Question

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

$\frac{dy}{dx} = \frac{xy}{x^{2}+y^{2}}$
Let $y = vx$
$\frac{dy}{dx} = v+x. \frac{dv}{dx}$
$v+x \frac{dv}{dx} = \frac{xvx}{x^{2}+v^{2}x^{2}} = \frac{v}{1+v^{2}}$
$x\frac{dv}{dx}= \frac{v}{1+v^{2}}-v = \frac{v-v-v^{3}}{1+v^{2}} = \frac{v^{3}}{1+v^{2}}$
$\int\frac{v}{1+v^{2}}. dv = \int-\frac{dx}{x}$
$\Rightarrow \int v^{-3}.dv+\int \frac{1}{v}dv = -\int \frac{dx}{x}$
$\Rightarrow \frac{v^{-2}}{-2}+\ell nv = -\ell nx + \lambda$
$\Rightarrow - \frac{1}{2v^{2}}+\ell n\left(\frac{y}{x}\right) = \ell nx + \lambda$
$\Rightarrow - \frac{1}{2} \frac{x^{2}}{y^{2}} +\ell ny - \ell nx = -\ell nx + \lambda $
$\Rightarrow -\frac{1}{2}+0 = \lambda \Rightarrow \lambda = -\frac{1}{2}$
$\Rightarrow - \frac{1}{2} \frac{x^{2}}{y^{2}}+\ell ny + \frac{1}{2} = 0$ at $y = e$
$\Rightarrow - \frac{1}{2} \frac{x^{2}}{y^{2}} +1+\frac{1}{2} = \Rightarrow \frac{x^{2}}{2e^{2}} = \frac{3}{2} \Rightarrow x^{2} = 3e^{2}$
$\therefore x = \sqrt{3}e$