Equation of curve passing through (1, 1) and satisfying the differential equation $\frac{dy}{dx}=\frac{2y}{x}$, (where x > 0, y > 0) is given by:

x2 = y x = y2 x = 2y y = 2x

x2 = y

$\int\frac{dy}{2y}=\int\frac{dx}{x}⇒\frac{1}{2}ln\,y=ln\,x+ln\,c⇒y=cx^2$ is passing through (1, 1). Hence, the curve is : y = x2