Integrating factors and nonhomogeneous differential equations
| English | Español |
|---|---|
| integrating factor/ˈɪntɪɡreɪtɪŋ ˈfæktə/ | factor integrante |
| particular solution/pəˈtɪkjʊlə səˈluːʃn/ | solución particular |
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?
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?
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: ____.
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.
Which answer fits this case?
Solve first-order linear equations using an integrating factor
Ke^x can be a particular solution of y″−3y′+2y=e^x for a suitable constant K.
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.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.