Question Number 100902 by bemath last updated on 29/Jun/20
$$\mathrm{solve}\:\mathrm{y}''−\mathrm{4y}'+\mathrm{4y}=\mathrm{0}\: \\ $$$$\mathrm{with}\:\mathrm{variation}\:\mathrm{method} \\ $$
Commented by bramlex last updated on 29/Jun/20
$${AE}\::\:\lambda^{\mathrm{2}} −\mathrm{4}\lambda+\mathrm{4}\:=\mathrm{0} \\ $$$$\left(\lambda−\mathrm{2}\right)^{\mathrm{2}} =\:\mathrm{0}\:\Rightarrow\lambda\:=\:\mathrm{2};\:\mathrm{2} \\ $$$${y}\:=\:{Ae}^{\mathrm{2}{x}} \:+\:{Bxe}^{\mathrm{2}{x}} \:\bigstar\: \\ $$