Question

Find the two values of k for which y(x) = e^kx is a solution of the differential equation y'' - 20y' + 91y = 0. Preview smaller value = larger value = Preview

Answer 1

The values of k are

1) k = 7.

2) k= 13

Explanation:

The given differential equation is

`y''-20y'+91y=0`

Now since it is given that
`y=e^(kx)` is a solution thus it must satisfy the given differential equation thus we have

`(d^2)/(dx^2)(e^(kx))-20(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`

This is a quadratic equation in 'k' thus solving it for k we get

`k=(20\pm √((-20)^2-4\cdot 1\cdot 91))/(2)\\\\\therefore k=7,k=13`