$ \ddot{y}+k^{2} \cdot y=0$: Book solution: $y_b(t)=A \cosh (k t)+B \sinh (k t)$ Mine: $y(t)=A e^{i k t}+B e^{-i k t}$. How to calculat $y_b$?

Question

My solution: $$ y^{\prime \prime}(t)=-y \cdot k^{2} \Leftrightarrow \ddot{y}+k^{2} \cdot y=0 \\ \Rightarrow \lambda^{2}+k^{2}=0 \Rightarrow \lambda_{1,2}= \pm i k \\ y(t)=A e^{i k t}+B e^{-i k t}$$

with $A, B \in \mathbb{R} $.

Books solution: $$ y_b(t)=A \cos (k t)+B \sin (k t) $$

The index $b$ just refers to the solution given in the book.

Both solution are correct but how do you calculate the $y_b$ solution?

Edit: There seems to be a printing error in the book, I corrected the function $y_b$. The question remains the same.