Question

Find the general solution to 2y ′′ − y ′ − y = 0.

Answer 1

Answer: y(x) =
`C_(1) e^(x) + C_(2) e^{(-x)/(2) }`

Explanation:

2y ′′ − y ′ − y = 0

The characteristic equation is:

`2r^(2) - r - 1 = 0`

`2r^(2) - 2r + r - 1 = 0`

2r(r-1) + 1(r-1) = 0

(r-1)(2r+1) = 0

`r_(1) = 1 , r_(2) = (-1)/(2)`

∴ there are two distinct roots

so the general equation is as follows:

y(x) =
`C_(1) e^{r_(1)x } + C_(2) e^{r_(2)x }`

y(x) =
`C_(1) e^(x) + C_(2) e^{(-x)/(2) }`