Question

If $\frac{dy}{dx}=\frac{xy}{x^2+y^2};y\left(1\right)=1;$ then a value of $x$ satisfying $y\left(x\right)=e$ is:

• A. $\sqrt{3}e$
• B. $\frac{1}{2}\sqrt{3}e$
• C. $\sqrt{2}e$
• D. $\frac{e}{\sqrt{2}}$

Answer 1

The correct option is A $\sqrt{3}e$

Let $x=vy$
$\frac{dx}{dy}=v+y\frac{dv}{dy}$
$\Rightarrow v+y\frac{dv}{dy}=\frac{y^2(1+v^2)}{v{y}^2}=\frac{1+v^2}{v}$
$\Rightarrow y\frac{dv}{dy}=\frac{1}{v}$
$\Rightarrow \frac{1}{y}dy=vdv$
$\Rightarrow \log y=\frac{v^2}{2}+\log c$
$\Rightarrow \log y=\frac{x^2}{2y^2}+\log c$
$\Rightarrow \log c+\frac{x^2}{2y^2}-\log y=0$
$y\left(1\right)=1\Rightarrow \log c+\frac{1}{2}-0=0$
$\log c=-\frac{1}{2}$
$y\left(x\right)=e$
$\Rightarrow -\frac{1}{2}+\frac{x^2}{2e^2}-1=0$
$\Rightarrow \frac{x^2}{e^2}=3$
$\Rightarrow x=\pm \sqrt{3e}$