How do I quickly identify the type of differential equation?

Check in order: separable, linear, homogeneous, Bernoulli, exact. Almost every JEE Main differential equation falls into one of these categories.

JEE Main Differential Equations 2027 — Order, Degree, Linear & Variable Separable: Complete Notes and 35 Practice Problems

Q: What is the difference between order and degree of an ODE?

Order is the highest derivative present. Degree is the exponent of that highest derivative once the equation is cleared of fractional powers and radicals of derivatives.

Q: Are partial differential equations in JEE Main syllabus?

No, JEE Main covers only ordinary differential equations, primarily first order and first degree, plus formation of ODEs from families of curves.

April 28, 2026 · · Blog

Last Updated: April 2026

JEE Main differential equations 2027 aspirants — Differential Equations carries an average of 2 questions (8 marks) in JEE Main every shift, per NTA’s January 2026 chapter-wise analysis. Combined with Calculus topics (limits, continuity, integration), it forms ~22% of the maths paper. The 2026 January session featured 4 problems on first-order ODEs across the 6 shifts. JEE Main cut-off (general) for 2026 was 93.23 percentile — every chapter matters. This guide covers order, degree, variable separable, homogeneous, linear, Bernoulli, exact ODEs, and 35 worked problems.

1. Definition, Order and Degree

A differential equation contains derivatives of a dependent variable with respect to one or more independent variables.

Order = the order of the highest derivative present.
Degree = the power of the highest-order derivative when the equation is expressed as a polynomial in derivatives (radicals/fractions of derivatives must be cleared first).

Example: (d²y/dx²)³ + (dy/dx)⁴ = sin x → order 2, degree 3.

Worked Example 1

Q. Find order and degree of (d²y/dx²) + 5(dy/dx) + 6y = 0.
Sol. Highest derivative = d²y/dx² → order 2. It appears to power 1 → degree 1. Linear, homogeneous, second-order.

2. Formation of Differential Equation

If a family of curves contains n arbitrary constants, the differential equation eliminating them has order n.

Worked Example 2

Q. Form ODE for y = a sin x + b cos x.
Sol. Two constants → order 2. dy/dx = a cos x − b sin x. d²y/dx² = −a sin x − b cos x = −y → d²y/dx² + y = 0.

3. Variable Separable ODEs

Form: dy/dx = f(x)·g(y). Separate: dy/g(y) = f(x) dx, then integrate.

Worked Example 3

Q. Solve dy/dx = (1 + x)(1 + y²).
Sol. dy/(1+y²) = (1+x) dx. Integrate: arctan y = x + x²/2 + c. y = tan(x + x²/2 + c).

Worked Example 4

Q. Solve dy/dx = e^(x−y).
Sol. dy/dx = e^x·e^(−y) → e^y dy = e^x dx → e^y = e^x + c.

4. Homogeneous ODE

Form: dy/dx = F(y/x) where F is homogeneous of degree 0. Substitute y = vx → dy/dx = v + x(dv/dx). Reduces to variable separable in v and x.

Worked Example 5

Q. Solve dy/dx = (x + y)/x.
Sol. dy/dx = 1 + y/x. Put y = vx → v + x(dv/dx) = 1 + v → x(dv/dx) = 1 → dv = dx/x → v = ln|x| + c → y = x ln|x| + cx.

5. Linear First-Order ODE

Standard form: dy/dx + P(x)y = Q(x). Integrating factor IF = e^∫P dx. Solution: y·IF = ∫(Q·IF) dx + c.

Worked Example 6

Q. Solve dy/dx + y/x = x².
Sol. P = 1/x → IF = e^(ln x) = x. Solution: y·x = ∫x²·x dx = x⁴/4 + c → y = x³/4 + c/x.

Worked Example 7

Q. Solve dy/dx + 2y = e^(−x), y(0) = 1.
Sol. IF = e^(2x). y·e^(2x) = ∫e^(−x)·e^(2x) dx = ∫e^x dx = e^x + c → y = e^(−x) + c·e^(−2x). y(0)=1 → 1 = 1 + c → c=0. y = e^(−x).

6. Bernoulli’s Equation

Form: dy/dx + Py = Qy^n (n ≠ 0,1). Divide by y^n, substitute v = y^(1−n) → dv/dx + (1−n)P·v = (1−n)Q (linear in v).

Worked Example 8

Q. Solve dy/dx + y = xy².
Sol. Bernoulli, n = 2. Divide by y²: y^(−2) dy/dx + y^(−1) = x. Put v = y^(−1) → dv/dx = −y^(−2) dy/dx. Equation becomes −dv/dx + v = x → dv/dx − v = −x. IF = e^(−x). v·e^(−x) = ∫−x·e^(−x) dx = (x+1)e^(−x) + c → 1/y = (x+1) + ce^x.

7. Exact Differential Equations

Form: M(x,y) dx + N(x,y) dy = 0 is exact iff ∂M/∂y = ∂N/∂x. Solution: ∫M dx (treating y constant) + ∫(terms of N free of x) dy = c.

8. Standard Forms — Master Table

Type	Standard Form	Method	General Solution
Variable Separable	dy/dx = f(x)g(y)	Separate & integrate	∫dy/g(y) = ∫f(x) dx + c
Homogeneous	dy/dx = F(y/x)	y = vx	Reduces to separable in v, x
Linear (1st order)	dy/dx + Py = Q	IF = e^∫P dx	y·IF = ∫(Q·IF) dx + c
Bernoulli	dy/dx + Py = Qy^n	v = y^(1−n)	Linear in v
Exact	M dx + N dy = 0, M_y = N_x	Direct integration	F(x,y) = c
Reducible to homogeneous	dy/dx = (a₁x+b₁y+c₁)/(a₂x+b₂y+c₂)	Shift origin	If a₁/a₂ ≠ b₁/b₂

9. 35 JEE Practice Problems (Selected with Solutions)

Q1. dy/dx = y/x → y = cx (variable separable).
Q2. dy/dx + y tan x = sec x. IF = sec x. y·sec x = ∫sec²x dx = tan x + c → y = sin x + c cos x.
Q3. Order and degree of [1 + (dy/dx)²]^(3/2) = d²y/dx². Square both sides: order 2, degree 2.
Q4. dy/dx = (x²+y²)/(2xy). Homogeneous. y=vx → v+x(dv/dx) = (1+v²)/(2v) → x(dv/dx) = (1−v²)/(2v) → 2v dv/(1−v²) = dx/x → −ln|1−v²| = ln|x| + c → x²−y² = c·x.
Q5. dy/dx + y cot x = 2 cos x. IF = sin x. y sin x = ∫2 cos x sin x dx = sin²x + c.
Q6. y = e^x(c₁ + c₂x) → ODE is y” − 2y’ + y = 0 (order 2).
Q7. Solve x dy − y dx = √(x²+y²) dx. Homogeneous: y = vx → x(dv·x + v dx) − vx dx = x√(1+v²) dx → x dv = √(1+v²) dx → ∫dv/√(1+v²) = ∫dx/x → sinh⁻¹(v) = ln|x| + c.
Q8. dy/dx = (3x+y+1)/(x−y+2). Lines intersect at (−3/4, 5/4)? Use shift x = X−3/4, y = Y+5/4 → reduces to homogeneous in X, Y.
Q9. Find IF of (1+y²) dx + (xy + y) dy = 0. Test exactness: M_y = 2y, N_x = y → not exact. Multiply by suitable IF (function of y).
Q10. y’ + 2y = sin x. IF = e^(2x). Solution: y = (2 sin x − cos x)/5 + ce^(−2x).
Q11. Population growth dN/dt = kN → N = N₀ e^(kt).
Q12. Newton’s law of cooling: dT/dt = −k(T − T_s) → T = T_s + (T₀ − T_s)e^(−kt).
Q13. dy/dx = sec(x+y). Put v = x+y → dv/dx = 1 + dy/dx = 1 + sec v → cos v dv/(1+cos v) = dx → integrate.
Q14. y dx − x dy + log x · dx = 0. Rewrite: dy/dx = y/x + (log x)/x. Linear. IF = 1/x. (y/x)’ = (log x)/x² → y/x = −(log x)/x − 1/x + c → y = −log x − 1 + cx.
Q15. Order of ODE of family y = (a + bx)e^x is 2 (two constants).
Q16. Slope of tangent = sum of coordinates → dy/dx = x + y → linear. IF = e^(−x). y e^(−x) = ∫x e^(−x) dx = −(x+1)e^(−x) + c → y = −(x+1) + ce^x.
Q17. dy/dx = e^(x+y) + e^(x)y² → after factoring: dy/dx = e^x(e^y + y²) — variable separable in some cases (this one isn’t strictly).
Q18. ODE for circles touching y-axis at origin: x² + y² = 2ax → 2x + 2y(dy/dx) = 2a; eliminate a: 2x + 2y·y’ = (x² + y²)/x → 2x²+2xy y’ = x² + y² → y’ = (y²−x²)/(2xy).
Q19. ODE of all parabolas y² = 4ax: differentiate → 2y y’ = 4a → a = y y’/2 → y² = 4·(y y’/2)·x = 2xy y’ → y = 2x y’.
Q20. Solve (1+x²) dy/dx + 2xy = 1/(1+x²). Linear. IF = (1+x²). y(1+x²) = ∫dx/(1+x²) = arctan x + c.
Q21. Family y = ce^(2x): dy/dx = 2ce^(2x) = 2y → dy/dx = 2y.
Q22. dy/dx = (y−x)/(y+x). Homogeneous; y=vx → integrate.
Q23. Mixing problem (salt in tank): dQ/dt = (rate in) − (rate out)·(Q/V). Linear ODE.
Q24. dy/dx = sin(x+y) + cos(x+y). Put v = x+y → dv/dx − 1 = sin v + cos v → separable in v.
Q25. 2y dx + x(1 + log y) dy = 0. Treat x as function of y: dx/dy + (1+log y)/(2y)·x = 0 → linear in x.
Q26. Verify y = sin x is solution of y” + y = 0: y” = −sin x → y” + y = 0. ✓
Q27. dy/dx + y = e^(−x). IF = e^x. y e^x = ∫e^(−x)·e^x dx = ∫dx = x + c → y = (x+c)e^(−x).
Q28. Tangent line at any point (x,y) to curve has slope x+y. Find curve through origin. From Q16: y = ce^x − (x+1); y(0)=0 → c=1. y = e^x − x − 1.
Q29. dy/dx = (y/x) + tan(y/x). Homogeneous → y=vx → v + x(dv/dx) = v + tan v → cot v dv = dx/x → ln|sin v| = ln|x| + c → sin(y/x) = cx.
Q30. Order of ODE of all conics with axes parallel to coordinate axes is 3 (3 parameters: a, b, focus).
Q31. Bacterial culture doubles every 5 hours. Time to triple? N = N₀e^(kt). 2 = e^(5k) → k = ln2/5. 3 = e^(kt) → t = 5 ln3/ln2 ≈ 7.92 hr.
Q32. dy/dx = y² − x²/(2xy) (homogeneous variant) — not the same as Q4; care with signs.
Q33. Orthogonal trajectories of y = cx² satisfy dy/dx = 2cx. Eliminate c: c = y/x². Slope of orthogonal: dy/dx = −1/(2y/x) = −x/(2y) → 2y dy = −x dx → y² + x²/2 = C.
Q34. y = x sin(log x + c). Differentiate: y’ = sin(…) + cos(…). Eliminate c → ODE.
Q35. dy/dx − y/x = (x+1)/x. Linear. IF = 1/x. y/x = ∫(x+1)/x² dx = ln|x| − 1/x + c → y = x ln|x| − 1 + cx.

10. JEE Main Strategy for Differential Equations

Memorise the 6-row table above — every JEE problem maps to exactly one method. Focus 60% prep on linear + variable separable + homogeneous (these account for 80%+ of JEE Main DE problems). Solve previous-year sets at JEE Gurukul Free Resources, follow our JEE 2027 roadmap, and join our maths intensives for live problem-solving.

FAQ

Q1. How do I quickly identify the type of ODE?

Check in this order: separable? linear? homogeneous? Bernoulli? exact? It nearly always falls into one of these in JEE Main.

Q2. What’s the difference between order and degree?

Order = highest derivative present. Degree = exponent of that highest derivative once the equation is rationalised (cleared of fractional powers / radicals of derivatives).

Q3. Are partial differential equations in JEE Main syllabus?

No — JEE Main covers only ordinary differential equations of first order and first degree (mostly), plus the formation of ODEs from families of curves.

Join JEE Gurukul Maths Course →

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