Question

Let $y = y(x)$ be the solution of the differential equation $(x + 1)y' – y = e^{3x}(x \+ 1)^2$, with $y(0)=\frac 13$. Then, the point $x=−\frac 43$ 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)

a point of local minima

Answer 2

$(x+1)\frac {dy}{dx}−y=e^{3x}(x+1)^2$

$\frac {dy}{dx} −\frac {y}{x+1}=e^{3x}(x+1)$

If $e^{−∫\frac {1}{x+1}x}=e^{−log (x+1)}=\frac {1}{x+1}$

∴ $y(\frac {1}{x+1})=∫\frac {e^{3x}(x+1)}{x+1}dx$

$\frac {y}{x+1} = ∫e^{3x}dx$

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

∵ $y(0)=\frac 13$

$\frac 13=\frac 13+c$
$c = 0$
$y=\frac {e^{3x}}{3}(x+1)$

$y' = e^{3x}(x+1)+\frac {e^{3x}}{3}$

$= e^{3x}(x+\frac 43)$

$y” = 3e^{3x}(x+\frac 43)+e^{3x}$

$=e^{3x}(3x+5)$
$y'=0$ at $x=−\frac 43$ & $y''=e^{−4}(1)>0$ at $x=−\frac 43$
⇒ $x=−\frac 43$ is point of local minima.

So, the correct option is (B): a point of local minima.