--- title: "First-Order Linear Equations and the Integrating Factor" description: "Multiply through by a factor chosen to turn the left side into the derivative of a product, then integrate once. The factor is built from the coefficient of y." canonical: https://lightmysky.com/learn/mathematics/first-order-linear-equations-and-the-integrating-factor-mt_BXmt2pbWp8 source: https://lightmysky.com/learn/mathematics/first-order-linear-equations-and-the-integrating-factor-mt_BXmt2pbWp8.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # First-Order Linear Equations and the Integrating Factor Multiply through by a factor chosen to turn the left side into the derivative of a product, then integrate once. The factor is built from the coefficient of y. Subject: Mathematics · Area: Differential Equations · Ages 19 to 20 Page: https://lightmysky.com/learn/mathematics/first-order-linear-equations-and-the-integrating-factor-mt_BXmt2pbWp8 ## Ready when they can - Identify first-order linear equations, put them in standard form and compute an integrating factor - Use the product rule to solve a linear equation, apply an initial condition and check the solution - Calculate an instantaneous salt mass balance in a constant-volume mixing tank and find the limiting mass of a supplied constant-coefficient tank equation ## Lesson: The factor that builds a product A first-order linear equation has the form a(x)y' + b(x)y = c(x). Its coefficients depend only on x. The unknown y and its derivative y' occur to the first power, without products such as yy'. Where a(x) ≠ 0, divide every term by a(x): y' + P(x)y = Q(x), with P = b/a and Q = c/a. Choose an interval where P and Q are continuous. For 3y' + 9y = 12, division gives y' + 3y = 4. Write exp(u) for e^u. Choose M = exp(∫P dx), so M' = PM. Separation gives the same result from M'/M = P. Set the integration constant to 0 when choosing M; any nonzero constant multiple also works. For constant P = k, M = e^(kx) and M' = kM. For P = ax, ∫P dx = ax²/2. For P = k/x on x > 0, ∫P dx = k ln x and M = x^k. An integrating factor must not vanish. *(drawing: The product rule gives (My)' = M(y' + Py). Thus multiplying y' + Py = Q by M gives (My)' = MQ.)* **Example.** Integrate (My)' = MQ and divide: y = (∫MQ dx + C)/M. Keep this C for the solution family. For y' + 3y = 2e^x, choose M = e^(3x). Multiplication gives (e^(3x)y)' = 2e^(4x). Integration gives e^(3x)y = e^(4x)/2 + C. Dividing gives y = e^x/2 + C e^(-3x), valid for all real x. An initial value fixes C after integration. In y = e^x/2 + C e^(-3x), y(0) = 2 gives 2 = 1/2 + C, so C = 3/2. Check the equation too: y' = e^x/2 - 3C e^(-3x), so y' + 3y = 2e^x. Checking y(0) alone cannot verify a solution. The function must satisfy the differential equation throughout its stated interval. Let y(t) be salt mass in a well-mixed tank of constant volume V. Equal inlet and outlet flows r keep V fixed. For inlet concentration c, salt enters at rc and leaves at ry/V. Thus y' + (r/V)y = rc. With V = 10 L, r = 1 L/min and c = 0.3 kg/L, y' + 0.1y = 0.3. The factor e^(0.1t) gives y = 3 + C e^(-0.1t). At any instant, subtract output mass rate from input mass rate. For constants k > 0 and b, y' + ky = b gives y = b/k + C e^(-kt). The exponential tends to zero as t increases, so each finite initial value approaches b/k. Setting y' = 0 finds the equilibrium; the decaying term proves convergence. Cooling with fixed room temperature Ts is modeled by T' = -k(T - Ts), or T' + kT = kTs. Its solution T = Ts + C e^(-kt) approaches Ts. **Recap.** Use a valid interval and standard form. Build M, use (My)' = MQ, integrate and divide. Apply any initial value and check the result. *Lesson, practice questions, worked solutions and flashcards written for this dataset. OpenStax Calculus Volume 2 was consulted for mathematical facts; no publisher prose is reproduced.* ## Practice 13 questions on this page, each with its working shown. ## Needs first - [Separable Equations](https://lightmysky.com/learn/mathematics/separable-equations-mt_itSTsLTkwQ) ## Optional connections - [The Product and Quotient Rules](https://lightmysky.com/learn/mathematics/the-product-and-quotient-rules-mt_NJ7INJ0jfv) ## Opens up - [Green's Functions for Boundary Value Problems](https://lightmysky.com/learn/mathematics/greens-functions-for-boundary-value-problems-mt_2ctFrfKIgW) - [The Logistic Equation and Saturating Growth](https://lightmysky.com/learn/mathematics/the-logistic-equation-and-saturating-growth-mt_7Y_48Ga33L)