Question

Find the two values of k for which y(x)=ekx is a solution of the differential equation

y″−9y′+18y=0.

Answer 1

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

$y''-9y'+18y=0$

Putiing $y=e^{kx}$ in the above differential euation we get

$\frac{d^2e^{kx}}{dx^2}-9\frac{de^{kx}}{dx}+18\cdot e^{kx}=0\\\\k^{2}e^{kx}-9k\cdot e^{kx}+18\cdot e^{kx}=0\\\\k^2-9k+18=0$

Solving the above uadratic euation with the standard formula we get

$k=\frac{-(-9)\pm \sqrt{(-9)^2-4\times 1\times 18} }{2}\\\\k_1=6\\\\k_2=3$

Answer 2

The solution of the equation for values of k is 6 and 3.

What is differentiation?

Differentiation is a process, in which we find the instantaneous rate of change in function based on one of its variables.

Given

Differential equation; $\rm y(x)=e^{kx}$

To find the derivative of the given equation differentiates the equation of both the sides with respect to x.

The first derivative is;

$\rm y'(x)=k. e^{kt}$

The second derivative is

$\rm y'(x)=k^2. e^{kt}$

Substitute all the values in the equation

$\rm y"-9y'+18y=0\\\\ k^2. e^{kt}-9(ke^{kt})+18(e^{kt})=0\\\\ k^2. e^{kt}-9ke^{kt}+18e^{kt}=0\\\\ e^{kt}(k^2-9k+18)=0\\\\ k^2-9k+18=0\\\\ k^2-6k-3k+18=0\\\\k(k-6)-3(k-6)=0\\\\(k-6)(k-3)=0\\\\k-6=0, \ k=6\\\\k-3=0, \ k=3$

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