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

The Correct Option is B

\((x+1)\frac {dy}{dx}−y=e^{3x}(x+1)^2\)

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

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.