Step 1: Concept Overview: The structure of the equation, specifically the expressions \(xdy-ydx\) and \(xdx+ydy\), suggests using polar coordinates for simplification.
Step 2: Key Substitution: Use the polar coordinate substitution: \(x = r\cos\theta\), \(y = r\sin\theta\). This leads to: - \(x^2+y^2 = r^2\) - \(xdx+ydy = rdr\) - \(xdy-ydx = r^2d\theta\)
Step 3: Detailed Solution:
Substitute the polar differential forms into the given equation: \[ \frac{r^2 d\theta}{r dr} = \sqrt{r^2} \] Simplify, assuming \(r>0\): \[ \frac{r d\theta}{dr} = r \] Divide by \(r\) (assuming \(r eq 0\)): \[ \frac{d\theta}{dr} = 1 \implies d\theta = dr \] Integrate both sides: \[ \int d\theta = \int dr \] \[ \theta = r + C' \] where \(C'\) is the constant of integration. Transform back to Cartesian coordinates: \( r = \sqrt{x^2+y^2} \) and \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \). \[ \tan^{-1}\left(\frac{y}{x}\right) = \sqrt{x^2+y^2} + C' \] Let \( C = -C' \), so: \[ \sqrt{x^2+y^2} = \tan^{-1}\left(\frac{y}{x}\right) - C \] This is equivalent to \( \sqrt{x^2+y^2} = \tan^{-1}\left(\frac{y}{x}\right) + C_1 \) where \(C_1\) is another constant. This matches option (B). Now, consider option (D): \( y = x\tan(\sqrt{x^2+y^2} + C) \) Divide by \(x\): \[ \frac{y}{x} = \tan(\sqrt{x^2+y^2} + C) \] Take the arctan of both sides: \[ \tan^{-1}\left(\frac{y}{x}\right) = \sqrt{x^2+y^2} + C \] This is the derived solution. Therefore, option (D) is also a correct solution.
Step 4: Final Answer:
Both options (B) and (D) are correct. The question's wording implies a single answer, but offers multiple valid choices. The correct multiple-choice selection is ""B and D only"".