Let $y(x)$ be a solution of the differential equation $\left(1+e^{x}\right) y^{\prime}+y e^{x}=1$. If $y(0)=2$, then which of the following statements is (are) true?

(A) $y(-4)=0$; (C) $y(x)$ has a critical point in the interval $(-1,0)$

(D) $y(x)$ has no critical point in the interval $(-1,0)$

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Differential Equations

Let

y (x)

be a solution of the differential equation

(1 + e^{x}) y^{'} + y e^{x} = 1

. If

y (0) = 2

, then which of the following statements is (are) true?

Answer: (A), (C)

Rewrite the differential equation:

(1 + e^{x}) y^{'} + y e^{x} = 1

y^{'} + \frac{e ^{x}}{1 + e ^{x}} y = \frac{1}{1 + e ^{x}}

This is a first-order linear ODE. The integrating factor is:

μ (x) = e^{\int \frac{e ^{x}}{1 + e ^{x}} d x} = e^{l n (1 + e^{x})} = 1 + e^{x}

Multiplying through:

(1 + e^{x}) y^{'} + y e^{x} = 1

\frac{d}{d x} [(1 + e^{x}) y] = 1

(1 + e^{x}) y = x + C

y = \frac{x + C}{1 + e ^{x}}

Using

y (0) = 2

2 = \frac{0 + C}{1 + 1} = \frac{C}{2}

C = 4

y (x) = \frac{x + 4}{1 + e ^{x}}

.

(A)

y (- 4) = \frac{- 4 + 4}{1 + e ^{- 4}} = 0

✓

(B)

y (- 2) = \frac{- 2 + 4}{1 + e ^{- 2}} = \frac{2}{1 + e ^{- 2}} \neq = 0

✗

(C) Critical points occur where

y^{'} (x) = 0

y^{'} = \frac{1 - y e ^{x}}{1 + e ^{x}} = 0

1 - y e^{x} = 0

y = e^{- x}

\frac{x + 4}{1 + e ^{x}} = e^{- x}

x + 4 = e^{- x} (1 + e^{x}) = e^{- x} + 1

x + 3 = e^{- x}

For

x \in (- 1, 0)

: LHS ranges from 2 to 3, RHS ranges from 1 to

e \approx 2.718

. They intersect, so there is a critical point. ✓

(D) Contradicts (C). ✗

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · One or more correct options