Question

If y = e2x is a solution to y''- 5y' + ky = 0, what is the value of k?

Answer 1

The value of k is 6

Explanation:

we need to find the value of k

Given : -
`y=e^(2x)` is the solution
`y''-5y'+ky=0`

`y=e^(2x)` ........(1)

differentiate
`y=e^(2x)` with respect to 'x'

`(dy)/(dx)=(d)/(dx)e^(2x)`

Since,
`(d)/(dx)e^(x) =e^(x)(d)/(dx)(x)`

`(dy)/(dx)=e^(2x)(d)/(dx)(2x)`

`(dy)/(dx)=e^(2x)* 2`

`(dy)/(dx)=2e^(2x)`

so,
`y'=2e^(2x)` ..........(2)

Again differentiation above with respect to 'x'

`(d)/(dx)(dy)/(dx)=(d)/(dx)2e^(2x)`

`(d^(2)y)/(dx^(2))=2e^(2x)(d)/(dx)(2x)`

`(d^(2)y)/(dx^(2))=2e^(2x)* 2`

`(d^(2)y)/(dx^(2))=4e^(2x)`

so,
`y''=4e^(2x)` ........(3)

Now, put the value of
`y\ ,y' \ \text{and} \ y''` in
`y''-5y'+ky=0`

`4e^(2x)-5(2e^(2x))+(e^(2x))k=0`

`4e^(2x)-10e^(2x)+e^(2x)k=0`

`-6e^(2x)+e^(2x)k=0`

add both the sides by
`6e^(2x)`

`e^(2x)k=6e^(2x)`

Cancel out the same terms from left and right sides

`k=6`

Hence, the value of k is 6