# 2nd ODE Homogeneous, Double roots

• Mar 27th 2011, 11:00 PM
Vamz
2nd ODE Homogeneous, Double roots
$y^{\prime \prime} + 6 y^{\prime} + 9 y = 0$

Initial Conditions:
$y(0)=3,\quad y^{\prime}(0)=-7$

So, I determined the general solution should look like this:

$Ae^{-3x}+Be^{-3x}=0$

When Y(0)=3
$$3=A+B$$

And when Y'(0)=-7
$
-7=-3A-3B
$

Im stuck here. Anytime I try and solve for one variable, I get
-7=-9 which is a totally false statement.

What do I do from here?
• Mar 27th 2011, 11:13 PM
Prove It
When there's a repeated root, the solution is actually $\displaystyle Ae^{-3x} + Bx\,e^{-3x}$.
• Mar 28th 2011, 04:08 PM
Vamz
Quote:

Originally Posted by Prove It
When there's a repeated root, the solution is actually $\displaystyle Ae^{-3x} + Bx\,e^{-3x}$.

So when Y(0)=3
3=A

and $Y'=-3Ae^{3x}+[Be^{-3x}-3Be^{-3x}]$
So when Y'(0)=-7,
B=3

So my final solution is
$3e^{-3x}+3xe^{-3x}$

This seems to be wrong! Where did I mess up?
• Mar 28th 2011, 05:21 PM
bkarpuz
Let
$y(x)=(a+bx)\mathrm{e}^{-3x}$,
then
$y'(x)=((b-3a)-3bx)\mathrm{e}^{-3x}$.
Applying the initial conditions, we get
$a=3$ and $b-3a=-7$,
which gives you
$a=3$ and $b=2$.
$y(x)=3\mathrm{e}^{-3x}+2x\mathrm{e}^{-3x}$.
You seem to compute $b$ wrong. :)
$\displaystyle y'(x) = -3Ae^{-3x} + Be^{-3x} - 3Bx\,e^{-3x}$.