Differential EquationsmediumPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Differential Equations: Let Solution Differential Equation Equal (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question
Let y=y(x)y=y(x) be the solution of the differential equation dydx=(1+x+x2)(1−y+y2)\dfrac{dy}{dx}=(1+x+x^2)(1-y+y^2), y(0)=12y(0)=\dfrac12. Then (2y(1)−1)(2y(1)-1) is equal to
A3tan⁡(1136)\sqrt3\tan\left(\dfrac{11\sqrt3}{6}\right)
B32tan⁡(11312)\dfrac{\sqrt3}{2}\tan\left(\dfrac{11\sqrt3}{12}\right)
C3tan⁡(11312)\sqrt3\tan\left(\dfrac{11\sqrt3}{12}\right)correct
D32tan⁡(1136)\dfrac{\sqrt3}{2}\tan\left(\dfrac{11\sqrt3}{6}\right)
Solution
Step 1: Separate variables:
dyy2−y+1=(x2+x+1) dx ⇒ dy(y−12)2+34=(x2+x+1) dx.\frac{dy}{y^2-y+1}=(x^2+x+1)\,dx\ \Rightarrow\ \frac{dy}{\left(y-\frac12\right)^2+\frac34}=(x^2+x+1)\,dx.
 Step 2: Integrate:
23tan⁡−1(y−1232)=x33+x22+x+C.\frac{2}{\sqrt3}\tan^{-1}\left(\frac{y-\frac12}{\frac{\sqrt3}{2}}\right)=\frac{x^3}{3}+\frac{x^2}{2}+x+C.
 Step 3: y(0)=12⇒0=0+C⇒C=0y(0)=\dfrac12\Rightarrow0=0+C\Rightarrow C=0. At x=1x=1, RHS =13+12+1=116=\dfrac13+\dfrac12+1=\dfrac{11}{6}:
23tan⁡−1(2y−13)=116 ⇒ tan⁡−1(2y−13)=11312.\frac{2}{\sqrt3}\tan^{-1}\left(\frac{2y-1}{\sqrt3}\right)=\frac{11}{6}\ \Rightarrow\ \tan^{-1}\left(\frac{2y-1}{\sqrt3}\right)=\frac{11\sqrt3}{12}.
 Step 4:
2y(1)−13=tan⁡(11312) ⇒ 2y(1)−1=3tan⁡(11312).\frac{2y(1)-1}{\sqrt3}=\tan\left(\frac{11\sqrt3}{12}\right)\ \Rightarrow\ 2y(1)-1=\sqrt3\tan\left(\frac{11\sqrt3}{12}\right).
 Correct answer: (3)
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