Question

For the differential equation \((x^2+y^2)\,dy = xy\,dx\), it is given that \(y(1) = 1\) and \(y(x_0) = e\), then the value of \(x_0\) is _____

• A. \(e\)
• B. \(\pm \sqrt{3}\,e\)
• C. \(3e^2\)
• D. \(e^2\)

Hint:

Substitute \(y=vx\) and separate the variables.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
\(\dfrac{dy}{dx}=\dfrac{xy}{x^2+y^2}\) is homogeneous of degree 0, so put \(y=vx\).

Step 2: Key Formula or Approach
\(v+x\dfrac{dv}{dx}=\dfrac{v}{1+v^2}\), so \(x\dfrac{dv}{dx}=-\dfrac{v^3}{1+v^2}\).

Step 3: Detailed Explanation
Separate: \(\dfrac{1+v^2}{v^3}dv=-\dfrac{dx}{x}\). Integrate: \(-\dfrac{1}{2v^2}+\log v=-\log x+C\).
Since \(v=y/x\): \(\log y-\dfrac{x^2}{2y^2}=C\).
Using \(y(1)=1\): \(0-\tfrac12=C\).
At \(y=e\): \(1-\dfrac{x_0^2}{2e^2}=-\dfrac12\), so \(x_0^2=3e^2\) and \(x_0=\pm\sqrt3\,e\).

Final Answer:
The value of \(x_0\) is \(\pm\sqrt3\,e\), option (B). \[ \boxed{\pm\sqrt3\,e\ \text{(B)}} \]