JEE Main 2026 — Mathematics Calculus

The given differential equation is $\dfrac{dy}{dx} = (1 + x + x^2)(1 - y + y^2)$dxdy​=(1+x+x2)(1−y+y2).

Separating the variables:

$$\dfrac{dy}{y^2 - y + 1} = (x^2 + x + 1) dx$$y2−y+1dy​=(x2+x+1)dx

Integrating both sides:

$$\int \dfrac{dy}{\left(y - \dfrac{1}{2}\right)^2 + \dfrac{3}{4}} = \int (x^2 + x + 1) dx$$∫(y−21​)2+43​dy​=∫(x2+x+1)dx

$$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{y - \dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right) = \dfrac{x^3}{3} + \dfrac{x^2}{2} + x + C$$3​2​tan−1​23​​y−21​​​=3x3​+2x2​+x+C

$$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{2y - 1}{\sqrt{3}}\right) = \dfrac{x^3}{3} + \dfrac{x^2}{2} + x + C$$3​2​tan−1(3​2y−1​)=3x3​+2x2​+x+C

Given $y(0) = \dfrac{1}{2}$y(0)=21​, substituting $x = 0$x=0 and $y = \dfrac{1}{2}$y=21​:

$$\dfrac{2}{\sqrt{3}} \tan^{-1}(0) = C \Rightarrow C = 0$$3​2​tan−1(0)=C⇒C=0

Substituting $x = 1$x=1 to find $y(1)$y(1):

$$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{2y(1) - 1}{\sqrt{3}}\right) = \dfrac{1^3}{3} + \dfrac{1^2}{2} + 1$$3​2​tan−1(3​2y(1)−1​)=313​+212​+1

$$\dfrac{2}{\sqrt{3}} \tan^{-1}\left(\dfrac{2y(1) - 1}{\sqrt{3}}\right) = \dfrac{11}{6}$$3​2​tan−1(3​2y(1)−1​)=611​

$$\tan^{-1}\left(\dfrac{2y(1) - 1}{\sqrt{3}}\right) = \dfrac{11\sqrt{3}}{12}$$tan−1(3​2y(1)−1​)=12113​​

$$2y(1) - 1 = \sqrt{3}\tan\left(\dfrac{11\sqrt{3}}{12}\right)$$2y(1)−1=3​tan(12113​​)

Answer: $\sqrt{3}\tan\left(\dfrac{11\sqrt{3}}{12}\right)$3​tan(12113​​)