Question

Given that y(t)=c1e4t+c2e−4ty(t)=c1e4t+c2e−4t is a solution to the differential equation y′′−16y=0y′′−16y=0, where c1c1 and c2c2 are arbitrary constants, find a function y(t)y(t) that satisfies the conditions

Answer 1

Solution :- Given differential equation

`y′′−16y=0`

So the characteristic equation will be

`image`

so, the required function will be

`y=c_(1)e^{r_(1)t}+c_(1)e^{r_(2)t}`

⇒
`y=c_(1)e^(4t)+c_(1)e^(-4t)` ia solution of the given differential equation

where
`c_(1) \\and \\c_(2)` are constants.