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.
What a learner can do afterwards
- 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
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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.
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.
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 questions, readings and flashcards written for this dataset. OpenStax Calculus Volume 2 was consulted for mathematical facts. No publisher prose or WeBWorK problem text is reproduced.
More to read
Why the factor works
Start with y' + P(x)y = Q(x). Multiplying by a function M gives My' + MPy = MQ. The derivative of My is My' + M'y. These expressions match when M' = PM.
For example, y' + 4y = 12 uses M = e^(4x). The multiplied equation is (e^(4x)y)' = 12e^(4x). Integration gives e^(4x)y = 3e^(4x) + C. Dividing gives y = 3 + C e^(-4x).
To check, differentiate: y' = -4C e^(-4x). Adding 4y leaves 12. The cancellation holds for every constant C.
Choosing a factor and finding a solution use constants differently. Any nonzero constant multiple of M works, so one chosen factor is enough. The constant in the solution must remain because it describes the solution family.
Check the interval and both conditions
An equation a(x)y' + b(x)y = c(x) can be divided by a(x) only where a(x) is nonzero. Work on an interval where the resulting P and Q are continuous. A formula obtained there does not automatically solve the original problem across an excluded point.
Consider y' + 4y = 12 with y(0) = 7. Its family is y = 3 + C e^(-4x). At zero, e^0 = 1, so 7 = 3 + C and C = 4.
The selected function is y = 3 + 4e^(-4x). Its derivative is -16e^(-4x), so y' + 4y = 12. Its value at zero is 7. Both conditions hold, and the formula is valid for all real x.
A proposed function can match an initial value without satisfying the differential equation. Check the derivative and substitution as well as the initial value.
From a physical model to a linear equation
Let y be dissolved salt mass in a tank. A well-mixed tank of volume V has salt concentration y/V at its outlet. With equal inlet and outlet volume flows r, the volume stays fixed. If inlet concentration is c, the input salt mass rate is rc and the output salt mass rate is ry/V. Therefore y' = rc - ry/V.
For constant positive V and r, this is y' + (r/V)y = rc. Its integrating factor is e^((r/V)t), and integration gives y = cV + C e^(-(r/V)t). An initial salt mass fixes C. The exponential decays, so the salt mass approaches cV under these assumptions.
Cooling uses a related model. If the surroundings remain at temperature Ts and k is a positive constant, T' = -k(T - Ts). Rearranging gives T' + kT = kTs. The solution T = Ts + C e^(-kt) approaches Ts. These conclusions depend on the stated model assumptions.
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