Question

The solution of the differential equation \( \frac{xdy - ydx}{xdx + ydy} = \sqrt{x^2+y^2} \) is:

• A. \( \frac{x}{y} = \sin^{-1}\sqrt{1-x^2} + C \); where C is a constant
• B. \( \sqrt{x^2+y^2} = \tan^{-1}\frac{y}{x} + C \); where C is a constant
• C. \( 1+x^2 = \tan^{-1}(y) + C \); where C is a constant
• D. \( y = x\tan(\sqrt{x^2+y^2}) + C \); where C is a constant

Hint:

Recognize standard differential forms! \(xdy-ydx\) screams "polar coordinates" or "division by \(x^2\) or \(y^2\)". \(xdx+ydy\) also suggests polar coordinates or anything involving \(x^2+y^2\). Seeing both together makes the polar coordinate transformation the most efficient path to the solution.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The structure of the numerator and denominator suggests a transformation to polar coordinates, as the combinations \(xdy-ydx\) and \(xdx+ydy\) have very simple forms in the polar system. 

Step 2: Key Formula or Approach: 
Use the polar coordinate substitution: \(x = r\cos\theta\), \(y = r\sin\theta\). This implies: - \(x^2+y^2 = r^2\) - \(xdx+ydy = rdr\) - \(xdy-ydx = r^2d\theta\) 

Step 3: Detailed Explanation: 
Substitute the polar differential forms into the given equation: \[ \frac{r^2 d\theta}{r dr} = \sqrt{r^2} \] Assuming \(r>0\), this simplifies to: \[ \frac{r d\theta}{dr} = r \] Dividing by \(r\) (assuming \(r \neq 0\)): \[ \frac{d\theta}{dr} = 1 \implies d\theta = dr \] Now, integrate both sides: \[ \int d\theta = \int dr \] \[ \theta = r + C' \] where \(C'\) is the constant of integration. Finally, substitute 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 \( C = -C' \), then we can write: \[ \sqrt{x^2+y^2} = \tan^{-1}\left(\frac{y}{x}\right) - C \] This is equivalent to \( \sqrt{x^2+y^2} = \tan^{-1}\left(\frac{y}{x}\right) + C_1 \) where \(C_1\) is just another constant. This matches option (B). 
Now let's check 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 same solution we derived. Therefore, option (D) is also a correct representation of the solution. 

Step 4: Final Answer: 
Both options (B) and (D) represent the solution to the differential equation. The question asks for ""the solution,"" implying a single choice, but provides multiple correct options. The corresponding multiple-choice option is ""B and D only"".