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

  1. (1) $\frac{1}{2} \sqrt{3} e$

  2. (2) $\frac{e}{\sqrt{2}}$

  3. (3) $\sqrt{2} e$

  4. (4) $\sqrt{3} e$

Correct Option: , 4


The given differential equation,

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

Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

Then, $v+x \frac{d v}{d x}=\frac{v x^{2}}{x^{2}+v^{2} x^{2}}=\frac{v}{1+v^{2}}$

$\Rightarrow \quad \frac{1+v^{2}}{v^{3}} d v=-\frac{1}{x} d x$

$\Rightarrow \quad \int\left(\frac{1}{v^{3}}+\frac{1}{v}\right) d v=\int \frac{-1}{x} d x$

$\Rightarrow \frac{-1}{2}\left(\frac{1}{v^{2}}\right)+\ln v=-\ln x+c$

$\Rightarrow-\frac{x^{2}}{2 y^{2}}=-\ln y+c$ $\left[\because v=\frac{y}{x}\right]$

When $x=1, y=1$, then $-\frac{1}{2}=c$

$\Rightarrow x^{2}=y^{2}(1+2 \ln y)$

At $y=e, x^{2}=e^{2}(3)$

$\Rightarrow \quad x=\pm \sqrt{3} e$

So, $x=\sqrt{3} e$

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