If y = y(x) is the solution of the differential equation 5 + e ^ x 2 + y ⋅ (dy)/(dx) + e ^ x = 0 satisfying y(0) = 1 then value of y ( _ e 13 ) is

Correct answer: B. - 1 Given (dy)/(2 + y) = - e ^ x dx 5 + e ^ x ⇒ ∫ (dy)/(2 + y) = - ∫ e ^ x dx 5 + e ^ x ⇒ _ e ( 2 + y ) = - _ e (5 + e ^ x ) + _ e C ⇒ _ e ( 2 + y ) = _ e ( C 5 + e ^ x ) ⇒ y = C 5 + e ^ x - 2 ∵ y ( 0 ) = 1 ∴ c = 18 y = 18 5 + e ^ x - 2 y ( _ e 13 ) = 18 5 + e ^ _ e (13) - 2 ⇒ y ( _ e 13 ) = (18)/(5 + 13) - 2 ∴ y( _ e 13) = - 1

JEE Main 20205 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 5 September
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2020 · 5 September · Morning Shift · Mathematics · Differential Equations

If y = y(x) is the solution of the differential equation 5 + e ^ x...

y = y\left(x\right) is the solution of the differential equation

\dfrac{5 + {e}^{x}}{2 + y} ⋅ \dfrac{dy}{dx} + {e}^{x} = 0

satisfying

y\left(0\right) = 1 then value of

y ( \log_{e}13 )

Options

A $1$
B $- 1$
C $0$
D $2$

Correct answer

B. $- 1$

Step-by-step solution

Given $\dfrac{dy}{2 + y} = \dfrac{- {e}^{x}dx}{5 + {e}^{x}}$

$⇒ ∫ \dfrac{dy}{2 + y} = - ∫ \dfrac{{e}^{x}dx}{5 + {e}^{x}}$

$⇒ \log_{e} ( 2 + y ) = - \log_{e}\left(5 + {e}^{x}\right) + \log_{e}C$

$⇒ \log_{e} ( 2 + y ) = \log_{e}\left(\dfrac{C}{5 + {e}^{x}}\right)$

$⇒ y = \dfrac{C}{5 + {e}^{x}} - 2$

$∵ y ( 0 ) = 1$

$∴ c = 18$

$y = \dfrac{18}{5 + {e}^{x}} - 2$

$y ( \log_{e}13 ) = \dfrac{18}{5 + {e}^{\log_{e}\left(13\right)}} - 2$

$⇒ y ( \log_{e}13 ) = \dfrac{18}{5 + 13} - 2$

$∴ y\left(\log_{e}13\right) = - 1$

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