Welcome to our community

Be a part of something great, join today!

Differential equation with eigenvector (complex number)

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
Solve the following system of linear differential equation
\(\displaystyle f'=f-g\)
\(\displaystyle g'=f+g\)

with bounded limit \(\displaystyle f(0)=0\), \(\displaystyle g(0)=1\)
could anyone check if My answer is correct? Just to make sure I understand correctly!
ps we get \(\displaystyle \lambda=1-i\) and \(\displaystyle \lambda=1+i\)

Regards,
\(\displaystyle |\pi\rangle\)
 
Last edited:

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Your solution is $f(t)=-\sin t,\;g(t)=\cos t.$ But the equality $f'(t)=f(t)-g(t)$ is not satisfied.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
Hello MHB,
Solve the following system of linear differential equation
\(\displaystyle f'=f-g\)
\(\displaystyle g'=f+g\)

with bounded limit \(\displaystyle f(0)=0\), \(\displaystyle g(0)=1\)
could anyone check if My answer is correct?
First thing to say here is that you never need to ask whether your answer to a differential equation is correct – you can check it for yourself! Your answer here (with $c_1=0$ and $c_1=1$) is $f(x) = -\sin t$ and $g(x) = \cos t$. Is that correct? To see whether it is, go back to the original equations. One of them says that $f' = f-g$. If your answer is correct then $f'(t) = -\cos t$ and $f(t) - g(t) = -\sin t - \cos t$. Those are not the same, so your answer cannot be right.

Your eigenvalues \(\displaystyle \lambda=1-i\) and \(\displaystyle \lambda=1+i\) are correct, but you must have gone wrong in calculating the eigenvectors.

Edit. Fernando Revilla got there first, but I'll leave my comment anyway, since it slightly amplifies what he said.