we are given
differential equation as
[tex]y'''-9y''+20y'=0[/tex]
we are given
[tex]y=e^{rx}[/tex]
Firstly, we will find y' , y'' and y'''
those are first , second and third derivative
First derivative is
[tex]y'=re^{rx}[/tex]
Second derivative is
[tex]y''=r*re^{rx}[/tex]
[tex]y''=r^2e^{rx}[/tex]
Third derivative is
[tex]y'''=r^2*re^{rx}[/tex]
[tex]y'''=r^3e^{rx}[/tex]
now, we can plug these values into differential equation
and we get
[tex]r^3 e^{rx}-9r^2 e^{rx}+20re^{rx}=0[/tex]
now, we can factor out common terms
[tex]e^{rx}(r^3 -9r^2 +20r)=0[/tex]
we can move that term on right side
[tex](r^3 -9r^2 +20r)=0[/tex]
now, we can factor out
[tex]r(r^2 -9r +20)=0[/tex]
[tex]r(r-5)(r-4)=0[/tex]
now, we can set them equal
[tex]r=0[/tex]
[tex]r-5=0[/tex]
[tex]r=5[/tex]
[tex]r-4=0[/tex]
[tex]r=4[/tex]
so, we will get
[tex]r=0,4,5[/tex]...............Answer
