First order system ODE, complex root

[kougou]

Hi.

For first order system ODE, complex root.
y'=Ay, where A is a 2by2 matrix. I am assuming the roots are complex. After finding the eigenvalue (complex conjugate) and their eigen-vectors (which come in a form of complex conjugate again), we plug into the solution y=ζexp(λt), where λ is egenvalue and ζ is its corresponding eigenvector. And Since λ is complex, we apply euler's formula.And at the end, we get to the part
y= u(t) + i*v(t).

I don't understand why both u(t) and v(t) are solution to the system. u(t) is clear. But v(t) is not. what happen to its imaginary number?

.

 

[lippyka]

The imaginary number in v(t) is taken care of by the imaginary part of the eigenvalue λ. The solution to the system is a combination of both u(t) and v(t). The real part of the eigenvalue λ determines the exponential growth or decay behavior, while the imaginary part determines the oscillatory behavior. Therefore, when the eigenvalue is complex, the solution will contain both exponential and oscillatory components, which is why both u(t) and v(t) are solutions to the system.