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Friday, 9 October

JEE Main Maths · Differential Equations

Initial Value Problem

Initial Value Problem (Differential Equations) was asked 7 times in 7 JEE Main shifts, 2017–2026, out of the 42 shifts Lumi analysed. 1 of the 7 had a numerical answer.

Times asked
7
in 7 shifts
Years
2017–2026
Usual format
Single correct
Multi-step
Pattern
Frequent
repeats across shifts

How it is asked

  • Solve initial value differential equation
  • Initial value problem for separable equation

Difficulty

Easy 0Medium 7Hard 0

Where it sits

Chapter: Differential Equations. Topic: Variable Separable Form, 12 questions over 6 ideas.

Every time it came

All 7 questions

Newest first, each with its options and official answer.

  1. 4 Apr 2026, Shift 1 · Q20Let y=y(x)y=y(x) be the solution of the differential equation dydx=(1+x+x2)(1−y+y2)\frac{dy}{dx}=(1+x+x^2)(1-y+y^2), y(0)=12y(0)=\frac12. Then (2y(1)−1)(2y(1)-1) is equal toMediumSingle correct
  2. 8 Apr 2024, Shift 1 · Q13Let y=y(x)y=y(x) be the solution of the differential equation (1+y2)etan⁡xdx+cos⁡2x (1+e2tan⁡x)dy=0(1+y^2)e^{\tan x}dx+\cos^2x\,(1+e^{2\tan x})dy=0, y(0)=1y(0)=1. Then…MediumSingle correct
  3. 9 Apr 2024, Shift 1 · Q11The solution curve, of the differential equation 2ydydx+3=5dydx2y\frac{dy}{dx}+3=5\frac{dy}{dx}, passing through the point (0,1)(0,1) is a conic, whose…MediumSingle correct
  4. 29 Jan 2024, Shift 1 · Q26If the solution curve y=y(x)y=y(x) of the differential equation (1+y2)(1+log⁡ex)dx+x dy=0(1+y^2)(1+\log_e x)dx+x\,dy=0, x>0x>0 passes through the point (1,1)(1,1) and…MediumNumerical value
  5. 8 Jan 2020, Shift 1 · Q62Let y=y(x)y=y(x) be a solution of the differential equation, 1−x2 dydx+1−y2=0\sqrt{1-x^2}\,\dfrac{dy}{dx}+\sqrt{1-y^2}=0, ∣x∣<1|x|<1. If…MediumSingle correct
  6. 2 Sep 2020, Shift 1 · Q63Let y=y(x)y=y(x) be the solution of the differential equation, 2+sin⁡xy+1⋅dydx=−cos⁡x, y>0, y(0)=1\frac{2+\sin x}{y+1}\cdot\frac{dy}{dx}=-\cos x,\ y>0,\ y(0)=1. If y(π)=ay(\pi)=a…MediumSingle correct
  7. 2 Apr 2017 (offline) · Q77If (2+sin⁡x)dydx+(y+1)cos⁡x=0(2+\sin x)\frac{dy}{dx}+(y+1)\cos x=0 and y(0)=1y(0)=1, then y(π2)y\left(\frac{\pi}{2}\right) is equal to :MediumSingle correct