Answer:
The values of k are
1) k = 7.
2) k= 13
Step-by-step explanation:
The given differential equation is
[tex]y''-20y'+91y=0[/tex]
Now since it is given that [tex]y=e^{kx}[/tex] is a solution thus it must satisfy the given differential equation thus we have
[tex]\frac{d^2}{dx^2}(e^{kx})-20\frac{d}{dx}e^{kx}+91e^{kx}=0\\\\k^{2}\cdot e^{kx}-20\cdot k\cdot e^{kx}+91e^{kx}=0\\\\e^{kx}(k^{2}-20k+91)=0\\\\k^{2}-20k+91=0[/tex]
This is a quadratic equation in 'k' thus solving it for k we get
[tex]k=\frac{20\pm \sqrt{(-20)^2-4\cdot 1\cdot 91}}{2}\\\\\therefore k=7,k=13[/tex]
