Question

Show that 𝑦 = 2𝑥 ln 𝑥 is a solution to the differential equation $x^2y''-xy'+y=0$

Answer 1

$$x^2y''-xy'+y=0$$

$$y=2x\ln x$$

$$y'=2\ln x+2$$

$$y''=2/x$$

$$x^2y''-xy'+y=\\ =x^2(2/x)-2x\ln x-2x+2x\ln x=0$$