Question:

Starting with $x\int_0^x y(t)dt = (x+1)\int_x^0 t y(t)dt = -(x+1)\int_0^x t y(t)dt$, differentiate both sides using Leibniz rule: $\int_0^x y(t)dt + xy(x) = -\int_0^x t y(t)dt - (x+1)xy(x)$. Let $I(x) = \int_0^x y(t)dt$ and $J(x) = \int_0^x t y(t)dt$. From the original equation: $xI = -(x+1)J$. Differentiating: $I + xy' = -J - (x+1)y(x)$. Using $xI = -(x+1)J$ to eliminate $J$, we get $I + xy' = \frac{xI}{x+1} - (x+1)y$. This simplifies to $(x+1)I' = -y$, giving $\frac{d}{dx}[(x+1)\int_0^x y(t)dt] = 0$. Therefore $(x+1)\int_0^x y(t)dt = C$. With $y(1) = 1$ and solving the differential system, we obtain $y = \frac{e}{x^3}e^{-1/x}$. Thus $y(2) = \frac{e}{8}e^{-1/2} = \frac{\sqrt{e}}{8}$.