Answer
$y= \sqrt{x^2+4}$
Work Step by Step
Let $y=f(x)$ be the equation of the curve passing through the point $(0,2)$ with the slope $\frac{x}{y}$ at $(x,y)$.
We have a differential equation: $\frac{dy}{dx}=\frac{x}{y}$, $y(0)=2$
Solve for $y$:
$\frac{dy}{dx}=\frac{x}{y}$ (Separate the variables)
$y dy =x dx$ (Integrate)
$\int y dy=\int x dx$
$\frac{y^2}{2}=\frac{x^2}{2}+\frac{C}{2}$
$y^2=x^2+C$ (Substitute $y(0)=2$)
$2^2=0^2+C$
$4=0+C$
$C=4$
$\therefore y^2=x^2+4$ or $y=\pm \sqrt{x^2+4}$
Only $y=\sqrt{x^2+4}$ fits as $y(0)=2>0$.
