Let y = y (x) be the solution of the differential equation dy = e ^ + y dx; α ∈ N . If y( _ e 2) = _ e 2 and y(0) = _ e ((1)/(2)), then the value of α is equal to ___.

Correct answer: 0 We have, ∫ e ^ - y dy = ∫ e ^ dx ⇒ - e ^ - y = e ^ α + c . . . (1) Put (x, y) = (ℓn2, ℓn2) - (1)/(2) = 2 ^ α α + c . . . (2) Put (x, y) ≡ (0, - ℓn2) - 2 = (1)/(α) + c From equation (1) - (2), we get 2 ^ α - 1 α = (3)/(2) ⇒ α = 2 ( as α ∈ N ) .

JEE Main 202127 JulyEvening Shift

Exam: JEE Main
Year: 2021
Date: 27 July
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2021 · 27 July · Evening Shift · Mathematics · Differential Equations

Let y = y (x) be the solution of the differential equation dy = e ^...

Let

y = y

\left(x\right)

be the solution of the differential equation

dy = {e}^{\alphax + y}dx; \alpha \in N .

If

y\left(\log_{e}2\right) = \log_{e}2 and

y\left(0\right) = \log_{e}\left(\dfrac{1}{2}\right),

then the value of

\alpha

is equal to ___.

Correct answer

Step-by-step solution

We have, $∫ {e}^{- y}dy = ∫ {e}^{\alphax}dx$

$⇒ - {e}^{- y} = \dfrac{{e}^{\alphax}}{\alpha} + c . . .$ \left(1\right) $Put$ \left(x, y\right) = $\left(ℓn2, ℓn2\right)$

$- \dfrac{1}{2} = \dfrac{{2}^{\alpha}}{\alpha} + c . . . \left(2\right)$

Put \left(x, y\right) ≡ \left(0, - ℓn2\right) $$ - 2 = \dfrac{1}{\alpha} + c $From equation$ \left(1\right) - $\left(2\right),$ we get

$\dfrac{{2}^{\alpha} - 1}{\alpha} = \dfrac{3}{2}$

$⇒ \alpha = 2 ($ as $\alpha \in N )$ .

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