# Thread: 2nd ODE Homogeneous, Double roots

1. ## 2nd ODE Homogeneous, Double roots

$\displaystyle y^{\prime \prime} + 6 y^{\prime} + 9 y = 0$

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

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

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

When Y(0)=3
[Math]3=A+B[/tex]

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

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?

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

3. Originally Posted by Prove It
When there's a repeated root, the solution is actually $\displaystyle \displaystyle Ae^{-3x} + Bx\,e^{-3x}$.
So when Y(0)=3
3=A

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

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

This seems to be wrong! Where did I mess up?

4. Let
$\displaystyle y(x)=(a+bx)\mathrm{e}^{-3x}$,
then
$\displaystyle y'(x)=((b-3a)-3bx)\mathrm{e}^{-3x}$.
Applying the initial conditions, we get
$\displaystyle a=3$ and $\displaystyle b-3a=-7$,
which gives you
$\displaystyle a=3$ and $\displaystyle b=2$.
$\displaystyle y(x)=3\mathrm{e}^{-3x}+2x\mathrm{e}^{-3x}$.
You seem to compute $\displaystyle b$ wrong.
$\displaystyle \displaystyle y'(x) = -3Ae^{-3x} + Be^{-3x} - 3Bx\,e^{-3x}$.