Euler against the integrating factor

You will test Euler's method on a linear equation whose exact answer comes from the integrating factor. The equation is $\dfrac{dy}{dx} + \dfrac{y}{x} = 3x^2$ for $x \ge 1$, with $y = 1$ at $x = 1$.

The idea. The integrating factor is $x$, so $\dfrac{d}{dx}(xy) = 3x^3$ and $xy = \dfrac{3x^4}{4} + C$. The point $(1, 1)$ gives $C = \dfrac{1}{4}$. The exact solution is $y = \dfrac{3x^4 + 1}{4x}$. Euler's method takes small steps along the slope, and its error should shrink when the steps get smaller.

Your tasks.

1. Finish f(x, y). It returns the slope $\dfrac{dy}{dx}$ written as a function of $x$ and $y$.
2. Finish euler(x0, y0, x_end, n). Take $n$ equal steps of size $h = \dfrac{x_{\text{end}} - x_0}{n}$. In each step, add $h \times f(x, y)$ to y, and add h to x.
3. Finish error(n). Return the size of the gap between Euler's answer and the exact answer at $x = 2$, with $n$ steps from $x = 1$.