First-Order Linear Equations and the Integrating Factor · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.
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The factor that builds a product

Mathematics · Differential Equations · ages 19-20
Name ______________________   Date ____________

  1. Lee proposes y = e^(-x) for y' + 2y = 0 with y(0) = 1. This function satisfies both parts of the initial-value problem.

    Circle one:   True   False

  2. Which equation is first-order linear in the unknown function y(x)?

    • y' + (x+2)y² = 1
    • y y' + (x+2)y = 1
    • y' + (x+2)y = 1
    • (y')² + (x+2)y = 1

  3. The equation y' + 2y = 6 has solution family y = 3 + C e^(-2x). What value of C makes y(0) = 8?

    Answer: ______________

  4. For y' + 2y = 5, which is an integrating factor?

    • M = e^(-2x)
    • M = e^(2x)
    • M = e^(x²)

  5. For y' + (2/x)y = 1 on x > 0, choose M = exp(∫(2/x) dx) with integration constant 0. What is M(3)?

    Answer: ______________

  6. A tank model is dy/dt + 0.5y = 2, with t in minutes and y in kg. From any finite initial amount, what value in kg does y approach as t increases?

    Answer: ______________

  7. Solve y' + 2y = e^x. Which family is the solution?

    • y = e^x/2 + C e^(-2x)
    • y = e^x/3 + C e^(2x)
    • y = e^x/3 + C e^(-2x)

  8. For y' + P(x)y = Q(x), choose a nonzero M with M' = PM. Multiplying by M makes the left side (My)'.

    Circle one:   True   False

  9. Pat uses M = e^(4x) for y' + 2y = 3. Which comparison shows that this factor does not give the required product derivative?

    • M' = 2M, and PM = 2M.
    • M' = 4M, but PM = 2M.
    • M' = 4M, and PM = 4M.

  10. To check a proposed solution y = (e^(2x))/7, calculate y' + 5y. Which expression results?

    • e^(2x)
    • (6/7)e^(2x)
    • 7e^(2x)
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Answer key

For grown-ups. Fold this page away before handing over the rest.

The factor that builds a product W1-mt_BXmt2pbWp8-s1

  1. False · The initial value holds, but y' + 2y = -e^(-x) + 2e^(-x) = e^(-x), which is not zero.
  2. y' + (x+2)y = 1 · In y' + (x+2)y = 1, the coefficients depend only on x, and y and y' occur to the first power without being multiplied together.
  3. 5 · At x = 0, the family gives 8 = 3 + C because e^0 = 1. Thus C = 5.
  4. M = e^(2x) · Standard form gives P = 2. Integrating gives 2x, so M = e^(2x) is an integrating factor.
  5. 9 · On x > 0, the chosen integral is 2 ln x. Thus M = x² and M(3) = 9.
  6. 4 · The family is y = 4 + C e^(-0.5t). The exponential tends to zero, so y approaches 4 kg.
  7. y = e^x/3 + C e^(-2x) · Multiplication by e^(2x) gives (e^(2x)y)' = e^(3x). Integrate and divide to obtain y = e^x/3 + C e^(-2x).
  8. True · The product rule gives (My)' = My' + M'y. Substituting M' = PM gives M(y' + Py).
  9. M' = 4M, but PM = 2M. · The factor must satisfy M' = PM. Here M' = 4M but PM = 2M, so the proposed factor does not match.
  10. e^(2x) · Differentiation gives y' = (2/7)e^(2x), while 5y = (5/7)e^(2x). Adding gives e^(2x).
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