Q45 / 75 · JEE Main 2026 (04 April Shift 1) · Mathematics
Let $y = y(x)$ be the solution of the differential equation $\dfrac{dy}{dx} = (1 + x + x^2)(1 - y + y^2)$, $y(0) = \dfrac{1}{2}$. Then $(2y(1) - 1)$ is equal to:
- A$\sqrt{3}\tan\left(\dfrac{11\sqrt{3}}{6}\right)$
- B$\dfrac{\sqrt{3}}{2}\tan\left(\dfrac{11\sqrt{3}}{12}\right)$
- Correct answerC$\sqrt{3}\tan\left(\dfrac{11\sqrt{3}}{12}\right)$
- D$\dfrac{\sqrt{3}}{2}\tan\left(\dfrac{11\sqrt{3}}{6}\right)$
Solution
The given differential equation is $\dfrac{dy}{dx} = (1 + x + x^2)(1 - y + y^2)$.
Separating the variables:
$\dfrac{dy}{y^2 - y + 1} = (x^2 + x + 1) dx$
Integrating both sides:
$\int \dfrac{dy}{\left(y - \dfrac{1}{2}\right)^2 + \dfrac{3}{4}} = \int (x^2 + x + 1) dx$
$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{y - \dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right) = \dfrac{x^3}{3} + \dfrac{x^2}{2} + x + C$
$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{2y - 1}{\sqrt{3}}\right) = \dfrac{x^3}{3} + \dfrac{x^2}{2} + x + C$
Given $y(0) = \dfrac{1}{2}$, substituting $x = 0$ and $y = \dfrac{1}{2}$:
$\dfrac{2}{\sqrt{3}} \tan^{-1}(0) = C \Rightarrow C = 0$
Substituting $x = 1$ to find $y(1)$:
$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{2y(1) - 1}{\sqrt{3}}\right) = \dfrac{1^3}{3} + \dfrac{1^2}{2} + 1$
$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{2y(1) - 1}{\sqrt{3}}\right) = \dfrac{11}{6}$
$\tan^{-1}\left(\dfrac{2y(1) - 1}{\sqrt{3}}\right) = \dfrac{11\sqrt{3}}{12}$
$2y(1) - 1 = \sqrt{3}\tan\left(\dfrac{11\sqrt{3}}{12}\right)$
Answer: $\sqrt{3}\tan\left(\dfrac{11\sqrt{3}}{12}\right)$
Separating the variables:
$\dfrac{dy}{y^2 - y + 1} = (x^2 + x + 1) dx$
Integrating both sides:
$\int \dfrac{dy}{\left(y - \dfrac{1}{2}\right)^2 + \dfrac{3}{4}} = \int (x^2 + x + 1) dx$
$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{y - \dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right) = \dfrac{x^3}{3} + \dfrac{x^2}{2} + x + C$
$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{2y - 1}{\sqrt{3}}\right) = \dfrac{x^3}{3} + \dfrac{x^2}{2} + x + C$
Given $y(0) = \dfrac{1}{2}$, substituting $x = 0$ and $y = \dfrac{1}{2}$:
$\dfrac{2}{\sqrt{3}} \tan^{-1}(0) = C \Rightarrow C = 0$
Substituting $x = 1$ to find $y(1)$:
$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{2y(1) - 1}{\sqrt{3}}\right) = \dfrac{1^3}{3} + \dfrac{1^2}{2} + 1$
$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{2y(1) - 1}{\sqrt{3}}\right) = \dfrac{11}{6}$
$\tan^{-1}\left(\dfrac{2y(1) - 1}{\sqrt{3}}\right) = \dfrac{11\sqrt{3}}{12}$
$2y(1) - 1 = \sqrt{3}\tan\left(\dfrac{11\sqrt{3}}{12}\right)$
Answer: $\sqrt{3}\tan\left(\dfrac{11\sqrt{3}}{12}\right)$
Hence, the correct answer is option: 'C'.