The solution curve of the differential equation, (1 + e ^ - x )(1 + y ^ 2 )(dy)/(dx) = y ^ 2 which passes through the point (0, 1) , is

Correct answer: C. y ^ 2 = 1 + y _ e ( 1 + e ^ x 2 ) (1 + e ^ - x )(1 + y ^ 2 )(dy)/(dx) = y ^ 2 Separate variables ⇒ (1 + y ^ 2 )dy y ^ 2 = dx 1 + e ^ - x Integrate both side ⇒ ∫ (1 + y ^ 2 )dy y ^ 2 = ∫ dx 1 + e ^ - x ⇒ ∫ (1 + 1 y ^ 2 )dy = ∫ e ^ x dx e ^ x + 1 ⇒ y - (1)/(y) = ( e ^ x + 1) + C Since curve pass through (0, 1) . ⇒ 1 - (1)/(1) = ( e ^ 0 + 1) + C ⇒ C = - (2) ⇒ y - (1)/(y) = ( e ^ x + 1) - (2) ⇒ y ^ 2 - 1 y = ( e ^ x + 1 2 ) ⇒ y ^ 2 = 1 + y _ e ( 1 + e ^ x 2 )

JEE Main 20203 SeptemberMorning Shift

Exam: JEE Main
Year: 2020
Date: 3 September
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2020 · 3 September · Morning Shift · Mathematics · Differential Equations

The solution curve of the differential equation, (1 + e ^ - x )(1 +...

The solution curve of the differential equation,

\left(1 + {e}^{- x}\right)\left(1 + {y}^{2}\right)\dfrac{dy}{dx} = {y}^{2}

which passes through the point \left(0, 1\right)$, is

Options

A ${y}^{2} + 1 = y\left(\log_{e}\left(\dfrac{1 + {e}^{- x}}{2}\right) + 2\right)$
B ${y}^{2} + 1 = y\left(\log_{e}\left(\dfrac{1 + {e}^{x}}{2}\right) + 2\right)$
C ${y}^{2} = 1 + y{\log}_{e}\left(\dfrac{1 + {e}^{x}}{2}\right)$
D ${y}^{2} = 1 + y{\log}_{e}\left(\dfrac{1 + {e}^{- x}}{2}\right)$

Correct answer

C. ${y}^{2} = 1 + y{\log}_{e}\left(\dfrac{1 + {e}^{x}}{2}\right)$

Step-by-step solution

\left(1 + {e}^{- x}\right)\left(1 + {y}^{2}\right)\dfrac{dy}{dx} = {y}^{2} $Separate variables$ ⇒ \dfrac{\left(1 + {y}^{2}\right)dy}{{y}^{2}} = \dfrac{dx}{1 + {e}^{- x}} $Integrate both side$ ⇒ ∫ \dfrac{\left(1 + {y}^{2}\right)dy}{{y}^{2}} = ∫ \dfrac{dx}{1 + {e}^{- x}} $$ ⇒ ∫ \left(1 + \dfrac{1}{{y}^{2}}\right)dy = ∫ \dfrac{{e}^{x}dx}{{e}^{x} + 1} $$ ⇒ y - \dfrac{1}{y} = \ln \left({e}^{x} + 1\right) + C $Since curve pass through \left(0, 1\right)$ .

$⇒ 1 - \dfrac{1}{1} = \ln \left({e}^{0} + 1\right) + C ⇒ C = - \ln \left(2\right)$ ⇒ y - $\dfrac{1}{y} =$ \ln \left({e}^{x} + 1\right) $- \ln \left(2\right)$

$⇒ \dfrac{{y}^{2} - 1}{y} = \ln \left(\dfrac{{e}^{x} + 1}{2}\right)$ ⇒ {y}^{2} = 1 + y{ $\log}_{e}\left(\dfrac{1 + {e}^{x}}{2}\right)$

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