Let y=y(x) be the solution of the differential equation (x+1) y^′-y= e^3 x(x+1)^2, with y(0)=1/3. Then, the point x=-4/3 for the curve y=y(x) is:

Let $\displaystyle y=y(x)$ be the solution of the differential equation $\displaystyle (x+1) y^{\prime}-y=\mathrm{e}^{3 x}(x+1)^2$, with $\displaystyle y(0)=\frac{1}{3}$. Then, the point $\displaystyle x=-\frac{4}{3}$ for the curve $\displaystyle y=y(x)$ is :