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Suppose we want to solve the differential equation \[\int_{0}^{x} f(y) \, dy = f(x) -1 \] We can rearrange and simplify this to \(1=f \big(1-\int \big)\), and dividing yields \(f=\frac{1}{1-\int}\). We can now take the geometric series expansion of the right hand side, from which we get \[f=1+\int+\int^2+\int^3+\ldots\] Recall \( \int=\int_{0}^{x} dx_0=x \) and exponentiation gives us \[ \Big(\int \Big)^n=\int_{0}^{x}\int_0^{x_{n-1}}\ldots\int_{0}^{x_1} 1 \, dx_0 \ldots dx_{n-1} = \frac{x^n}{n!} \] Therefore, we have our solution \[ f(x) = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+ \frac{x^4}{24} +\dots = e^x\]