Problem

(a) Show that an appropriate integrating factor for

$$\sqrt{x^2 + 16}\frac{\mathrm{d}y}{\mathrm{d}x} + y = x\sqrt{x^2 + 16}$$

is

$$\frac{1}{4}x + \frac{1}{4}\sqrt{x^2 + 16}.$$

(b) Hence find the solution of the differential equation

$$\sqrt{x^2 + 16}\frac{\mathrm{d}y}{\mathrm{d}x} + y = x\sqrt{x^2 + 16}$$

for which $y = 6$ when $x = 3$.