Problem

(a) Show that

$$\frac{\mathrm{d}}{\mathrm{d}x}\big(\tanh^{-1}x\big) = \frac{1}{1 - x^2}.$$

(b) Find the solution of the differential equation

$$x\frac{\mathrm{d}y}{\mathrm{d}x} - y = x^2\tanh^{-1}x,$$

for $0 < x < 1$, given that $y = 0$ when $x = \frac{1}{2}$. Give your answer in an exact form.