Answer
$y(x)=x^4+C_2x^{3}+C_1$
Work Step by Step
Suppose that $a= \dfrac{dy}{dx}$ and $ \dfrac{da}{dx}= \dfrac{d^2y}{dx^2}$
Integrating factor; $I.P.=e^{\int-2x^{-1} dx}=e^{-2 \ln x}=x^{-2}$
Now, $\dfrac{d}{dx}(ax^{-2})=4$
Integrate to obtain: $ae^{-2}=4x+C \implies \dfrac{dy}{dx}=4x^3+cx^2$
Therefore, the general solution is: $y(x)=x^4+C_2x^{3}+C_1$