Question

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

• A. \(\sqrt{3} \tan \left( \frac{11\sqrt{3}}{6} \right)\)
• B. \(\frac{\sqrt{3}}{2} \tan \left( \frac{11\sqrt{3}}{12} \right)\)
• C. \(\sqrt{3} \tan \left( \frac{11\sqrt{3}}{12} \right)\)
• D. \(\frac{\sqrt{3}}{2} \tan \left( \frac{11\sqrt{3}}{6} \right)\)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a variable separable differential equation. We group terms involving \(y\) on the left and terms involving \(x\) on the right, then integrate both sides. Both sides will result in \(\tan^{-1}\) forms after completing the square.

Step 2: Key Formula or Approach:
1. \(\int \frac{dy}{y^2 - y + 1} = \int (x^2 + x + 1) \, dx\). 2. LHS: \(\int \frac{dy}{(y-1/2)^2 + 3/4} = \frac{2}{\sqrt{3}} \tan^{-1}\left( \frac{y-1/2}{\sqrt{3}/2} \right) = \frac{2}{\sqrt{3}} \tan^{-1}\left( \frac{2y-1}{\sqrt{3}} \right)\). 3. RHS: \(\frac{x^3}{3} + \frac{x^2}{2} + x + C\).

Step 3: Detailed Explanation:
1. Apply initial condition \(y(0) = 1/2\): \(\frac{2}{\sqrt{3}} \tan^{-1}(0) = 0 + 0 + 0 + C \implies C = 0\). 2. Now, at \(x = 1\): \(\frac{2}{\sqrt{3}} \tan^{-1}\left( \frac{2y(1)-1}{\sqrt{3}} \right) = \frac{1}{3} + \frac{1}{2} + 1 = \frac{2 + 3 + 6}{6} = \frac{11}{6}\). 3. Rearrange to find \(2y(1)-1\): \(\tan^{-1}\left( \frac{2y(1)-1}{\sqrt{3}} \right) = \frac{11}{6} \cdot \frac{\sqrt{3}}{2} = \frac{11\sqrt{3}}{12}\). \(\frac{2y(1)-1}{\sqrt{3}} = \tan\left( \frac{11\sqrt{3}}{12} \right)\). 4. \(2y(1)-1 = \sqrt{3} \tan\left( \frac{11\sqrt{3}}{12} \right)\).

Step 4: Final Answer:
The expression is equal to \(\sqrt{3} \tan \left( \frac{11\sqrt{3}}{12} \right)\).