Question

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