Question

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

• A. not a critical point
• B. a point of local minima
• C. a point of local maxima
• D. a point of inflection

Answer 1

Answer: (B)

$(x+1)dy-ydx=e^{3x}(x+1{)}^2$
$\frac{(x+1)dy-ydx}{(x+1{)}^2}=e^{3x}$
$d\left(\frac{y}{x+1}\right)=e^{3x}\Rightarrow \frac{y}{x+1}=\frac{e^{3x}}{3}+C$
$\left(0,\frac{1}{3}\right)\Rightarrow C=0\Rightarrow y=\frac{(x+1)e^{3x}}{3}$
$\frac{dy}{dx}=\frac{1}{3}\left((x+1)3e^{3x}+e^{3x}\right)=\frac{3^{3x}}{3}(3x+4)$

Clearly, $x=\frac{-4}{3}$ is point of local minima