Why do we drop $i$ in euler ode's?

Question

Consider

$$y''(t) + \lambda y(t) = 0$$

Setting $y=e^{rt}$

$$y = c_1e^{i\sqrt{\lambda}t} + c_2e^{-i\sqrt{\lambda}t}$$

Using Euler's identity

$$y= c_1\left[\cos\sqrt{\lambda} t + i \sin\sqrt{\lambda} t \right] + c_2\left[\cos\sqrt{\lambda} t - i \sin\sqrt{\lambda} t \right]$$

How do we get rid of the $i$ to get our "standard-looking" general solution?

Answer 1

Note that $$y_1= \left[\cos\sqrt{\lambda} t + i \sin\sqrt{\lambda} t \right]$$

and $$y_2 =\left[\cos\sqrt{\lambda} t - i \sin\sqrt{\lambda} t \right]$$

are solutions to your linear differential equation.

Thus the sum and the difference of $y_1$ and $y_2$ are also solutions. $$ \frac { y_1+y_2}{2} = cos \sqrt{\lambda} t $$ and $$ \frac {y_1-y_2}{2i} = sin\sqrt{\lambda} t $$ Now the general solution is a linear combination of these two solutions.

That is $$y= C_1 cos\sqrt{\lambda} t + C_2 \sin\sqrt{\lambda} t $$

Answer 2

We aren't "dropping" anything. $$y = (c_1 + c_2) \cos (\sqrt{\lambda} t) + (c_1 - c_2) i \sin(\sqrt{\lambda} t)$$ Note that $c_1+c_2$ and $(c_1-c_2) i$ can be any two complex numbers.