match each term with its precise meaning in this lesson.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
The order follows the stated process; check each stage before the next.