Question

If $(x^2+y^2)dy=2xydx$ and $y\left(0\right)=2e^2,y\left(x_0\right)=1$, then $x_0=$

• A. $\sqrt{1-2e^2}$
• B. $\sqrt{2e^2-1}/\sqrt{2}$
• C. $\sqrt{e^2-1}/\sqrt{2}$
• D. $\sqrt{e^2+1}/\sqrt{2}$

Answer 1

The correct option is A $\sqrt{1-2e^2}$

$(x^2+y^2)dy=2xydx$

$y^{2}dy=2xydx-x^{2}dy$

$dy=\frac{yd\left(x^2\right)-x^{2}dy}{y^2}$

$dy=d\left(\frac{x^2}{y}\right)$

By integrating, we get

$y=\frac{x^2}{y}+C$

$y\left(0\right)=2e^2$

$c=2e^2$

$y=\frac{x^2}{y}+2e^2$

$y\left(x_0\right)=1$

$1=\frac{x_0^2}{1}+2e^2$

$x_0=\sqrt{1-2e^2}$