Let f(x) = x – 1 and for . If , y(0) = 0, then y(1) is [2025]

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Solution

Q.
Let f(x) = x – 1 and $g (x) = e^{x}$ for $x \in R$ . If $\frac{d y}{d x} = (e^{- 2 \sqrt{x}} g (f (f (x))) - \frac{y}{\sqrt{x}})$ , y(0) = 0, then y(1) is [2025]

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

2 $\frac{2 e - 1}{e^{3}}$

3 $\frac{1 - e^{3}}{e^{4}}$

4 $\frac{1 - e^{2}}{e^{4}}$

Ans.
(1)

f(x) = x – 1 and $g (x) = e^{x}$

f(f(x)) = f(x – 1) = (x – 1) – 1 = x – 2

$g (f (f (x))) = g (x - 2) = e^{x - 2}$

Now, $\frac{d y}{d x} + \frac{y}{\sqrt{x}} = e^{- 2 \sqrt{x}} e^{x - 2} \Rightarrow \frac{d y}{d x} + \frac{y}{\sqrt{x}} = e^{x - 2 - 2 \sqrt{x}}$

Which is a linear differential equation

$∴ I . F . = e^{\int \frac{1}{\sqrt{x}} d x} = e^{2 \sqrt{x}}$

$∴$ Solution of differential equation is given by

$y e^{2 \sqrt{x}} = \int e^{x - 2 - 2 \sqrt{x}} e^{2 \sqrt{x}} d x$

$\Rightarrow y e^{2 \sqrt{x}} = \int e^{x - 2} d x \Rightarrow y e^{2 \sqrt{x}} = e^{x - 2} + C$

Now, y(0) = 0

$\Rightarrow 0 = e^{- 2} + C \Rightarrow C = - e^{- 2}$

$∴ y (x) = e^{x - 2 - 2 \sqrt{x}} - e^{- 2 - 2 \sqrt{x}}$

Now, $y (1) = e^{1 - 2 - 2} - e^{- 2 - 2} = e^{- 3} - e^{- 4} = \frac{e - 1}{e^{4}}$

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