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

\(x^{2}=y\) \(x=y^{2}\) \(x=2y\) \(y=2x\)

\(x^{2}=y\)

\(\begin{aligned}\frac{dy}{y}&=\frac{2dx}{x}\\ \text{So, }\log y&=2\log x+\log c\\ y&=x^{2}c\\ \text{Setting }x=1,y&=1\text{ we get }c=1\\ \text{Thus, }y&=x^{2}\end{aligned}\)