Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=(y+1)\left((y+1)e^{x^2/2}-x\right), 0<x<2.1$, with $y(2)=0$. Then the value of $\frac{dy}{dx}$ at $x=1$ is equal to:

(A) $\frac{-\mathrm{e}^{3/2}}{\left(\mathrm{e}^{2}+1\right)^{2}}$

(B) $-\frac{2\mathrm{e}^{2}}{\left(1+\mathrm{e}^{2}\right)^{2}}$

(C) $\frac{e^{5/2}}{\left(1+e^{2}\right)^{2}}$

(D) $\frac{5\mathrm{e}^{1/2}}{\left(\mathrm{e}^{2}+1\right)^{2}}$

Differential Equations — JEE Main · 2021 · 18 Mar 2021 · Shift 2 Question with Solution

Let

y = y (x)

be the solution of the differential equation

\frac{d y}{d x} = (y + 1) ((y + 1) e^{x^{2} /2} - x), 0 < x < 2.1

, with

y (2) = 0

. Then the value of

\frac{d y}{d x}

x = 1

is equal to:

Answer: (A)

Putting

u = y + 1

turns the equation into

\frac{d u}{d x} = u^{2} e^{x^{2} /2} - ux

. Dividing by

u^{2}

and setting

v = \frac{1}{u}

, so that

\frac{d v}{d x} = - \frac{1}{u ^{2}} \frac{d u}{d x}

, gives the linear equation

\frac{d v}{d x} - xv = - e^{x^{2} /2}

With integrating factor

e^{- x^{2} /2}

this reads

\frac{d}{d x} (v e^{- x^{2} /2}) = - 1

, so

v e^{- x^{2} /2} = C - x

and

v = e^{x^{2} /2} (C - x)

Since

y (2) = 0

gives

u (2) = 1

and

v (2) = 1

1 = e^{2} (C - 2)

, hence

C = 2 + e^{- 2}

. At

x = 1

v (1) = e^{1/2} (1 + e^{- 2}) = \frac{e ^{1/2} ( e ^{2} + 1 )}{e ^{2}}, u (1) = \frac{1}{v ( 1 )} = \frac{e ^{3/2}}{e ^{2} + 1}

Substituting into

\frac{d y}{d x} = u (u e^{x^{2} /2} - x)

x = 1

, where

u (1) e^{1/2} = \frac{e ^{2}}{e ^{2} + 1}

\frac{d y}{d x} = \frac{e ^{3/2}}{e ^{2} + 1} (\frac{e ^{2}}{e ^{2} + 1} - 1) = - \frac{e ^{3/2}}{( e ^{2} + 1 ) ^{2}}

so the derivative at

x = 1

- \frac{e ^{3/2}}{( e ^{2} + 1 ) ^{2}}

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Single correct option