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

Step 1: Simplify the composite function

First, we need to simplify the composite function $f(f(f(x)))$. We apply $f(x)$ three times. $f(x) = x - 1$ $f(f(x)) = f(x-1) = (x-1) - 1 = x - 2$ $f(f(f(x))) = f(x-2) = (x-2) - 1 = x - 3$.

Step 2: Substitute into g(x) and the differential equation

Now we substitute $f(f(f(x))) = x - 3$ into $g(x) = e^x$, which gives us $g(f(f(f(x)))) = e^{x-3}$. Then, we substitute this back into the given differential equation.

Step 3: Rearrange into a linear differential equation

We rearrange the differential equation into the standard form of a linear first-order differential equation, which is $\frac{dy}{dx} + P(x)y = Q(x)$. Here, $P(x) = \frac{1}{\sqrt{x}}$ and $Q(x) = e^{x-3-2\sqrt{x}}$.

Step 4: Find the integrating factor

The integrating factor (I.F.) for a linear differential equation is given by $e^{\int P(x) dx}$. In this case, $\int \frac{1}{\sqrt{x}} dx = \int x^{-1/2} dx = 2x^{1/2} = 2\sqrt{x}$. So, the integrating factor is $e^{2\sqrt{x}}$.

Step 5: Solve the differential equation

Multiply both sides of the differential equation by the integrating factor. The left side becomes $\frac{d}{dx}(y \cdot I.F.)$. The right side simplifies to $\int e^{x-3} dx = e^{x-3} + C$.

Step 6: Apply the initial condition

We use the initial condition $y(0) = 0$ to find the value of the constant $C$. Substituting $x=0$ and $y=0$ into the solution gives $0 = e^{-3} + C$, so $C = -e^{-3}$.

Step 7: Find y(1)

Substitute the value of $C$ back into the general solution to get the particular solution. Then, substitute $x=1$ to find $y(1)$.

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