Let f(x)=x-1 and g(x)=e^x for x ∈ ℝ . If (d y)/(d x)=(e^ -2 x g(f(f(x)))- y x ), y(0)=0 , then y(1) is :-

Correct answer: C. (e-1)/(e^4) aligned & f(x)=x-1 & f(f(x))=f(x)-1=x-1-1=x-2 & g(f(f(x)))=e^ x-2 & (d y)/(d x)=e^ -2 x × e^ x-2 - 1 x y & (d y)/(d x)+ 1 x y=e^ x-2 x -2 which is L.D.E & I.F. =e^ ∫ d y x =e^ 2 x aligned Its solution is aligned & y × e^ 2 x =∫ e^ 2 x × e^ x-2 x -2 d x+c & y × e^ 2 x =∫ e^ x-2 d x+c & y × e^ 2 x =e^ x-2 +c aligned Given x=0, y=0 0=e^ -2 +c ; c=-e^ -2 y × e^ 2 x =e^ x-2 -e^ -2 when x=1, y × e^2=e^ -1 -e^ -2 y= e^ -1 -e^ -2 e^2 =((1)/(e)-(1)/(e^2))/(e^2)=(e^2-e)/(e^5)=(e-1)/(e^4) Option (1) is correct

JEE Main 20258 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Differential Equations

Let f(x)=x-1 and g(x)=e^x for x ∈ ℝ . If (d y)/(d x)=(e^ -2 x...

Let

f(x)=x-1

and

g(x)=e^x

for

x \in \mathbb{R}

. If

\frac{d y}{d x}=\left(e^{-2 \sqrt{x}} g(f(f(x)))-\frac{y}{\sqrt{x}}\right), y(0)=0

, then

y(1)

is :-

Options

A $\frac{1-e^2}{e^4}$
B $\frac{2 e-1}{e^3}$
C $\frac{e-1}{e^4}$
D $\frac{1-e^3}{e^4}$

Correct answer

C. $\frac{e-1}{e^4}$

Step-by-step solution

\begin{aligned}<br/>& f(x)=x-1 \\ & f(f(x))=f(x)-1=x-1-1=x-2 \\ & g(f(f(x)))=e^{x-2} \\ & \therefore \frac{d y}{d x}=e^{-2 \sqrt{x}} \times e^{x-2}-\frac{1}{\sqrt{x}} y \\ & \frac{d y}{d x}+\frac{1}{\sqrt{x}} y=e^{x-2 \sqrt{x}-2} \text{ which is L.D.E } \\ & \text{ I.F. }=e^{\int \frac{d y}{\sqrt{x}}}=e^{2 \sqrt{x}}<br/>\end{aligned}

Its solution is

\begin{aligned}<br/>& y \times e^{2 \sqrt{x}}=\int e^{2 \sqrt{x}} \times e^{x-2 \sqrt{x}-2} d x+c \\ & y \times e^{2 \sqrt{x}}=\int e^{x-2} d x+c \\ & y \times e^{2 \sqrt{x}}=e^{x-2}+c<br/>\end{aligned}

\text{ Given } \mathrm{x}=0, \mathrm{y}=0 \Rightarrow 0=\mathrm{e}^{-2}+\mathrm{c} \quad ; \mathrm{c}=-\mathrm{e}^{-2}

\therefore \mathrm{y} \times \mathrm{e}^{2 \sqrt{\mathrm{x}}}=\mathrm{e}^{\mathrm{x}-2}-\mathrm{e}^{-2}

\text{ when } \mathrm{x}=1, \mathrm{y} \times \mathrm{e}^2=\mathrm{e}^{-1}-\mathrm{e}^{-2}

y=\frac{e^{-1}-e^{-2}}{e^2}=\frac{\frac{1}{e}-\frac{1}{e^2}}{e^2}=\frac{e^2-e}{e^5}=\frac{e-1}{e^4}

Option (1) is correct

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