JEE Maths — Differential Equations Strategy

Chapter Overview & Weightage

Differential Equations carries 6-8% weightage in JEE Main and shows up in 1-2 questions per paper. The chapter is mostly mechanical — recognise the form, apply the standard technique, and you’re done in under 2 minutes. JEE Advanced occasionally includes a tougher conceptual question on order/degree or modelling.

Year	JEE Main Qs	JEE Advanced Qs
2024	2	1
2023	2	1
2022	1	1
2021	2	1

Key Concepts You Must Know

Order: highest derivative present in the equation.
Degree: power of the highest-order derivative (after rationalisation).
General solution: family of solutions with arbitrary constants ( $n$ constants for an $n$ -th order ODE).
Particular solution: specific member of the family obtained by applying initial conditions.
Variable separable: $f(x)dx = g(y)dy$ . Just integrate both sides.
Homogeneous: $\dfrac{dy}{dx} = F(y/x)$ . Substitute $y = vx$ .
Linear (first order): $\dfrac{dy}{dx} + Py = Q$ . Use integrating factor $e^{\int P\,dx}$ .
Bernoulli’s equation: $\dfrac{dy}{dx} + Py = Qy^n$ . Substitute $v = y^{1-n}$ to linearise.
Exact equations: $M dx + N dy = 0$ with $\partial M/\partial y = \partial N/\partial x$ .

Important Formulas

For $\dfrac{dy}{dx} + P(x)y = Q(x)$ :

$IF = e^{\int P\,dx}$

$y \cdot IF = \int Q \cdot IF\,dx + C$

When to use: ODE is linear in $y$ with variable coefficient.

For $\dfrac{dy}{dx} = F(y/x)$ , set $y = vx$ , so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$ .

The equation becomes separable in $v$ and $x$ .

When to use: numerator and denominator are homogeneous of the same degree.

For $\dfrac{dy}{dx} + Py = Qy^n$ , divide by $y^n$ and substitute $v = y^{1-n}$ .

The result is linear in $v$ :

$\frac{dv}{dx} + (1-n)Pv = (1-n)Q$

Solved Previous Year Questions

PYQ 1 (JEE Main 2024 Shift 1)

Solve $\dfrac{dy}{dx} = \dfrac{x+y}{x-y}$ .

Solution: Homogeneous (both numerator and denominator are degree 1). Substitute $y = vx$ :

$v + x\frac{dv}{dx} = \frac{x + vx}{x - vx} = \frac{1+v}{1-v}$

$x\frac{dv}{dx} = \frac{1+v}{1-v} - v = \frac{1 + v - v + v^2}{1-v} = \frac{1+v^2}{1-v}$

Separate:

$\frac{1-v}{1+v^2}dv = \frac{dx}{x}$

Integrate: $\tan^{-1}v - \dfrac{1}{2}\ln(1+v^2) = \ln|x| + C$ .

Substitute back $v = y/x$ :

$\tan^{-1}(y/x) - \frac{1}{2}\ln(1 + y^2/x^2) = \ln|x| + C$

PYQ 2 (JEE Main 2023)

If $y(x)$ satisfies $\dfrac{dy}{dx} + 2y = e^{-x}$ and $y(0) = 1$ , find $y(1)$ .

Solution: Linear with $P = 2$ , $Q = e^{-x}$ . $IF = e^{2x}$ .

$y \cdot e^{2x} = \int e^{-x} \cdot e^{2x}\,dx + C = \int e^x\,dx + C = e^x + C$

$y = e^{-x} + Ce^{-2x}$

At $x=0$ : $1 = 1 + C \implies C = 0$ . So $y = e^{-x}$ , and $y(1) = e^{-1} = 1/e$ .

PYQ 3 (JEE Advanced 2022)

A curve passes through $(2, 0)$ and the slope at $(x, y)$ is $\dfrac{(x+1)^2 + y - 3}{x+1}$ . Find the equation of the curve.

Solution: Rewrite as

$\frac{dy}{dx} - \frac{1}{x+1}y = (x+1) - \frac{3}{x+1}$

Linear with $P = -1/(x+1)$ . $IF = e^{-\ln(x+1)} = 1/(x+1)$ .

Multiply through and integrate. Apply $y(2) = 0$ to find the constant. Final form: $y = (x+1)^2 \ln|x+1| - 3 + (x+1)\cdot k$ , with $k$ determined by initial condition.

Difficulty Distribution

Easy (50%): Variable separable, basic linear ODEs with simple $P, Q$ .
Medium (35%): Homogeneous, Bernoulli, linear with trickier integrating factors.
Hard (15%): Exact equations, ODEs requiring substitution, modelling problems.

Expert Strategy

Recognition order: Check forms in this sequence — variable separable → homogeneous → linear → Bernoulli → exact. Most JEE questions fall into the first three.

For linear ODEs, the rule " $IF = e^{\int P\,dx}$ " works only when the equation is in standard form $\dfrac{dy}{dx} + Py = Q$ . If you see something like $x\dfrac{dy}{dx} + 2y = x^3$ , divide by $x$ first to get $P = 2/x$ , then $IF = e^{\int 2/x\,dx} = x^2$ .

Order/degree questions are 1-mark giveaways. The catch: if the equation contains $\sqrt{}$ or fractional powers, rationalise first before computing degree. After rationalisation, the degree may be different from what it looked like initially.

Common Traps

Trap 1: Wrong sign in IF. For $\dfrac{dy}{dx} - 3y = e^x$ , $P = -3$ (not $+3$ ). $IF = e^{-3x}$ . Sign errors here cost full marks.

Trap 2: Solving non-linear equations as linear. $\dfrac{dy}{dx} + Py^2 = Q$ is not linear (it’s Bernoulli with $n = 2$ ). Substitute $v = 1/y$ to linearise.

Trap 3: Applying initial conditions before integrating. Always solve for the general solution first, then apply $y(x_0) = y_0$ to find the constant. Reverse order leads to wrong answers.

JEE Main almost always includes one “growth/decay” word problem: bacteria doubling, radioactive decay, Newton’s law of cooling, etc. These reduce to simple separable or linear ODEs. Practice 5-10 of these and the pattern becomes obvious.