Question

If \( x = f(y) \) is the solution of the differential equation \[ (1 + y^2) + (x - 2e^{\tan^{-1}y}) \frac{dy}{dx} = 0, \quad y \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right), \] with \( f(0) = 1 \), then \( f\left( \frac{1}{\sqrt{3}} \right) \) is equal to:

• A. \( e^{\frac{\pi}{3}} \)
• B. \( e^{\frac{\pi}{12}} \)
• C. \( e^{\frac{\pi}{6}} \)
• D. \( e^{\frac{\pi}{4}} \)

Hint:

When solving differential equations with separable variables: - Rearrange terms to separate \( x \) and \( y \) on opposite sides. - Integrate both sides carefully and apply any initial conditions provided to solve for constants of integration.

Answer 1

The Correct Option is C

The differential equation \((1 + y^2) + (x - 2e^{\tan^{-1}y}) \frac{dy}{dx} = 0\) is rewritten as \(\frac{dy}{dx} = -\frac{1+y^2}{x-2e^{\tan^{-1}y}}\).

Given \(x = f(y)\), the equation can be expressed in terms of \(y\). Substituting \(\frac{dx}{dy} = -\frac{x-2e^{\tan^{-1}y}}{1+y^2}\) and simplifying yields:

\[\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{2e^{\tan^{-1}y}}{1+y^2}.\]

This is a linear first-order differential equation in the form \(\frac{dx}{dy} + P(y)x = Q(y)\), with \(P(y) = \frac{1}{1+y^2}\) and \(Q(y) = \frac{2e^{\tan^{-1}y}}{1+y^2}\).

The integrating factor \(I(y)\) is calculated as:

\[I(y) = e^{\int P(y) \, dy} = e^{\int \frac{1}{1+y^2} \, dy} = e^{\tan^{-1}y}.\]

Multiplying the equation by \(I(y)\) results in:

\[e^{\tan^{-1}y}\frac{dx}{dy} + \frac{e^{\tan^{-1}y}x}{1+y^2} = \frac{2e^{2\tan^{-1}y}}{1+y^2}.\]

The left side is the derivative of \(xe^{\tan^{-1}y}\). Integrating both sides gives:

\[\frac{d}{dy}\left(xe^{\tan^{-1}y}\right) = \frac{2e^{2\tan^{-1}y}}{1+y^2}.\]

The integral of the right side is:

\[xe^{\tan^{-1}y} = \int \frac{2e^{2\tan^{-1}y}}{1+y^2} \, dy = e^{2\tan^{-1}y} + C.\]

The general solution is therefore:

\[x = e^{\tan^{-1}y} + Ce^{-\tan^{-1}y}.\]

Using the initial condition \(f(0) = 1\), we get \(1 = e^0 + Ce^0\), which implies \(C = 0\).

The specific solution is \(x = e^{\tan^{-1}y}\).

Evaluating \(f\left(\frac{1}{\sqrt{3}}\right)\) yields:

\[f\left(\frac{1}{\sqrt{3}}\right) = e^{\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)}.\]

Since \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}\), the final result is:

\[f\left(\frac{1}{\sqrt{3}}\right) = e^{\frac{\pi}{6}}.\]