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

To solve the differential equation \((1 + x^2)\frac{dy}{dx} + (1 - \tan^{-1}x) = 0\)with the initial condition \(y(0) = 1\), we start by rewriting it as: \((1 + x^2)dy + (1 - \tan^{-1}x)dx = 0\).

This can be rearranged to: \(dy = -\frac{(1 - \tan^{-1}x)}{(1 + x^2)}dx\). 

The next step is to integrate both sides. The integral on the left is straightforward, while the right-hand side needs integration by parts. Let's solve the integral on the right:

Using integration by parts: \(\int (1 - \tan^{-1}x) \cdot \frac{1}{1 + x^2} \, dx\).

Let \(u = 1 - \tan^{-1}x\)and \(dv = \frac{1}{1 + x^2} \, dx\).

Then, \(du = -\frac{1}{1 + x^2} \, dx\)and \(v = \tan^{-1}x\).

The formula for integration by parts is: \(\int u \, dv = uv - \int v \, du\).

Applying this, we get: \(\int (1 - \tan^{-1}x) \cdot \frac{1}{1 + x^2} \, dx = (1 - \tan^{-1}x) \cdot \tan^{-1}x - \int -\tan^{-1}x \cdot \left(-\frac{1}{1 + x^2}\right) \, dx = (1 - \tan^{-1}x)\tan^{-1}x - \int \frac{(\tan^{-1}x)^2}{1 + x^2} \, dx \approx x(1 - \tan^{-1}x)\).

Finally, substitute back: \(y = \int -\frac{(1 - \tan^{-1}x)}{(1 + x^2)}dx + C\).

Use the initial condition \(y(0) = 1\):

• When \(x = 0\), \(y = 1\).
• Solving gives: \(1 = 0 + C\).
• Thus, \(C = 1\).

So, \(y = -\left(\int -\frac{(1 - \tan^{-1}x)}{(1 + x^2)}dx\right) + 1\).

Evaluate at \(x = 1\):

This results in: \(y(1) = \dfrac{\pi^2}{32} - \dfrac{\pi}{4} + 1\).

Therefore, the correct answer is \(\dfrac{\pi^2}{32} - \dfrac{\pi}{4} + 1\).