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First order equations and integrating factors questions

Multiply a stubborn first order equation by exactly the right function and its left side folds into one derivative, ready to integrate whole.

6 original questions · 23 marks · the first order equations and integrating factors notes · Differential equations

Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.

• 2 marks · warm-up

  For an equation in the form dy/dx + P(x)y = Q(x), write down the integrating factor and state what multiplying by it achieves.

  Worked answer

  IF = e^{∫P dx}. After multiplying through, the left side is exactly the derivative of (IF × y), so one integration solves the equation. B1 for the integrating factor, B1 for the product-rule statement.

• 2 marks · write · warm-up

  Write down the integrating factor for dy/dx + y tan x = cos x, simplifying it as far as possible.

  Worked answer

  ∫tan x dx = ln(sec x), so IF = e^{ln sec x} = sec x. M1 for ∫tan x dx = ln(sec x), A1 for sec x. A log-form integral collapses when it is exponentiated. Leaving the factor as e^{ln sec x} gets the method mark but not the accuracy mark, and it makes every later line harder to handle.

• 5 marks

  Solve dy/dx + y/x = 6x for x > 0, given that y = 3 when x = 1.

  Worked answer

  Here P = 1/x, so IF = e^{ln x} = x. Multiplying through, d(xy)/dx = 6x².
  Integrating gives xy = 2x³ + c. The constant belongs on this line, before dividing by x; introducing it afterwards produces cx instead of c and loses the accuracy mark.
  Then y = 3 at x = 1 gives 3 = 2 + c, so c = 1 and y = 2x² + 1/x.
  Check: dy/dx + y/x = (4x − 1/x²) + (2x + 1/x²) = 6x.
  M1 A1 for the integrating factor, M1 for xy = 2x³ + c, A1 for c = 1, A1 for the final solution.

• 4 marks · find

  Find the general solution of dy/dx − 2y = e^5x.

  Worked answer

  P = −2, so IF = e^−2x and d(e^−2xy)/dx = e^3x. The sign of P travels into the exponent unchanged; e^2x here would leave a left side that is nobody's derivative.
  Integrating: e^−2xy = e^3x/3 + C, so y = e^5x/3 + Ce^2x. M1 A1 for the integrating factor, M1 for integrating, A1 for the general solution.

• 4 marks

  Solve dy/dx + 2xy = 4x, and describe the behaviour of every solution as x grows.

  Worked answer

  P = 2x is not constant, which changes nothing about the method and only the shape of the factor. IF = e^x², so d(e^x²y)/dx = 4x e^x².
  The right side integrates by recognition, since 4x e^x² is the derivative of 2e^x². Hence e^x²y = 2e^x² + C and y = 2 + Ce^−x².
  As x grows the transient Ce^−x² dies away very quickly, so every solution settles onto the steady state y = 2, whatever value C takes. M1 for the integrating factor, A1 for e^x², A1 for the general solution, B1 for the limit.

• 6 marks · stretch

  Solve dy/dx − 2y/x = x² ln x for x > 0, given that y = 0 when x = 1.

  Worked answer

  P = −2/x, so ∫P dx = −2 ln x and IF = e^{−2 ln x} = x⁻². Reading the factor as e^{−2 ln x} without simplifying to a power of x is what makes the next integration look impossible.
  Multiplying through, d(x⁻²y)/dx = ln x.
  Now integrate ln x by parts, taking u = ln x and dv = dx: ∫ln x dx = x ln x − ∫1 dx = x ln x − x + c. Writing ln x as 1 × ln x is the step worth spotting; there is no integral of ln x by inspection.
  So x⁻²y = x ln x − x + c, and y = 0 at x = 1 gives 0 = 0 − 1 + c, so c = 1.
  Hence y = x²(x ln x − x + 1), that is x³ ln x − x³ + x².
  Check at x = 2: y = 8 ln 2 − 8 + 4 ≈ 1.545, and differentiating the answer returns the original equation.
  M1 A1 for the integrating factor, M1 for reaching d(x⁻²y)/dx = ln x, M1 A1 for the parts integration, A1 for the particular solution.

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