# Differential equation with eigenvector (complex number)

#### Petrus

##### Well-known member
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,
$$\displaystyle |\pi\rangle$$

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#### Fernando Revilla

##### Well-known member
MHB Math Helper
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
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.