JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Differential Equations

If the solution curve

f (x, y) = 0

of the differential equation

(1 + lo g_{e} x) \frac{d x}{d y} - x lo g_{e} x = e^{y}

x > 0

, passes through the points

(1, 0)

and

(α, 2)

then

α^{α}

is equal to

Answer: (D)

Solution

Rewrite the differential equation as

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

. Let

t = x ln x

, so

\frac{d t}{d x} = 1 + ln x

, hence

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

. The equation becomes

\frac{d t}{d y} - t = e^{y}

, a first-order linear ODE. The integrating factor is

e^{- y}

. Multiplying through:

\frac{d}{d y} (t e^{- y}) = 1

, so

t e^{- y} = y + C

. Thus

t = y e^{y} + C e^{y}

, i.e.,

x ln x = y e^{y} + C e^{y}

. Using the initial condition

(1, 0)

1 \cdot ln 1 = 0 + C

, so

C = 0

. The solution is

x ln x = y e^{y}

. At

(α, 2)

α ln α = 2 e^{2}

. Then

α^{α} = e^{α l n α} = e^{2 e^{2}}

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Single correct option