JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Differential Equations

\frac{d y}{d x} = \frac{2 ^{x} y + 2 ^{y} \cdot 2 ^{x}}{2 ^{x} + 2 ^{x + y} l o g _{e} 2}, y (0) = 0

, then for

y = 1

, the value of

x

lies in the interval:

Answer: (A)

Solution

The right side factors as

\frac{2 ^{x} ( y + 2 ^{y} )}{2 ^{x} ( 1 + 2 ^{y} ln 2 )} = \frac{y + 2 ^{y}}{1 + 2 ^{y} ln 2}

, so

\frac{( 1 + 2 ^{y} ln 2 ) d y}{y + 2 ^{y}} = d x

. With

u = y + 2^{y}

d u = (1 + 2^{y} ln 2) d y

, hence

\frac{d u}{u} = d x

and

ln (y + 2^{y}) = x + C

. The condition

y (0) = 0

gives

C = 0

, so

ln (y + 2^{y}) = x

. At

y = 1

x = ln 3 \approx 1.10

, which lies in

(1, 2)

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Single correct option

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