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JEE Mains · Maths · STD 12 - 9. differential equations

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:

  1. A \(\sqrt{3}\tan\left(\dfrac{11\sqrt{3}}{6}\right)\)
  2. B \(\dfrac{\sqrt{3}}{2}\tan\left(\dfrac{11\sqrt{3}}{12}\right)\)
  3. C \(\sqrt{3}\tan\left(\dfrac{11\sqrt{3}}{12}\right)\)
  4. D \(\dfrac{\sqrt{3}}{2}\tan\left(\dfrac{11\sqrt{3}}{6}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\sqrt{3}\tan\left(\dfrac{11\sqrt{3}}{12}\right)\)

Step-by-step Solution

Detailed explanation

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: \(\int \dfrac{dy}{\left(y - \dfrac{1}{2}\right)^2 + \dfrac{3}{4}} = \int (x^2 + x + 1) dx\)…
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