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JEE Mains Maths · Differential Equations

Equations Hidden in Integrals and Limits

Questions that never print a differential equation: a function is defined by an integral of itself, by a limit, or by a rule on its derivatives, and differentiating is what produces the equation.

All Differential Equations notesJEE Mains Maths strategy

Why this matters

Fourteen PYQs, and five of them are from 2026 alone. Differentiating the given relation produces an equation from the earlier pages, and putting the lower limit into the original relation gives the starting value for free. Two ideas cover the page.

Concept 1 of 2: Differentiating an integral equation

If ff appears inside ∫ax\int_a^x, differentiate both sides: by the fundamental theorem, ddx∫axg(t) dt=g(x)\frac{d}{dx}\int_a^xg(t)\,dt=g(x). The relation becomes a differential equation, and setting x=ax=a in the original relation gives f(a)f(a), because the integral vanishes there. When the integral has constant limits, it is just a number — call it kk and solve for it at the end.

Definition

  • ddx∫axg(t) dt=g(x)\frac{d}{dx}\int_a^xg(t)\,dt=g(x).
  • Initial value: put x=ax=a in the original relation.
  • ∫0xex−tf(t) dt=ex∫0xe−tf(t) dt\int_0^xe^{x-t}f(t)\,dt=e^x\int_0^xe^{-t}f(t)\,dt: take exe^x out before differentiating.
  • ∫02f(t) dt=k\int_0^2f(t)\,dt=k is a constant.

Fundamental theorem

f(x)=h(x)+∫axg(t,f(t)) dt ⇒ f′(x)=h′(x)+g(x,f(x)), f(a)=h(a)f(x)=h(x)+\int_a^xg\big(t,f(t)\big)\,dt\ \Rightarrow\ f'(x)=h'(x)+g\big(x,f(x)\big),\ f(a)=h(a)

Worked example

f(x)=1+∫0x2t f(t) dtf(x)=1+\int_0^x2t\,f(t)\,dt. Find f(1)f(1).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 26 · Q79Moderate

Example 1 · Differential Equations · Equations Hidden in Integrals and Limits

Let f(x)=∫0xetf(t)dt+exf(x) =\int_{0}^{x} e^{t}f(t)dt +e^{x} be a differentiable function for all x∈Rx \in R. Then f(x)f(x) equals.

The starting value is free

Differentiating throws away the constant, so the equation alone cannot give ff. Put the lower limit into the original relation to get f(a)f(a) — the question almost never states it.
Drill 9 more on differentiating an integral equation

Concept 2 of 2: Equations from limits and derivative rules

A limit like lim⁡t→xt2f(x)−x2f(t)t−x\lim_{t\to x}\frac{t^2f(x)-x^2f(t)}{t-x} is a 00\frac00 form; differentiating the numerator in tt (L'Hôpital) gives 2xf(x)−x2f′(x)2xf(x)-x^2f'(x), and the stated value turns this into a linear equation. A rule such as f′′=ff''=f is solved by multiplying by f′f' and integrating: (f′)2=f2+C(f')^2=f^2+C.

Definition

  • lim⁡t→xg(t,x)t−x=∂g∂t∣t=x\lim_{t\to x}\frac{g(t,x)}{t-x}=\frac{\partial g}{\partial t}\Big|_{t=x} when g(x,x)=0g(x,x)=0.
  • lim⁡t→xt2f(x)−x2f(t)t−x=2xf(x)−x2f′(x)\lim_{t\to x}\frac{t^2f(x)-x^2f(t)}{t-x}=2xf(x)-x^2f'(x).
  • f′′=ff''=f: f′f′′=f′f⇒(f′)2=f2+Cf'f''=f'f\Rightarrow(f')^2=f^2+C.
  • 2ff′=f2+f′2⇒(f′−f)2=0⇒f′=f2ff'=f^2+f'^2\Rightarrow(f'-f)^2=0\Rightarrow f'=f.

The limit as a derivative

lim⁡t→xt2f(x)−x2f(t)t−x=2xf(x)−x2f′(x)\lim_{t\to x}\frac{t^2f(x)-x^2f(t)}{t-x}=2xf(x)-x^2f'(x)

Worked example

lim⁡t→xtf(x)−xf(t)t−x=1\lim_{t\to x}\frac{tf(x)-xf(t)}{t-x}=1 for all x>0x>0, and f(1)=0f(1)=0. Find ff.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 2 · Q66Moderate

Example 2 · Differential Equations · Equations Hidden in Integrals and Limits

Let y=y(x)y = y(x) be a differentiable function in the interval (0,∞)(0,\infty) such that y(1)=2y(1) = 2. and lim⁡t→x(t2y(x)−x2y(t)x−t)=3\lim_{t \rightarrow x} \left( \frac{t^{2}y(x) -x^{2}y(t)}{x - t} \right)= 3 for each x>0x > 0. Then 2y(2)2y(2) is equal to

Watch the order in the denominator

…x−t\frac{\dots}{x-t} and …t−x\frac{\dots}{t-x} differ by a sign. Differentiate in tt and divide by the derivative of the denominator in tt, which is −1-1 for x−tx-t.
Drill 3 more on equations from limits and derivative rules

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Differentiating an integral equation

    Fundamental theorem

    f(x)=h(x)+∫axg(t,f(t)) dt ⇒ f′(x)=h′(x)+g(x,f(x)), f(a)=h(a)f(x)=h(x)+\int_a^xg\big(t,f(t)\big)\,dt\ \Rightarrow\ f'(x)=h'(x)+g\big(x,f(x)\big),\ f(a)=h(a)
  • Equations from limits and derivative rules

    The limit as a derivative

    lim⁡t→xt2f(x)−x2f(t)t−x=2xf(x)−x2f′(x)\lim_{t\to x}\frac{t^2f(x)-x^2f(t)}{t-x}=2xf(x)-x^2f'(x)

Watch out for (2)

Test yourself on Differential Equations

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Differential Equations chapter testDrill all 14 questions