ChiralSuperfields said:
Thank you for your reply
@Orodruin!
I've been in touch with the Prof and there is no typo in the solution (at least the complex case part you said there was). Do you please know why you think this is a typo (I am getting the same solution as you by the way)?
Thanks!
Yes there are. There are
obvious typos in the complex solution as pointed out in red in my previous post.
Just try to insert the given solution and you will see that it is, in fact, not a solution. Insert the corrected solution from post #4 and you will see that it
is a solution.
Failure to recognise this is a worrying sign. Did you point out exactly where the typos were?
A similar issue, by the way, holds for the real solution, where the ##b/2## is not multiplied by ##t## as it should be. It is furthermore unclear from that solution whether the ##t## after the square root is a multiplication or a division. Your professor should use parentheses to make this clear. The easiest remedy is to put parentheses around the full root expression:
$$
\exp\left([b/2 + \sqrt{b^2 - 4}/2] t\right)
$$
Also note how the root was messed up: Using ##4 - b^2## instead of ##b^2 - 4## in the real (##b < -2##) case would make the square root argument negative and therefore make the square root imaginary.
The correct solutions are:
Complex case:
$$
u_h = e^{b\color{Red}{t}/2} \left[c_1 \cos\left(\frac{t\sqrt{4-b^2}}{\color{Red}{2}}\right) + c_2 \sin\left(\frac{t\sqrt{4-b^2}}{\color{Red}{2}}\right)\right]
$$
Real case:
$$
u_h = c_1 \exp\left( \frac{\color{Red}{t}}{2}[b + \sqrt{\color{Red}{+}b^2 \color{Red}{-} 4}] \right) + c_2 \exp\left( \frac{\color{Red}{t}}{2}[b - \sqrt{\color{Red}{+}b^2 \color{Red}{-} 4}] \right)
$$
I have marked in red where typos of the provided solution have been corrected. The repeated root solutions is free of typos.
Your professor is welcome here to try to argue their piece if they still think that they do not have typos (they would be wrong, but they are free to present it and be corrected).