Answer:
The 2 values of k are 3 and 6.
Step-by-step explanation:
If the given function of y is a solution of the differential equation then it shall be able to satisfy it
Thus we have
[tex]y''-9y'+18y=0[/tex]
Putiing [tex]y=e^{kx}[/tex] in the above differential euation we get
[tex]\frac{d^2e^{kx}}{dx^2}-9\frac{de^{kx}}{dx}+18\cdot e^{kx}=0\\\\k^{2}e^{kx}-9k\cdot e^{kx}+18\cdot e^{kx}=0\\\\k^2-9k+18=0[/tex]
Solving the above uadratic euation with the standard formula we get
[tex]k=\frac{-(-9)\pm \sqrt{(-9)^2-4\times 1\times 18} }{2}\\\\k_1=6\\\\k_2=3[/tex]
The solution of the equation for values of k is 6 and 3.
Differentiation is a process, in which we find the instantaneous rate of change in function based on one of its variables.
Given
Differential equation; [tex]\rm y(x)=e^{kx}[/tex]
To find the derivative of the given equation differentiates the equation of both the sides with respect to x.
The first derivative is;
[tex]\rm y'(x)=k. e^{kt}[/tex]
The second derivative is
[tex]\rm y'(x)=k^2. e^{kt}[/tex]
Substitute all the values in the equation
[tex]\rm y"-9y'+18y=0\\\\ k^2. e^{kt}-9(ke^{kt})+18(e^{kt})=0\\\\ k^2. e^{kt}-9ke^{kt}+18e^{kt}=0\\\\ e^{kt}(k^2-9k+18)=0\\\\ k^2-9k+18=0\\\\ k^2-6k-3k+18=0\\\\k(k-6)-3(k-6)=0\\\\(k-6)(k-3)=0\\\\k-6=0, \ k=6\\\\k-3=0, \ k=3[/tex]
Hence, the solution of the equation for values of k is 6 and 3.
To know more about differential equations click the link given below.
https://brainly.com/question/52461
