Let y = y (x) be the solution of the differential equation (x + 1) y ^ ' - y = e ^ 3x (x + 1) ^ 2 , with y(0)=(1)/(3) . Then, the point x = - (4)/(3) for the curvey = y(x) is

Correct answer: B. a point of local minima (x + 1)dy - ydx = e ^ 3x (x + 1) ^ 2 dx (x + 1)dy - ydx ( x + 1 ) ^ 2 = e ^ 3x dx d((y)/(x + 1)) = e ^ 3x dx On integration, we get (y)/(x + 1) = e ^ 3x 3 + C Given y(0) = (1)/(3) So C = 0 ⇒ y = (x + 1) e ^ 3x 3 (dy)/(dx) = e ^ 3x 3 (3x + 4) d ^ 2 y d x ^ 2 = e ^ 3x (3x + 5) Clearly, x = (- 4)/(3) is a point of local minima as ( d ^ 2 y d x ^ 2 ) _ x = (- 4)/(3)

JEE Main 202225 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 25 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2022 · 25 June · Morning Shift · Mathematics · Differential Equations

Let y = y (x) be the solution of the differential equation (x + 1)...

Let

y = y \left(x\right) be

the

solution of the differential equation \left(x + 1\right){y}^{'} - y = {e}^{3x}{\left(x + 1\right)}

^{2}

, with

y\left(0\right)=\dfrac{1}{3}

. Then, the point

x = - \dfrac{4}{3} for the curvey = y\left(x\right) is

Options

Anot a critical point
Ba point of local minima
Ca point of local maxima
Da point of inflection

Correct answer

B. a point of local minima

Step-by-step solution

\left(x + 1\right)dy - ydx = {e}^{3x}{\left(x + 1\right)}^{2}dx $$ \dfrac{\left(x + 1\right)dy - ydx}{( x + 1{)}^{2}} = {e}^{3x}dx $$ d\left(\dfrac{y}{x + 1}\right) = {e}^{3x}dx $On integration, we get$ \dfrac{y}{x + 1} = \dfrac{{e}^{3x}}{3} + C $Given$ y\left(0\right) = \dfrac{1}{3} $So$ C = 0 $$ ⇒ y = \dfrac{\left(x + 1\right){e}^{3x}}{3} $$ \dfrac{dy}{dx} = \dfrac{{e}^{3x}}{3}\left(3x + 4\right) $$ \dfrac{{d}^{2}y}{d{x}^{2}} = {e}^{3x}\left(3x + 5\right)

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

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