Let y = y(x) be the solution of the differential equation dy/dx = (1 + x + x^2)(1 - y + y^2), y(0) = 1/2. Then (2y(1) - 1) is equal to

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q20 – JEE-MAIN 2026

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Home Maths Differential EquationsQuestion #175

Let

y = y (x)

be the solution of the differential equation

\frac{d y}{d x} = (1 + x + x^{2}) (1 - y + y^{2})

y (0) = \frac{1}{2}

. Then

(2 y (1) - 1)

is equal to

Clue 1 of 3: Recognize the Equation Type

The given differential equation has its variables separated. The terms involving $y$ can be grouped on one side with $d y$ , and terms involving $x$ can be grouped on the other side with $d x$ .

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Ninja Strategy•Structural Analysis

Quickly determine the structure of the answer by tracking the constants from integration. The integral of the polynomial at $x = 1$ is $\frac{11}{6}$ , and the $\tan^{- 1}$ integral form gives factors of $\frac{2}{\sqrt{3}}$ and $\frac{1}{\sqrt{3}}$ , which dictate the final argument and pre-factor.

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Step 1: Separate Variables and Integrate✦ Active

The differential equation is separable. Rearrange it and integrate both sides. The left side requires completing the square in the denominator: $1 - y + y^{2} = y^{2} - y + 1 = (y - \frac{1}{2})^{2} + \frac{3}{4}$ .

\int \frac{d y}{(y - \frac{1}{2})^{2} + (\frac{\sqrt{3}}{2})^{2}} = \int (1 + x + x^{2}) d x

Evaluating the integrals gives:

\frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} (\frac{y - \frac{1}{2}}{\frac{\sqrt{3}}{2}}) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + C

\frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 y - 1}{\sqrt{3}}) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + C

Step 2: Apply Initial Condition to Find the Constant○ Expand

Use the given condition $y (0) = 1 / 2$ to find the constant of integration $C$ . Substitute $x = 0$ and $y = 1 / 2$ into the general solution.

\frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 (\frac{1}{2}) - 1}{\sqrt{3}}) = 0 + 0 + 0 + C

This simplifies to $\frac{2}{\sqrt{3}} \tan^{- 1} (0) = C$ , which means $C = 0$ . The particular solution is:

\frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 y - 1}{\sqrt{3}}) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3}

Step 3: Evaluate the Expression at x=1○ Expand

The question asks for the value of $2 y (1) - 1$ . Substitute $x = 1$ into the particular solution.

\frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 y (1) - 1}{\sqrt{3}}) = 1 + \frac{1^{2}}{2} + \frac{1^{3}}{3} = 1 + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}

Now, solve for the expression $2 y (1) - 1$ .

\tan^{- 1} (\frac{2 y (1) - 1}{\sqrt{3}}) = \frac{11}{6} \cdot \frac{\sqrt{3}}{2} = \frac{11 \sqrt{3}}{12}

\frac{2 y (1) - 1}{\sqrt{3}} = \tan (\frac{11 \sqrt{3}}{12})

2 y (1) - 1 = \sqrt{3} \tan (\frac{11 \sqrt{3}}{12})

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Maths Question 20 – JEE-MAIN 2026

Clue 1 of 3: Recognize the Equation Type

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