Question

If \( y = y(x) \) and \[ (1 + x^2)\,dy + (1 - \tan^{-1}x)\,dx = 0 \] and \( y(0) = 1 \), then \( y(1) \) is equal to:

• A. \( \dfrac{\pi^2}{32} + \dfrac{\pi}{4} + 1 \)
• B. \( \dfrac{\pi^2}{32} - \dfrac{\pi}{4} + 1 \)
• C. \( \dfrac{\pi^2}{32} - \dfrac{\pi}{2} - 1 \)
• D. \( \dfrac{\pi^2}{32} - \dfrac{\pi}{2} + 1 \)

Hint:

For equations of the form \( f(x)\,dy + g(x)\,dx = 0 \): \begin{itemize} \item Separate variables carefully \item Use standard integrals involving \( \tan^{-1}x \) \end{itemize}

Answer 1

The Correct Option is B

Step 1: Rewrite the given differential equation: \[ (1 + x^2)\,dy = - (1 - \tan^{-1}x)\,dx \]
Step 2: Separate the variables: \[ dy = -\frac{1 - \tan^{-1}x}{1 + x^2}\,dx \]
Step 3: Integrate both sides: \[ \int dy = -\int \frac{1}{1 + x^2}\,dx + \int \frac{\tan^{-1}x}{1 + x^2}\,dx \] \[ y = -\tan^{-1}x + \frac{1}{2}(\tan^{-1}x)^2 + C \]
Step 4: Apply the initial condition \( y(0) = 1 \): Since \( \tan^{-1}(0) = 0 \), \[ 1 = C \Rightarrow C = 1 \]
Step 5: Evaluate \( y(1) \): \[ \tan^{-1}(1) = \frac{\pi}{4} \] \[ y(1) = -\frac{\pi}{4} + \frac{1}{2}\left(\frac{\pi}{4}\right)^2 + 1 = \frac{\pi^2}{32} - \frac{\pi}{4} + 1 \]