The solution of the differentiable equation $2x\frac{dy}{dx}+y=14x^3, x>0$, is

$y=2x^3+cx^{-\frac{1}{2}}$

$2x\frac{dy}{dx}+y=14x^3$ dividing by leading 2x

$⇒\frac{dy}{dx}\underbrace{\frac{1}{2x}}_{p}y=\underbrace{7x^2}_{q}$

Intergrating factor = $e^{\int pdx}=e^{\int\frac{1}{2x}dx}$

$=e^{\frac{1}{2}ln\,x}=e^{ln\sqrt{x}}=\sqrt{x}$

multiplying eqn with integrating factor

$\sqrt{x}\frac{dy}{dx}+\frac{1}{2\sqrt{x}}y=7x^{5/2}$

integrating both side w.r.t. x

$\int\sqrt{x}\frac{dy}{dx}+\frac{1}{2\sqrt{x}}ydx=7\int x^{5/2}dx$

$=\sqrt{x}y=7\frac{x^{7/2}}{7}×2+c$

$y=2x^3+cx^{-\frac{1}{2}}$