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The integrating factor method is the most reliable tool in the differential equations chapter: once you recognise a linear equation, it always works. Marks are lost in the set-up, not the idea: reading P before the equation is in standard form, mishandling a logarithm in the exponent, or adding the constant in the wrong place. This post explains why the method works, simplifies the common integrating factors, and works three examples, including one from physics.
What does the syllabus include?
The JEE (Main) 2026 syllabus lists linear differential equations of the type dy/dx + p(x)y = q(x) in Unit 9. The JEE (Advanced) 2026 brochure lists "linear first order differential equations". Section 9.4.3 of the current NCERT Class 12 chapter on differential equations covers the method. Check the documents for your exam year.
Why does the integrating factor work?
NCERT's derivation is short. Multiply dy/dx + Py = Q by some function g(x). You want the left side to become the derivative of y·g(x), which by the product rule is g dy/dx + y g′. That needs g′ = Pg, so g′/g = P, and integrating gives log g = ∫P dx. So g(x) = e^(∫P dx), the integrating factor (I.F.).
Then d/dx[y × I.F.] = Q × I.F., and integrating gives
y × I.F. = ∫ Q × I.F. dx + C.
NCERT's steps are: write the equation in the form dy/dx + Py = Q, find the I.F., then write the solution in this form. If the equation is linear in x instead, dx/dy + P₁x = Q₁, the same steps work with the roles of x and y swapped.
Which integrating factors come up most?
The exponent is often a logarithm, and e^(log u) = u. Work it out before you multiply:
P = 1/x gives I.F. = e^(log x) = x (for x > 0).
P = −2/x gives e^(−2 log x) = 1/x².
P = tan x gives e^(log sec x) = sec x.
P = 2x/(1 + x²) gives e^(log(1 + x²)) = 1 + x².
P = a constant k gives e^(kx).
Worked example 1: a trigonometric P
Solve dy/dx + y tan x = sec x, for 0 < x < π/2.
Here P = tan x, so I.F. = sec x. Multiply through: d/dx(y sec x) = sec²x.
Integrate: y sec x = tan x + C, so y = sin x + C cos x.
Check. dy/dx = cos x − C sin x, and y tan x = sin²x/cos x + C sin x. Adding gives cos x + sin²x/cos x = 1/cos x = sec x. Correct.
Worked example 2: an initial value problem
Solve (1 + x²) dy/dx + 2xy = 4x², with y = 0 when x = 0.
First, standard form: divide by (1 + x²) to get dy/dx + [2x/(1 + x²)]y = 4x²/(1 + x²).
I.F. = 1 + x². Multiplying back gives d/dx[y(1 + x²)] = 4x², so y(1 + x²) = 4x³/3 + C.
Use the condition: at x = 0, 0 = 0 + C, so C = 0. The solution is y = 4x³/[3(1 + x²)].
Notice that the left side of the original equation was already the derivative of y(1 + x²). Spotting that saves a step.
Worked example 3: current in an RL circuit
The same equation describes current growing in a circuit with an inductor L, a resistor R and a battery of emf E: L di/dt + Ri = E. With sample values for a textbook-style problem, E = 12 V, R = 4 Ω and L = 2 H, and i = 0 at t = 0:
Standard form: di/dt + 2i = 6. I.F. = e^(2t). Then d/dt(i e^(2t)) = 6e^(2t), so i e^(2t) = 3e^(2t) + C. At t = 0, 0 = 3 + C, so C = −3.
So i = 3(1 − e^(−2t)) amperes. The current rises towards E/R = 3 A. At t = 0.5 s it is 3(1 − e^(−1)), about 1.90 A.
Which mistakes cost the most marks?
Reading P before dividing by the coefficient of dy/dx.
Losing a minus sign: for dy/dx − 3y = x, P = −3 and I.F. = e^(−3x).
Leaving the I.F. as e^(log sec x) and then integrating with it.
Forgetting to multiply Q by the I.F.
Not checking. Substitute the answer back; it takes a minute.
Why check by substituting?
Arslan (2010) tested 77 candidate mathematics teachers after a traditional course on differential equations. 85% answered the procedural questions correctly, but only 30% answered the conceptual ones correctly. Knowing the steps is not the same as knowing what a solution means. Substituting your answer back reminds you that a solution is a function that makes the equation true, and it catches most errors.
A practice routine
I.F. drill (10 minutes, three times a week). For ten equations, put each in standard form and simplify its integrating factor. Do not solve.
Solve and substitute. Solve five fully and check each by substitution.
One initial value problem a week, and one from physics or chemistry, such as RC charging or first-order kinetics.
Mix types with separable and homogeneous equations, as in our guide to spotting the type of differential equation.
If you can solve these but do not see why the method works, Fermi's AI tutor works out what you already know and builds a plan for the session, so you can go back to the product rule step at your own pace.
Today, solve dy/dx + 3y = e^(−2x) and check your answer by substitution.
