Question

If $\frac{d y}{d x}=\frac{x y}{x^2+y^2}, y(1)=1$, then one of the values of $x_0$ satisfying $y\left(x_0\right)=e$ is given by

• A. $e \sqrt{2}$
• B. $e \sqrt{3}$
• C. $e \sqrt{5}$
• D. $\frac{e}{\sqrt{2}}$

Answer 1

Answer: (B)

The given differential equation is
$\frac{d y}{d x} =\frac{x y}{x^2+y^2}.....$(i)
$ =\frac{y / x}{1+\left(\frac{y}{x}\right)^2}=g\left(\frac{y}{x}\right) \left[\because \frac{d y}{d x}=g\left(\frac{y}{x}\right)\right]$
$\therefore$ Eq. (i) is the homogeneous differential equation
So, put $y =v x$
$\Rightarrow \frac{d y}{d x} =v+x \frac{d v}{d x}$
$ \therefore v+x \frac{d v}{d x}=\frac{x^2 v}{x^2\left(1+v^2\right)} $
$ \Rightarrow \int \frac{1+v^2}{v^3} d v=-\int \frac{d x}{x} $
$ \Rightarrow -\frac{1}{2 v^2}+\log v=-\log x+\log C $
$ \Rightarrow -\frac{1}{2} \cdot \frac{x^2}{y^2}+\log |y|=\log C $
$\because y(1)=1 \Rightarrow-\frac{1}{2}=\log C$
$\therefore -\frac{1}{2} \cdot \frac{x^2}{y^2}+\log |y|=-\frac{1}{2} $
$\Rightarrow \log _e|y|+\frac{1}{2}=\frac{x^2}{2 y^2}$
Again, when $x=x_0, y=e$
$1+\frac{1}{2} =\frac{x_0^2}{2 e^2}$
$x_0 =\sqrt{3} e$