12 min readPathway sheet1-mark MCQ on the integrating factor; 3-mark answers, often with an initial condition; also a JEE favourite

Linear equations: dy/dx + Py = Q

For dy/dx + Py = Q, multiply by the integrating factor e^(∫P dx). Then the solution is y × IF = ∫(Q × IF) dx + C.

Standard formdy/dx + Py = Q, where P and Q are functions of x only (or constants)

Integrating factor (IF)IF = e^(∫P dx)

Solutiony × IF = ∫(Q × IF) dx + C

Key simplificatione^(log f(x)) = f(x); so P = 1/x gives IF = x, and P = tan x gives IF = sec x

Why the integrating factor works

1
A first order DE is linear when y and dy/dx appear only to the first power and are not multiplied together.
2
Multiply dy/dx + Py = Q by g(x) = e^(∫P dx). Note that g′(x) = P·g(x).
3
The left side becomes g(dy/dx) + gPy = g(dy/dx) + g′y = d/dx (y·g), by the product rule.
4
So d/dx (y·g) = Q·g. Integrate both sides with respect to x.
5
So, y·g = ∫ Q·g dx + C. This g is the integrating factor.

The method

1
Write the DE in standard form dy/dx + Py = Q. Divide by the coefficient of dy/dx first.
2
Read off P (the coefficient of y, with its sign) and Q.
3
Find IF = e^(∫P dx). Simplify using e^(log f) = f.
4
Write the solution: y × IF = ∫(Q × IF) dx + C.
5
Integrate the right side. Divide by IF if you want y alone.
6
Use any given condition to find C.

Example 1: direct

Problem

Solve dy/dx + 2y = e⁻ˣ.

Working

Standard form already: P = 2, Q = e⁻ˣ.
IF = e^(∫2 dx) = e²ˣ.
y·e²ˣ = ∫ e⁻ˣ·e²ˣ dx + C = ∫ eˣ dx + C.
y·e²ˣ = eˣ + C.
Divide by e²ˣ: y = e⁻ˣ + Ce⁻²ˣ.

Answer

y = e⁻ˣ + Ce⁻²ˣ

Example 2: board-style, with a condition

Problem

Find the particular solution of x(dy/dx) + y = x log x (x > 0), given that y = 1/4 when x = 1.

Working

Divide by x: dy/dx + (1/x)y = log x. So P = 1/x, Q = log x.
IF = e^(∫(1/x) dx) = e^(log x) = x.
Solution: y·x = ∫ x log x dx + C.
By parts (log x first, x second): ∫ x log x dx = (x²/2) log x − ∫ (x²/2)(1/x) dx = (x²/2) log x − x²/4.
So xy = (x²/2) log x − x²/4 + C.
Put x = 1, y = 1/4: 1/4 = 0 − 1/4 + C, so C = 1/2.
xy = (x²/2) log x − x²/4 + 1/2.

Answer

y = (x/2) log x − x/4 + 1/(2x)

Ready reckoner for the IF

P	∫P dx	IF = e^(∫P dx)
k (a constant)	kx	e^(kx)
1/x	log x	x
−1/x	−log x	1/x
2/x	2 log x	x²
tan x	log sec x	sec x
cot x	log sin x	sin x
2x/(1 + x²)	log(1 + x²)	1 + x²

Mistakes that cost marks

Not dividing by the coefficient of dy/dx first. For x(dy/dx) + 2y = x³, P = 2/x, not 2.
Losing the sign of P. dy/dx − 3y = Q has P = −3, so IF = e⁻³ˣ.
Forgetting + C, or adding C after dividing by the IF (then C must be divided too).
Writing e^(log x) as log x. It is x.

In real life

A pressure cooker cooling on the kitchen counter follows a linear DE of the form dy/dt + Py = Q. So does the charge on a capacitor charging through a resistor in an electronic circuit.

Exam angle

1 marks: “Find the integrating factor of x(dy/dx) − y = x³ (x > 0).” Standard form: dy/dx − (1/x)y = x². IF = e^(−log x) = 1/x.

3 marks: “Solve dy/dx + y = cos x.” P = 1, IF = eˣ. yeˣ = ∫ eˣ cos x dx = eˣ(sin x + cos x)/2 + C. So y = (sin x + cos x)/2 + Ce⁻ˣ.

3 marks: “Find the particular solution of (1 + x²)(dy/dx) + 2xy = 6x², given y = 1 when x = 0.” dy/dx + [2x/(1 + x²)]y = 6x²/(1 + x²). IF = 1 + x². y(1 + x²) = ∫ 6x² dx = 2x³ + C. x = 0, y = 1 gives C = 1. y = (2x³ + 1)/(1 + x²).

Remember it

Standard form, P, IF, then y × IF = ∫ Q × IF dx + C.