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 $x=3$.