x' =
[ -4 2
2 -1] x
with x(0) =
[1
3]
solve for the differential equation where x =
[x(t)
y(t)]
i solved for the eigenstuff and got eigenvectors:
[-2
1]
and
[1
2]
so general solution I got was
C1e^(-5t) [-2,1] + C2 [1.2]
i tried the whole RREF augmented matrix thing with the initial condition and i just cant seem to get the answer. in frustration, i am now on yahoo answers. please explain steps! thanks
[ -4 2
2 -1] x
with x(0) =
[1
3]
solve for the differential equation where x =
[x(t)
y(t)]
i solved for the eigenstuff and got eigenvectors:
[-2
1]
and
[1
2]
so general solution I got was
C1e^(-5t) [-2,1] + C2 [1.2]
i tried the whole RREF augmented matrix thing with the initial condition and i just cant seem to get the answer. in frustration, i am now on yahoo answers. please explain steps! thanks
-
Okay, you got the first part correct!
After doing that part, make a system of equations:
2c1 + c2 = 1
-c1 + 2c2 = 3
You might be wondering why I didn't write -2c1 but that is because I multiplied your vector (-2 1) by -1 so I have (2 -1) instead. I try not to have negatives or fractions on the first part in the vector so I multiply by -1 or the reciprocal.
Okay, after you have your system of equations, you solve for c1 and c2. c1 is -1/5 and c2 is 7/5.
You end up with -1/5 e^(-5t) (2 -1) + 7/5 (1 2)
The first answer is found by: -1/5e^(-5t) * 2 + 7/5 * 1
and the second answer is found by:
-1/5 e^(-5t) * -1 + 7/5 * 2
The answers are:
x(t) =-2/5e^(-5t) + 7/5
y(t) = 1/5e^(-5t) + 14/5
If you don't understand my explanations (it's hard to explain without proper formatting) then you can check out the website listed in the source section. It's what helped me understand this. Good luck!
After doing that part, make a system of equations:
2c1 + c2 = 1
-c1 + 2c2 = 3
You might be wondering why I didn't write -2c1 but that is because I multiplied your vector (-2 1) by -1 so I have (2 -1) instead. I try not to have negatives or fractions on the first part in the vector so I multiply by -1 or the reciprocal.
Okay, after you have your system of equations, you solve for c1 and c2. c1 is -1/5 and c2 is 7/5.
You end up with -1/5 e^(-5t) (2 -1) + 7/5 (1 2)
The first answer is found by: -1/5e^(-5t) * 2 + 7/5 * 1
and the second answer is found by:
-1/5 e^(-5t) * -1 + 7/5 * 2
The answers are:
x(t) =-2/5e^(-5t) + 7/5
y(t) = 1/5e^(-5t) + 14/5
If you don't understand my explanations (it's hard to explain without proper formatting) then you can check out the website listed in the source section. It's what helped me understand this. Good luck!