# Thread: x'' + 3x' + 2x = 0

1. ## x'' + 3x' + 2x = 0

Show that for the system x'' + 3x' + 2x = 0
x(0) = 0 x'(0) dosen't equal 0
it holds that x(t) dosen't equal 0 for all t.

I've got as far as y = Ae^-2x + Be^-x
and the differential of this.

Now im really stuck.

2. Originally Posted by ardam
Show that for the system x'' + 3x' + 2x = 0
x(0) = 0 x'(0) dosen't equal 0
it holds that x(t) dosen't equal 0 for all t.

I've got as far as y = Ae^-2x + Be^-x
and the differential of this.

Now im really stuck.
OK, first off, clearly x(t) DOES equal 0 at a value of t, namely t = 0. So I assume you meant that it holds that x(t) doesn't equal 0 for all OTHER t.

You've gotten as far as

$x = Ae^{-2t} + Be^{-t}$.

Using the initial condition

$0 = A + B \implies B = -A$.

So $x = Ae^{-2t} - Ae^{-t}$

$x = A(e^{-2t} - e^{-t})$.

If $x = 0$ then either $A = 0$ or $e^{-2t} - e^{-t} = 0$. The latter only occurs at $t = 0$, as was already stated by the initial condition.

So the only other way that $x = 0$ is if $A = 0$.

Recall that $\frac{dx}{dt} \neq 0$ as stated originally.

$\frac{dx}{dt} = A(-2e^{-2t} + e^{-t})$

So $A(-2e^{-2t} + e^{-t}) \neq 0$.

This implies that $A \neq 0$.

Therefore, $x \neq 0$ for any $t \neq 0$.

3. Originally Posted by Prove It
OK, first off, clearly x(t) DOES equal 0 at a value of t, namely t = 0. So I assume you meant that it holds that x(t) doesn't equal 0 for all OTHER t.

You've gotten as far as

$x = Ae^{-2t} + Be^{-t}$.

Using the initial condition

$0 = A + B \implies B = -A$.

So $x = Ae^{-2t} - Ae^{-t}$

$x = A(e^{-2t} - e^{-t})$.

If $x = 0$ then either $A = 0$ or $e^{-2t} - e^{-t} = 0$. The latter only occurs at $t = 0$, as was already stated by the initial condition.

So the only other way that $x = 0$ is if $A = 0$.

Recall that $\frac{dx}{dt} \neq 0$ as stated originally.

$\frac{dx}{dt} = A(-2e^{-2t} + e^{-t})$

So $A(-2e^{-2t} + e^{-t}) \neq 0$.

This implies that $A \neq 0$.

Therefore, $x \neq 0$ for any $t \neq 0$.
Thanks now i understand what i should have done
i have just reread the question and i should have written holds for all t>0
does this change the question too much?

4. Originally Posted by ardam
Thanks now i understand what i should have done
i have just reread the question and i should have written holds for all t>0
does this change the question too much?
It doesn't change the question at all.

I showed that $x \neq 0$ for all $t \neq 0$.

All $t > 0$ is a part of all $t \neq 0$.