Question

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

• A. $\sqrt{2\left(e^2-1\right)}$
• B. $\sqrt{2\left(e^2+1\right)}$
• C. $\sqrt{3}\cdot e$
• D. $\sqrt{\frac{e^2+1}{2}}$

Answer 1

Answer: (C)

We have, $\frac{dy}{dx}=\frac{xy}{x^2+y^2}$
Substitute $y=vx$
$\Rightarrow \frac{dy}{dx}=\frac{xdv}{dx}+v$
Now, equation becomes
$\frac{xdv}{dx}+v=\frac{v}{1+v^2}$
$\Rightarrow \frac{xdv}{dx}=\frac{v}{1+v^2}-v$
$\Rightarrow \frac{dx}{x}+\left(\frac{1+v^2}{v^3}\right)dv=0$
On integration, we get
$lnx+lnv-\frac{1}{2v^2}=c$
$\Rightarrow lny=\frac{x^2}{2y^2}+c$
$x=1$,
$y=1$
$c=-\frac{1}{2}$
$\therefore x=x_0$,
$y=e$
$\Rightarrow 1=\frac{x_0^2}{2e^2}-\frac{1}{2}$
$\Rightarrow x_0^2=3e^2$,
$x_0=\sqrt{3}\cdot e$.