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:
The given differential equation allows variables to be separated. We arrange all \(y\) terms on one side and all \(x\) terms on the other, then integrate both sides.
Step 2: Key Formula or Approach:
\[ \frac{dy}{1 - y + y^2} = (1 + x + x^2) dx \] We complete the square in the denominator:
\(y^2 - y + 1 = \left(y - \frac{1}{2}\right)^2 + \frac{3}{4}\).
The integral form \(\int \frac{du}{u^2 + a^2} = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) + C\) is used.
Step 3: Detailed Explanation:
Integrate both sides:
\[ \int \frac{dy}{\left(y - \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \int (1 + x + x^2) dx \] \[ \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\left(\frac{y - \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = x + \frac{x^2}{2} + \frac{x^3}{3} + C \] \[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2y - 1}{\sqrt{3}}\right) = x + \frac{x^2}{2} + \frac{x^3}{3} + C \] Use the initial condition \(y(0) = \frac{1}{2}\) to find \(C\):
\[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2(\frac{1}{2}) - 1}{\sqrt{3}}\right) = 0 + 0 + 0 + C \implies \frac{2}{\sqrt{3}} \tan^{-1}(0) = C \implies C = 0 \] The particular solution is:
\[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2y - 1}{\sqrt{3}}\right) = x + \frac{x^2}{2} + \frac{x^3}{3} \] We need to find the value at \(x = 1\):
\[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2y(1) - 1}{\sqrt{3}}\right) = 1 + \frac{1^2}{2} + \frac{1^3}{3} = 1 + \frac{1}{2} + \frac{1}{3} = \frac{11}{6} \] Step 4: Final Answer:
Solve for \(2y(1) - 1\):
\[ \tan^{-1}\left(\frac{2y(1) - 1}{\sqrt{3}}\right) = \frac{\sqrt{3}}{2} \times \frac{11}{6} = \frac{11\sqrt{3}}{12} \] \[ \frac{2y(1) - 1}{\sqrt{3}} = \tan\left(\frac{11\sqrt{3}}{12}\right) \] \[ 2y(1) - 1 = \sqrt{3}\tan\left(\frac{11\sqrt{3}}{12}\right) \]