find the general solution

Find the general solution using the method of elimination
x'=x+2y+z
y'=6x-y
z'=-x-2y-z

I used a matrix and found
D^3 + D^2 - 12D=0
and then I think
x(t)= A + Be^(-4t) +Ce^(3t)

But then I don't know where to go, or how to solve it,

Please Help

Thanks

Quote:

Originally Posted by mandy123

Find the general solution using the method of elimination
x'=x+2y+z
y'=6x-y
z'=-x-2y-z

I used a matrix and found
D^3 + D^2 - 12D=0
and then I think
x(t)= A + Be^(-4t) +Ce^(3t)

But then I don't know where to go, or how to solve it,

Please Help

Thanks

Let's do it a little different. First solve the second equation for $x$ so $x = \frac{y'+y}{6}\;\;(1)$ . Substitute this into the first equation. This gives
$\frac{y''+y'}{6} = \frac{y'+y}{6} + 2y + z$ . Solving for $z$ gives $z = \frac{y'' - 13y}{6}\;\;(2)$ . Now substitute this into the first equation giving (after simplying gives

$y''' + y'' - 12y' = 0$ (same equation as for x)

The solution as you said is $y = c_1 + c_2 e^{-4t} + c_3e^{3t}$ (I've chosen different constants not to be confusing with your x(t) solution). Now substitute this into (1) and (2) above to get $x\; \text{and}\; z$ .

wow, this makes a lot more sense than what the book was telling me to do, thank you so much!!!!(Nod)