Let $y=y(x)$ be the solution of the differential equation $\frac{d y}{d x}=1+x e^{y-x},-\sqrt{2}<x<\sqrt{2}, y(0)=0$ then, the minimum value of $y(x), x \in(-\sqrt{2}, \sqrt{2})$ is equal to :

(C) $(1-\sqrt{3})-\log _{e}(\sqrt{3}-1)$

(A) $(1+\sqrt{3})-\log _{e}(\sqrt{3}-1)$

(B) $(2+\sqrt{3})+\log _{e} 2$

(D) $(2-\sqrt{3})-\log _{e} 2$

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

\frac{d y}{d x} = 1 + x e^{y - x}, - 2 < x < 2, y (0) = 0

then, the minimum value of

y (x), x \in (- 2, 2)

is equal to :

Answer: (C)

Putting

v = y - x

turns the equation into

\frac{d v}{d x} = x e^{v}

, which separates:

e^{- v} d v = x d x \Rightarrow - e^{- v} = \frac{x ^{2}}{2} + C

With

y (0) = 0

one has

v (0) = 0

and

C = - 1

, so

e^{- v} = 1 - \frac{x ^{2}}{2}

and

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

Setting the derivative to zero,

\frac{d y}{d x} = 1 + \frac{x}{1 - \frac{x ^{2}}{2}} = 0 \Rightarrow x^{2} - 2 x - 2 = 0 \Rightarrow x = 1 \pm 3

and only

x = 1 - 3

lies in

(- 2, 2)

. On that interval

1 - \frac{x ^{2}}{2} > 0

, so

\frac{d y}{d x} \to - \infty

x \to - 2^{+}

while

\frac{d y}{d x} = 1

x = 0

; the stationary point is therefore a minimum. Since

1 - \frac{( 1 - 3 ) ^{2}}{2} = 3 - 1

, the least value of

y

(1 - 3) - lo g_{e} (3 - 1)

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option