Eigenvalue method for solving system of differential equations

[theBEAST]

Homework Statement

Here is the question along with my work. I attempted to solve for the actual solution using both eigenvectors. From what I have been taught it should yield the same answer... But as you can see (circled in red) the solutions are clearly different. Is this normal or maybe because I am making an algebra mistake?

I know for sure the solution in attempt 1 is correct because that is what wolfram alpha gives me. But attempt 2 for some reason has a different answer.

 

[HallsofIvy]

I really don't understand what you mean by "solving with V1" and "solving with V2". The general solution, from which you get the solution with the given initial values, must be a linear combination of V1 and[ V2. You cannot get a solution from either one alone.

 

[STEMucator]

I really don't understand what you mean by "solving with V1" and "solving with V2". The general solution, from which you get the solution with the given initial values, must be a linear combination of V1 and[ V2. You cannot get a solution from either one alone.

This is not true. You can retrieve a real valued solution from either $x^{(1)}$ or $x^{(2)}$ by taking the real parts of either one by using Euler's formula.

 

[theBEAST]

This is not true. You can retrieve a real valued solution from either $x^{(1)}$ or $x^{(2)}$ by taking the real parts of either one by using Euler's formula.

Yup, take a look at the first example from Pauls notes. They said you only need one eigenvector:

http://tutorial.math.lamar.edu/Classes/DE/ComplexEigenvalues.aspx

So am I doung something wrong or am I suppose to get a slightly different answer?

 

[pasmith]

In your solution with $v_2$, you should have substituted $\mu = -3\sqrt 3$ into the your formula. Since $\sin(-3\sqrt3 t) = - \sin(3 \sqrt 3 t)$, the result is that every term multiplied by $\sin (3 \sqrt3 t)$ has the wrong sign.

Make that correction and you should get the same result as with $v_1$.

 

[theBEAST]

This is not true. You can retrieve a real valued solution from either $x^{(1)}$ or $x^{(2)}$ by taking the real parts of either one by using Euler's formula.

Wow! After doing these for such a long time I did not know that. I always thought you just took the magnitude of the eigenvalue... At least that's what it looked like in the generic formula I wrote and copied from the professor.

Thanks!