Problem

(a) Show that an appropriate integrating factor for

$$\sqrt{x^2-1}\frac{\mathrm{d}y}{\mathrm{d}x}+y=x^2-x\sqrt{x^2-1}$$

is $x+\sqrt{x^2-1}$.

(b) Hence find the solution of the differential equation

$$\sqrt{x^2-1}\frac{\mathrm{d}y}{\mathrm{d}x}+y=x^2-x\sqrt{x^2-1}$$

for which $y=1$ when $x=\frac{5}{4}$. Give your answer in the form $y=f(x)$.