Question

If $\left(x^{2}+y^{2}\right)dy=xy\,dx$, $y\left(1\right)=1$, 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
$ln\,x+ln\,v-\frac{1}{2v^{2}}=c$
$\Rightarrow ln\,y=\frac{x^{2}}{2y^{2}}+c$
$x=1$,
$y=1$
$c=-\frac{1}{2}$
$\therefore x=x_{0}$,
$y=e$
$\Rightarrow 1=\frac{x^{2}_{0}}{2e^{2}}-\frac{1}{2}$
$\Rightarrow x^{2}_{0}=3e^{2}$,
$x_{0}=\sqrt{3}\cdot e$.