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Q.
Let $y = y ( x )$ be the solution of the differential equation $(x+1) y^{\prime}-y=e^{3 x}(x+1)^{2}$, with $y(0)=\frac{1}{3}$. Then, the point $x=-\frac{4}{3}$ for the curve $y = y ( x )$ is:
$(x+1) d y-y d x=e^{3 x}(x+1)^{2}$
$\frac{(x+1) d y-y d x}{(x+1)^{2}}=e^{3 x}$
$d\left(\frac{y}{x+1}\right)=e^{3 x} \Rightarrow \frac{y}{x+1}=\frac{e^{3 x}}{3}+C$
$\left(0, \frac{1}{3}\right) \Rightarrow C=0 \Rightarrow y=\frac{(x+1) e^{3 x}}{3}$
$\frac{d y}{d x}=\frac{1}{3}\left((x+1) 3 e^{3 x}+e^{3 x}\right)=\frac{3^{3 x}}{3}(3 x+4)$
Clearly, $x=\frac{-4}{3}$ is point of local minima
