Let y = y (x) be the solution of the differential equation, (2 + x)/(y + 1) . (dy)/(dx) =- x , y > 0, y(0) = 1 . If y(π) = a and (dy)/(dx) at x = π is b, then the ordered pair (a, b) is equal to

Correct answer: C. (1, 1) (dy)/(1 + y) = (- x)/(2 + x)dx array ll (1 + y) = - (2 + x) + lnc array (1 + y) (2 + x) = c We know y(0) = 1 So array ll c = 4 array array ll 1 + y = (4)/(2 + x) ⇒ y = (4)/(2 + x) -1 array y(π) = 2 - 1 = 1 = a (dy)/(dx) = - 4 (2 + x) ^ 2 · x = 1 At x = π ⇒ b = 1 (a, b) = (1, 1)

JEE Main 20202 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 2 September
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2020 · 2 September · Morning Shift · Mathematics · Differential Equations

Let y = y (x) be the solution of the differential equation, (2 +...

Let

y = y \left(x\right) be

the

solution of the differential equation, \dfrac{2 + \sin x}{y + 1}

\dfrac{dy}{dx} =- \cos x

y > 0, y\left(0\right) = 1

. If

y\left(\pi\right)

= a and

\dfrac{dy}{dx} at

x = \pi

is

b, then the

ordered pair \left(a, b\right) is equal to

Options

A $\left(2, \dfrac{3}{2}\right)$
B $\left(1, - 1\right)$
C $\left(1, 1\right)$
D $\left(2, 1\right)$

Correct answer

C. $\left(1, 1\right)$

Step-by-step solution

\dfrac{dy}{1 + y} = \dfrac{- \cos x}{2 + \sin x}dx $$ $\begin{array}{ll}\ln \left(1 + y\right) = - \ln \left(2 + \sin x\right) + lnc\end{array}$ $\left(1 + y\right)$ \left(2 + \sin x\right) = c $We know$ y\left(0\right) = 1 $So$ $\begin{array}{ll}c = 4\end{array}$ $$ $\begin{array}{ll}1 + y = \dfrac{4}{2 + \sin x} ⇒ y = \dfrac{4}{2 + \sin x} -1\end{array}$ $$ y\left(\pi\right) $= 2 - 1 = 1 = a$

\dfrac{dy}{dx} =\dfrac{- 4}{{\left(2 + \sin x\right)}^{2}} \cdot \cos x = 1 $At$ x = \pi ⇒ b = 1 $\left(a, b\right)$ = \left(1, 1\right)$

Practice Differential Equations on Quantrex Academy →

More from Differential Equations