# Is this a correct solution ? to ax'' = 0

1. Jan 18, 2010

### Cypeq

$ax^{''} = 0$

characteristic eqation
$ar^{2} = 0$

$r_{1,2} = 0$

since we have double result

$x(t) = c_{1}e^{0t} + c_{2}t e^{0t}$

so $$e^0 = 1$$

$$x(t) = c_{1} + c_{2}t$$

is this a correct result :S ? i'm learning integration by myself and i'm not shure about this case it's not cover in my book if results can be zero or it doesn't matter :S ?

Last edited: Jan 18, 2010
2. Jan 18, 2010

### HallsofIvy

Staff Emeritus
Yes, that's correct. The simpler way to see that is to first divide by a to get just x"= 0. That is the same as saying d(x')/dt= 0 so x' is a constant: x'(t)= c2. Now integrate that: the integral of a constant, c1 is, of course, x(t)= c2t+ c1.

(I'm surprised you would know about "characteristic equations" for linear differential equations with constant coefficients if you are just learning integration!)