Solving a System of Differential Equations with Complex Eigenvalues

[Temp0]

1. Homework Statement
http://puu.sh/cSK1u/62e2f1c74d.png olve the system:
x' = [-4, -4
4, -4]
with x(0) = [ 2, 3]

Find x1 and x2 and give your solution in real form.2. Homework Equations 3. The Attempt at a Solution
Just a note here, I'm basically forced to self-learn this course because my professor doesn't seem to make sense. So I just tried to imitate this website's way of solving one of these problems.
http://tutorial.math.lamar.edu/Classes/DE/ComplexEigenvalues.aspx

So I start by finding my complex eigenvalues, which are -4 \pm 4i

I only need one of them, so I take -4 + 4i, and I plug it into my matrix to solve for the eigenvectors, which I only need one of. I find:

v1 = [i, 1]

Which I then plug into my first equation,
x_1 (t) = e^{-4t}(cos(4t) + isin(4t)) [i, 1]

Then I multiply in, separate the imaginary from the real, and I have:
[-e^{-4t} sin(4t), e^{-4t} cos(4t)] + i[e^{-4t} cos(4t), e^{-4t}sin(4t)]

Next, as the website suggests, I find the constants using the initial values provided, and I get that C1 = 2 and C2 = 3.

Plugging in the constants, I end up with the equation:
2 [-e^{-4t} sin(4t), e^{-4t} cos(4t)] + 3[e^{-4t} cos(4t), e^{-4t}sin(4t)]

However, I don't understand what the question means by "Real form". Could anyone explain this to me? thank you in advance.

 

[ehild]

1. Homework Statement
http://puu.sh/cSK1u/62e2f1c74d.png olve the system:
x' = [-4, -4
4, -4]
with x(0) = [ 2, 3]

Find x1 and x2 and give your solution in real form.2. Homework Equations 3. The Attempt at a Solution
Just a note here, I'm basically forced to self-learn this course because my professor doesn't seem to make sense. So I just tried to imitate this website's way of solving one of these problems.
http://tutorial.math.lamar.edu/Classes/DE/ComplexEigenvalues.aspx

So I start by finding my complex eigenvalues, which are -4 \pm 4i

I only need one of them, so I take -4 + 4i, and I plug it into my matrix to solve for the eigenvectors, which I only need one of. I find:

v1 = [i, 1]

You need both in principle. And write the general solution as linear combination of these two independent "basic" solutions, belonging to the different eigenvalues. Fitting the linear combination to the initial conditions, you get a real solution of the differential equation.

 

[Temp0]

I tried that and I end up with:

x = C_1 e^{-4t} (cos(4t) + isin(4t)) [i,1] + C_2e^(-4t) (cos(4t) - isin(4t)) [i, -1]

Then after fitting it to the initial conditions, I end up with :
C_1 = -i - \frac {3}{2} , C_2 = -i + \frac{3} {2}

However I don't see how this solution is real, since the i is still present.

 

[ehild]

I tried that and I end up with:

x = C_1 e^{-4t} (cos(4t) + isin(4t)) [i,1] + C_2e^(-4t) (cos(4t) - isin(4t)) [i, -1]

Then after fitting it to the initial conditions, I end up with :
C_1 = -i - \frac {3}{2} , C_2 = -i + \frac{3} {2}

However I don't see how this solution is real, since the i is still present.

Keeping the complex form, the general solution is \vec x = C_1 e^{(-4+4i)t} [i,1] + C_2e^{(-4-4i)t} [i, -1]
Fitting to the initial conditions, C1=1.5-i; C2=-1.5-i. Your constants does not seem correct.
Substitute the constants back into the expression for x and changing the exponential form to the trigonometric ones , you get a real expression for both components of x.

 

[Orodruin]

Inserting $t = 0$ into your expression, I get [3,2] instead of [2,3] ...

 

[Temp0]

Oh, yeah I think I got the two numbers swapped. I recalculated and got

C1 = 3/2 - i, C2 = -3/2 - i

However, even after substituting in the constants and changing the exponential forms to trigonometric, I am still left with:

( cos(4t) + isin(4t) ) [e^(-4t) ( (3/2) i + 1), e^(-4t) (3/2 - i) ] - (cos(4t) - isin(4t)) [e^(-4t) (3/2 i - 1), e^(-4t) ( (-3/2) - i )]

When I collect the terms for x1 and x2, I don't see how I am supposed to get rid of the i that makes the expression imaginary.

 

[Orodruin]

Did you try simply multiplying in the trigonometric factors and adding the two vectors together? (Obviously, $e^{-4t}$ can be factored out of the expression so you do not have to care about it.)

 

[Temp0]

Oh wow! I didn't expect it to work out like that ._.

So I got x1 = e^(-4t) (-3 sin(4t) + 2cos(4t))
x2 = e^(-4t) (3cos(4t) + 2 sin(4t))

Initial condition checks out, did I make any other mistakes? I didn't want to type the entire solution that I did since I expanded the entire thing and had an equation stretch two lines on my piece of paper.

 

[Orodruin]

The final check is of course just plugging your solution back into your differential equation to see that it works out. Once you have done that, you're home safe.

 

[Temp0]

Alright, thanks guys, I was really confused about this but I'm starting to understand more. Thank you for all the help.