Question:

<p><strong>Step 1:</strong> Rearrange the given equation into standard form.</p><p>Given: $\frac{xdy}{x^2 + y^2} - \frac{y}{x^2 + y^2}dx = 0$</p><p>Rewrite as: $-\frac{y}{x^2 + y^2}dx + \frac{x}{x^2 + y^2}dy = 0$</p><p><strong>Step 2:</strong> Identify M and N.</p><p>Here, $M = -\frac{y}{x^2 + y^2}$ and $N = \frac{x}{x^2 + y^2}$</p><p><strong>Step 3:</strong> Verify exactness by checking $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.</p><p>$\frac{\partial M}{\partial y} = -\frac{(x^2 + y^2) - y(2y)}{(x^2 + y^2)^2} = -\frac{x^2 - y^2}{(x^2 + y^2)^2}$</p><p>$\frac{\partial N}{\partial x} = \frac{(x^2 + y^2) - x(2x)}{(x^2 + y^2)^2} = \frac{y^2 - x^2}{(x^2 + y^2)^2} = -\frac{x^2 - y^2}{(x^2 + y^2)^2}$</p><p>Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact.</p><p><strong>Step 4:</strong> Find the solution by integrating M with respect to x (treating y as constant).</p><p>$\int M\,dx = \int -\frac{y}{x^2 + y^2}dx = -y \cdot \frac{1}{y}\tan^{-1}\left(\frac{x}{y}\right) + g(y) = -\tan^{-1}\left(\frac{x}{y}\right) + g(y)$</p><p><strong>Step 5:</strong> Differentiate with respect to y and compare with N to find g(y).</p><p>$\frac{\partial}{\partial y}\left[-\tan^{-1}\left(\frac{x}{y}\right) + g(y)\right] = -\frac{-1}{1 + (x/y)^2} \cdot \frac{-x}{y^2} + g'(y) = -\frac{x}{x^2 + y^2} + g'(y)$</p><p>This must equal $N = \frac{x}{x^2 + y^2}$, so $g'(y) = 0$, giving $g(y) = c_1$ (constant).</p><p><strong>Step 6:</strong> Write the solution.</p><p>$-\tan^{-1}\left(\frac{x}{y}\right) = c_1$</p><p>$\tan^{-1}\left(\frac{x}{y}\right) = -c_1 = c$ (where c is an arbitrary constant)</p><p><strong>∴ Answer: A</strong></p>