Chapter 1 - First-Order Differential Equations - 1.3 The Geometry of First-Order Differential Equations - Problems - Page 32: 11

$y^2=4x$

We are given: $$2xdy-ydx=0$$ $$2xdy=ydx$$ $$\frac{dy}{dx}=\frac{y}{2x}$$ $$2\frac{dy}{y}=\frac{dx}{x}$$ Intergrating both sides with respect to x. $$2\ln (y)=\ln (x) + c$$ $$\ln (y^2)=\ln C(x)$$ Since $y(1)=2$ $$\rightarrow 2^2=C.1$$ $$C=4$$ Therefore, the equation is equal to $y^2=4x$.