I am currently reading Differential equations in 24 hours with solutions and historical notes. by Scott Imhoff, PhD. In the fifth hour we are given: $$\ddot{y} + y = 0$$ $$y = c_1 \cos t + c_2 \sin t$$ $$ \vec{C} = <c_1, c_2> $$ When $ y(0) = 1 $ and $\dot{y}(0) = 0 $ then $ \vec{C} = <1, 0> $ and $ y = \cos t $
When $ y(0) = 0 $ and $\dot{y}(0) = 1 $ then $ \vec{C} = <0, 1> $ and $ y = \sin t $
I observe, in general, we have $ \vec{C} = r <\cos\theta, \sin\theta > $ where (r) is the magnitude and $\theta$ is the phase angle and we can write $ y = r \cos\theta \cos t + r \sin\theta \sin t $ as a general solution.
If we let $\vec{C}$ be the complex vector $<1, i> $ then a particular solution is $ y = \cos t + i \sin t $ and we note that this is equal to exponential form $ e^{ti}$. To verify this we can do: $$ \frac{d}{dt} e^{ti} = ie^{ti}$$ $$ \frac{d^2}{dt^2} e^{ti} = -e^{ti}$$ and see $ -e^{ti} + e^{ti} = 0$ is a particular solution to the given differential equation.
My question is: How do I generalize the exponential solution? That is, how do I introduce the $\vec{C}$ into the exponential form while maintaining the nice interpretation of phase angle and magnitude?
I do not know if this is what you are looking for, but here we go.
You can multiply both sides by $y'$: \begin{align*} y'' + y = 0 & \Longleftrightarrow y''y' + y'y = 0\\\\ & \Longleftrightarrow (y')^{2} + y^{2} = k^{2}\\\\ & \Longleftrightarrow y' = \pm\sqrt{k^{2} - y^{2}} \end{align*}
Now you can apply the integration by substitution in order to get $y$.
Hopefully this helps!
EDIT
In order to answer your question, we can proceed as follows: \begin{align*} y(t) & = c_{1}\cos(t) + c_{2}\sin(t)\\\\ & = \sqrt{c^{2}_{1} + c^{2}_{2}}\left(\frac{c_{1}\cos(t)}{\sqrt{c^{2}_{1} + c^{2}_{2}}} + \frac{c_{2}\sin(t)}{\sqrt{c^{2}_{1} + c^{2}_{2}}}\right)\\\\ & = k\sin(t + \varphi) \end{align*} where $k = \sqrt{c^{2}_{1}+c^{2}_{2}}$ and
$$\varphi = \arcsin\left(\frac{c_{1}}{\sqrt{c^{2}_{1} + c^{2}_{2}}}\right)$$
If you still have any questions, please let me know.