Let f(x)=x-1 and g(x)=e^x for x in R. If frac | vedclass

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(C) Given $f(x)=x-1$ and $g(x)=e^x$. First,find $g(f(f(x)))$: $f(f(x)) = f(x-1) = (x-1)-1 = x-2$. $g(f(f(x))) = g(x-2) = e^{x-2}$. The differential eq

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

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(C) Given $f(x)=x-1$ and $g(x)=e^x$. First,find $g(f(f(x)))$: $f(f(x)) = f(x-1) = (x-1)-1 = x-2$. $g(f(f(x))) = g(x-2) = e^{x-2}$. The differential equation is $\frac{d y}{d x} + \frac{y}{\sqrt{x}} = e^{-2 \sqrt{x}} \cdot e^{x-2} = e^{x-2 \sqrt{x}-2}$. This is a linear differential equation of the form $\frac{d y}{d x} + P(x)y = Q(x)$,where $P(x) = \frac{1}{\sqrt{x}}$ and $Q(x) = e^{x-2 \sqrt{x}-2}$. The integrating factor ($I$.$F$.) is $e^{\int P(x) dx} = e^{\int \frac{1}{\sqrt{x}} dx} = e^{2 \sqrt{x}}$. The solution is $y \cdot e^{2 \sqrt{x}} = \int Q(x) \cdot (I.F.) dx + C$. $y \cdot e^{2 \sqrt{x}} = \int e^{x-2 \sqrt{x}-2} \cdot e^{2 \sqrt{x}} dx + C = \int e^{x-2} dx + C = e^{x-2} + C$. Using $y(0)=0$: $0 \cdot e^0 = e^{0-2} + C \implies 0 = e^{-2} + C \implies C = -e^{-2}$. So,$y \cdot e^{2 \sqrt{x}} = e^{x-2} - e^{-2}$. At $x=1$: $y \cdot e^2 = e^{1-2} - e^{-2} = e^{-1} - e^{-2} = \frac{1}{e} - \frac{1}{e^2} = \frac{e-1}{e^2}$. Therefore,$y(1) = \frac{e-1}{e^4}$.

Let $f(x)=x-1$ and $g(x)=e^x$ for $x \in R$. If $\frac{d y}{d x}=\left(e^{-2 \sqrt{x}} g(f(f(x)))-\frac{y}{\sqrt{x}}\right)$ and $y(0)=0$,then $y(1)$ is:

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