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: Switch the variables
Write $\dfrac{dx}{dy}=\dfrac{x^2+y^2}{xy}$ and set $x=uy$.

Step 2: Solve
$u+y\,u'=u+\dfrac1u$, so $u\,du=\dfrac{dy}{y}$ and $\dfrac{u^2}{2}=\log y+C$, i.e. $\dfrac{x^2}{2y^2}=\log y+C$. With $(1,1)$: $C=\tfrac12$. At $y=e$: $\dfrac{x_0^2}{2e^2}=1+\tfrac12$, so $x_0=\pm\sqrt3\,e$, option (B).

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