Given that y=1y = 1y=1 at x=0x = 0x=0, solve the differential equation
dydx=12xy12e3x,y≥0 \frac{\text{d}y}{\text{d}x} = \frac{12xy^{\frac{1}{2}}}{\text{e}^{3x}}, \quad y \ge 0 dxdy=e3x12xy21,y≥0giving your answer in the form y12=g(x)y^{\frac{1}{2}} = g(x)y21=g(x).
Hence find the equation of the horizontal asymptote to the curve with equation y12=g(x)y^{\frac{1}{2}} = g(x)y21=g(x).