Differential Equations: Types & Methods JEE Main Guide

Differential Equations (DEs) is the final chapter in JEE Main Calculus and contributes 2–3 questions per session. It sits at the boundary between Calculus and Applied Mathematics — the methods are mechanical once identified, but identifying the correct method from the form of the equation requires practice. This guide classifies every DE type the exam uses and gives the solution method for each, with the identification criteria you need to classify instantly.

Order, Degree, and Formation of Differential Equations

Order = the highest derivative present. Degree = the power of the highest-order derivative (after the equation is rationalised — square roots cleared). Formation: differentiate the given family of curves and eliminate the arbitrary constants. If a family has n arbitrary constants, you need to differentiate n times to form the DE of order n. JEE asks: "Form the DE of the family of circles x²+y²=r²" — differentiate once (2x+2y·dy/dx = 0 → x+y·y' = 0) and the constant r is eliminated.

JEE tests order and degree identification as direct MCQ options. Common traps: (1) degree is undefined when the highest derivative appears under a square root (irrational DE); (2) degree is the power of the HIGHEST-order derivative, not the highest power anywhere in the equation. If dy/dx appears squared and d²y/dx² appears once: order = 2, degree = 1 (not 2). Take a free Differential Equations mock to verify your classification skills. For the Calculus context, see our differential equations guide.

Variable Separable and Homogeneous Equations

Variable Separable: dy/dx = f(x)g(y). Rewrite as dy/g(y) = f(x)dx and integrate both sides directly. Recognition: the right side factors into a function of x only times a function of y only. Substitution for quasi-separable: dy/dx = f(ax+by+c) — substitute v = ax+by+c, dv/dx = a+b·dy/dx, giving dv/dx − a = b·f(v), now separable.

Homogeneous DE: dy/dx = f(y/x). Recognition: RHS is a function of y/x only (every term in numerator and denominator has the same total degree in x and y). Method: substitute v = y/x (so y = vx, dy/dx = v + x·dv/dx): the equation becomes v + x·dv/dx = f(v) → x·dv/dx = f(v) − v → now separable in v and x. After integrating, back-substitute v = y/x. JEE uses homogeneous DEs in about one question per session — the y = vx substitution is the mandatory recognition-and-method pair. For integration techniques used in solving DEs, see our integration substitution guide.

Linear First-Order Differential Equations

Standard form: dy/dx + P(x)y = Q(x). Integrating factor (IF) = e^(∫P(x)dx). Multiply both sides by IF: d/dx[y·IF] = Q(x)·IF. Integrate: y·IF = ∫Q(x)·IF dx + C. The method always works for linear first-order DEs. Recognition: the y term is linear (not squared, not inside a function) and its coefficient (after normalisation) depends only on x. Example: dy/dx + y/x = x² (P = 1/x, Q = x²). IF = e^(∫1/x dx) = e^(lnx) = x. Multiply: d/dx[xy] = x³ → xy = x⁴/4 + C → y = x³/4 + C/x.

Bernoulli's equation: dy/dx + P(x)y = Q(x)·yⁿ. Divide by yⁿ: y⁻ⁿ·dy/dx + P(x)·y¹⁻ⁿ = Q(x). Substitute v = y¹⁻ⁿ: dv/dx = (1−n)y⁻ⁿ·dy/dx → dv/dx + (1−n)P(x)v = (1−n)Q(x). Now linear in v — apply the IF method. JEE tests Bernoulli's equation in about one question per two sessions. Recognise it by the yⁿ term on the right (n ≠ 0, 1). For related applications of DEs in Physics problems, see our separable differential equations guide.

Geometric Interpretation and Orthogonal Trajectories

The slope of a curve y = f(x) at any point is dy/dx = slope of the tangent. The normal at (x,y) has slope −dx/dy. JEE uses DEs to find curves with given geometric properties: "Find the curve where the tangent at (x,y) makes equal intercepts on the axes" → x-intercept = x − y/(dy/dx), y-intercept = y − x·dy/dx; set equal → after algebra: dy/dx = y/x → separable → y = kx. "Find the curve passing through (1,2) such that the area of the triangle formed by the tangent, x-axis, and the ordinate is always 2" → set up the area from the tangent equation and derive a DE. Orthogonal trajectories: replace dy/dx by −dx/dy in the DE of the given family → solve for the orthogonal family. This appears once per three sessions in JEE. For the complete Calculus chapter map, see our calculus complete guide.