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Linear first-order differential equations: Class 12 practice

dy/dx + Py = Q is solved with the integrating factor IF = e^(∫P dx): y × IF = ∫Q × IF dx + C.. Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Integration

The idea behind linear first-order differential equations

dy/dx + Py = Q is solved with the integrating factor IF = e^(∫P dx): y × IF = ∫Q × IF dx + C.

Step-by-step method

  1. Identify P and Q.
  2. IF = e^(∫P dx).
  3. y·IF = ∫Q·IF dx + C; use any initial condition.

Worked example 1 (Basic)

Solve dy/dx + 1y = 9 with y(0) = -4.

  1. IF = e^(1x)
  2. y e^(1x) = (9/1) e^(1x) + C
  3. C = -13
  4. y = 9 − 13e^(−1x)
  5. Answer: y = 9 − 13e^(−1x)

Worked example 2 (Standard)

Solve dy/dx + 1y = 4 with y(0) = -4.

  1. IF = e^(1x)
  2. y e^(1x) = (4/1) e^(1x) + C
  3. C = -8
  4. y = 4 − 8e^(−1x)
  5. Answer: y = 4 − 8e^(−1x)

Worked example 3 (Challenge)

Solve dy/dx + 3y = 6 with y(0) = -2.

  1. IF = e^(3x)
  2. y e^(3x) = (6/3) e^(3x) + C
  3. C = -4
  4. y = 2 − 4e^(−3x)
  5. Answer: y = 2 − 4e^(−3x)

Common mistakes and misconceptions

  • Wrong sign of P
  • Forgetting C

Practice questions

  1. (Basic) Solve dy/dx + 3y = 4 with y(0) = 1.
  2. (Basic) Solve dy/dx + 1y = 4 with y(0) = 5.
  3. (Standard) Solve dy/dx + 3y = 3 with y(0) = 3.
  4. (Standard) Solve dy/dx + 4y = 9 with y(0) = 0.
  5. (Challenge) Solve dy/dx + 1y = 9 with y(0) = -1.
  6. (Challenge) Solve dy/dx + 3y = 3 with y(0) = 4.

Answer key with full working

  • Q1: IF = e^(3x) → y e^(3x) = (4/3) e^(3x) + C → C = -0.33 → y = 4/3 − 0.33e^(−3x) → Answer: y = 4/3 − 0.33e^(−3x)
  • Q2: IF = e^(1x) → y e^(1x) = (4/1) e^(1x) + C → C = 1 → y = 4 + 1e^(−1x) → Answer: y = 4 + 1e^(−1x)
  • Q3: IF = e^(3x) → y e^(3x) = (3/3) e^(3x) + C → C = 2 → y = 1 + 2e^(−3x) → Answer: y = 1 + 2e^(−3x)
  • Q4: IF = e^(4x) → y e^(4x) = (9/4) e^(4x) + C → C = -2.25 → y = 9/4 − 2.25e^(−4x) → Answer: y = 9/4 − 2.25e^(−4x)
  • Q5: IF = e^(1x) → y e^(1x) = (9/1) e^(1x) + C → C = -10 → y = 9 − 10e^(−1x) → Answer: y = 9 − 10e^(−1x)
  • Q6: IF = e^(3x) → y e^(3x) = (3/3) e^(3x) + C → C = 3 → y = 1 + 3e^(−3x) → Answer: y = 1 + 3e^(−3x)

Syllabus note

This skill is listed under "Differential Equations" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

Frequently asked questions

Which chapter is linear first-order differential equations in?

"Differential Equations", Class 12 Maths (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

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