Question

Consider the differential equation $\frac{dy}{dx} = (x^2 + x + 1)(y^2 - y + 1)$. If $y(0) = \frac{1}{2}$, then the value of $2(y(1)) - 1$ is

• A. $\sqrt{3} \tan \left( \frac{11\sqrt{3}}{12} \right)$
• B. $\sqrt{3} \tan \left( \frac{12\sqrt{3}}{17} \right)$
• C. $\sqrt{2} \tan \left( \frac{11\sqrt{3}}{12} \right)$
• D. $\sqrt{2} \tan \left( \frac{12\sqrt{3}}{11} \right)$

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a variable separable differential equation. We rearrange terms and integrate both sides.
Step 2: Key Formula or Approach:
Separating variables:
\[ \int \frac{dy}{y^2 - y + 1} = \int (x^2 + x + 1) dx \]
Step 3: Detailed Explanation:
1. Integrate the LHS:
$y^2 - y + 1 = (y - 1/2)^2 + 3/4$.
\[ \int \frac{dy}{(y - 1/2)^2 + (\sqrt{3}/2)^2} = \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{y - 1/2}{\sqrt{3}/2} \right) = \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2y-1}{\sqrt{3}} \right) \]
2. Integrate the RHS:
\[ \int (x^2 + x + 1) dx = \frac{x^3}{3} + \frac{x^2}{2} + x + C \]
3. Use $y(0) = 1/2$:
$\frac{2}{\sqrt{3}} \tan^{-1}(0) = 0 + 0 + 0 + C \implies C = 0$.
4. Find $y(1)$:
$\frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2y(1)-1}{\sqrt{3}} \right) = \frac{1}{3} + \frac{1}{2} + 1 = \frac{11}{6}$.
$\tan^{-1} \left( \frac{2y(1)-1}{\sqrt{3}} \right) = \frac{11\sqrt{3}}{12}$.
$2y(1)-1 = \sqrt{3} \tan \left( \frac{11\sqrt{3}}{12} \right)$.
Step 4: Final Answer:
The value is $\sqrt{3} \tan \left( \frac{11\sqrt{3}}{12} \right)$.