Question Number 93513 by Shakhzod last updated on 13/May/20
$${y}^{''} +{y}^{'} −\mathrm{2}{y}=\mathrm{0} \\ $$
Answered by i jagooll last updated on 13/May/20
$$\mathrm{homogenous}\:\mathrm{solution}\: \\ $$$$\lambda^{\mathrm{2}} +\lambda−\mathrm{2}=\mathrm{0} \\ $$$$\left(\lambda+\mathrm{2}\right)\left(\lambda−\mathrm{1}\right)=\mathrm{0} \\ $$$$\lambda\:=\:−\mathrm{2}\:;\:\mathrm{1} \\ $$$$\mathrm{y}_{\mathrm{h}} \:=\:\mathrm{Ae}^{−\mathrm{2x}} \:+\:\mathrm{Be}^{\mathrm{x}} \: \\ $$
Commented by Shakhzod last updated on 13/May/20
$${Thank}\:{you}\:{but}\:{I}\:{also}\:{just}\:{solved}\:{this} \\ $$$${problem}.\:{Please}\:{help}\:{the}\:{rest}\:{of}\:{the}\:{tasks}. \\ $$
Commented by mr W last updated on 13/May/20
$${your}\:{question}\:{is}\:{completely}\:{solved}. \\ $$$${what}\:{is}\:{the}\:{rest}\:{of}\:{the}\:{task}? \\ $$