Mathematics - Differential Equation Question with Solution | TestHub

$\left(x+1\right)dy-ydx=e^{3x}{\left(x+1\right)}^{2}dx$

$\frac{\left(x+1\right)dy-ydx}{(x+1{)}^2}=e^{3x}dx$

$d\left(\frac{y}{x+1}\right)=e^{3x}dx$

On integration, we get

$\frac{y}{x+1}=\frac{e^{3x}}{3}+C$

Given $y\left(0\right)=\frac{1}{3}$

So $C=0$

$\Rightarrow y=\frac{\left(x+1\right)e^{3x}}{3}$

$\frac{dy}{dx}=\frac{e^{3x}}{3}\left(3x+4\right)$

$\frac{d^{2}y}{d{x}^2}=e^{3x}\left(3x+5\right)$

Clearly, $x=\frac{-4}{3}$ is a point of local minima as ${\left(\frac{d^{2}y}{d{x}^2}\right)}_{x=\frac{-4}{3}}<0$