Integrating factor method
Some locks open only with the right key. The integrating factor is the key: multiply both sides by it and the left side becomes the derivative of a single product.
An equation of the form dy/dx + P y = Q, where P and Q are functions of x only, is linear. Multiply by the integrating factor IF = e^(∫P dx). Then the left side becomes d/dx (y · IF), and the solution is y · IF = ∫ Q · IF dx + C.
Formula: IF = e^(∫P dx), y · IF = ∫ Q · IF dx + C
Rule 1: Find IF, then integrate — Write the equation in standard form first.
- Easy: dy/dx + y = eˣ has IF = eˣ, so y eˣ = e^(2x)/2 + C and y = eˣ/2 + C e^(−x).
- Medium: dy/dx + 2y = 4x has IF = e^(2x), so y = 2x − 1 + C e^(−2x).
- Exam-type: dy/dx + y/x = x² has IF = x, so xy = x⁴/4 + C and y = x³/4 + C/x.
Rule 2: Divide to standard form — The coefficient of dy/dx must be 1.
- Easy: dy/dx + 3y = 6 has IF = e^(3x) and y = 2 + C e^(−3x).
- Medium: x dy/dx + 2y = x² becomes dy/dx + 2y/x = x, with IF = x², giving y = x²/4 + C/x².
- Exam-type: dy/dx + y tan x = sin x has IF = sec x.
Always check that the standard form dy/dx + Py = Q is reached before taking the integrating factor.
Forgetting to divide the full equation by the coefficient of dy/dx, or dropping the constant C.
Make IF from P, multiply, integrate.
Practice questions
- 1Find the integrating factor of dy/dx + 2y = 3.
- 2Find the integrating factor of dy/dx + y/x = x².
- 3Find the integrating factor of dy/dx + y tan x = sin x.
- 4Solve dy/dx + y = eˣ.
- 5Solve dy/dx + 2y = 4x.
- 6Solve dy/dx + y/x = x².
- 7Solve dy/dx + y = 1.
- 8Solve x dy/dx + 2y = x² for x > 0.
Show answers (try them yourself first)
- 1e^(2x)
- 2e^(ln x) = x
- 3e^(∫tan x dx) = e^(ln|sec x|) = sec x
- 4y = eˣ/2 + C e^(−x)
- 5y = 2x − 1 + C e^(−2x)
- 6y = x³/4 + C/x
- 7y = 1 + C e^(−x)
- 8y = x²/4 + C/x²
Linear equations with conditions and in x
A savings account earns interest and also gets monthly deposits. The balance follows a linear equation, and the opening balance picks the right solution.
After finding the general solution, use the given condition to fix C. When the roles of x and y are exchanged, the equation dx/dy + P₁ x = Q₁ is also linear, with IF = e^(∫P₁ dy), and the solution is x · IF = ∫ Q₁ · IF dy + C.
Formula: dx/dy + P₁ x = Q₁, IF = e^(∫P₁ dy)
Rule 1: Use the condition — Find C after finding the full solution.
- Easy: dy/dx + y = 1, y(0) = 3 gives y = 1 + 2 e^(−x).
- Medium: dy/dx + y tan x = cos x, y(0) = 0 gives y = x cos x.
- Exam-type: dy/dx + y/x = x², y(1) = 1 gives y = x³/4 + 3/(4x).
Rule 2: Equation in x as the unknown — Look for dx/dy.
- Easy: dx/dy + x = y has IF = e^y and the solution x = y − 1 + C e^(−y).
- Medium: y dx + (x − y²) dy = 0 becomes dx/dy + x/y = y, with IF = y and the solution x = y²/3 + C/y.
- Exam-type: dx/dy + 2x/y = y has IF = y² and the solution x = y²/4 + C/y².
If dy/dx has a messy form but dx/dy is linear in x, solve for x as a function of y.
Using x as the independent variable when the equation is linear in x, with y as the independent variable.
Condition fixes C, and dx/dy swaps the roles.
Practice questions
- 1Solve dy/dx + y = 1 with y(0) = 3.
- 2Solve dy/dx + y tan x = cos x with y(0) = 0.
- 3Solve dy/dx + y/x = x² with y(1) = 1.
- 4Solve dy/dx + 2y = e^(−x) with y(0) = 0.
- 5Solve dx/dy + x = y.
- 6Solve y dx + (x − y²) dy = 0.
- 7Solve dx/dy + 2x/y = y.
- 8Find the general solution of dy/dx + 3y = 6.
Show answers (try them yourself first)
- 1y = 1 + 2 e^(−x)
- 2IF = sec x, y sec x = x + C, and y(0) = 0 gives C = 0, so y = x cos x
- 3y = x³/4 + 3/(4x)
- 4y e^(2x) = eˣ + C, and y(0) = 0 gives C = −1, so y = e^(−x) − e^(−2x)
- 5x = y − 1 + C e^(−y)
- 6x = y²/3 + C/y
- 7x = y²/4 + C/y²
- 8y = 2 + C e^(−3x)