Solving the harmonic oscillator $y'' + \omega^2y = 0$ and uniqueness of the solution for a general second order ODE My first question is about the final form of the solution for the harmonic oscillator (w.o. initial conditions) and my second question is about the classification of all solutions of second order ODEs.
1.) My lecture notes claim that $y''(t) = -\omega^2y(t)$ implies that $y(t) = C\sin(\omega t + \theta)$ for some $C, \theta$. One solution method for this sort of second order ODE is to use the ansatz $y(t) = e^{\lambda t}$ and to find the roots of the characteristic polynomial of the original equation. In our case the characteristic equation turns out to be $\lambda^2 + \omega^2 = 0$ from which we see that $\lambda = \pm i \omega$ and $y(t) = C_1e^{i\omega t} + C_2e^{-i\omega t} = \cos(\omega t)(C_1 + C_2) + i\sin(\omega t)(C_1 - C_2)$ is a general solution with some constants $C_1, C_2$. If one allows $C$ and $\theta$ to be complex valued, I suppose it is then justified to write the general solution as $y(t) = C\sin(\omega t + \theta)$. So is the final form given in my notes just a preference thing?
2.) I cannot for the life of me find a reference which proves that all the predescribed solution method gives all possible solutions in the homogeneous case, i.e. I am missing a uniqueness proof. Do you happen to know some good reference for this or can you produce the said proof yourself? Thank you!
 A: You want to show that the range of both general solutions are the same. First let's look at the formula you found: $y_1(t)=\cos(\omega t)(C_1+C_2)+i\sin(\omega t)(C_1-C_2)$. To make this a little more readable we introduce two new constants.
$\eta_1=C_1+C_2,\ \eta_2=i(C_1-C_2)$ By rewriting we can obtain relations between $C_1, C_2$ and $\eta_1, \eta_2$:
$C_1=\eta_1-C_2,\ C_2=\frac{\eta_1+i\eta_2}{2}$ From this we can see that we can choose any $\eta_1, \eta_2 \in \mathbb{C}$ and obtain $C_1, C_2$ so that $y_1(t)=\eta_1\cos(\omega t)+\eta_2\sin(\omega t)$. In other words: $\eta_1, \eta_2$ span $\mathbb{C^2}$
Now we look at the solution from your lecture notes:
$y_2(t)=A\sin(\omega t + B)$
Using equality $\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)$ we rewrite the solution:
$y_2(t)=A\sin(\omega t)\cos(B)+A\cos(\omega t)\sin(B)$ Again we define two new variables:
$\xi_1=A\cos(B),\ \xi_2=A\sin(B)$. This would make the solution $y_2(t)=\xi_1\sin(\omega t) + \xi_2\cos(\omega t)$. Since we know that sine and cosine are linearly independent, in order for $y_1$ to have the same range as $y_2$ we must be able to find $A, B$ for any $\xi_1, \xi_2 \in \mathbb{C}$.
Let us derive formulas for $A,B$ in terms of $\xi_1, \xi_2$:
$A^2\cos^2(B)+A^2\sin^2(B)=A^2(\cos^2(B)+\sin^2(B))=A^2=\xi_1^2+\xi_2^2 \\
A=\pm\sqrt{\xi_1^2+\xi_2^2} \\
\xi_1=A\cos(B) \rightarrow B = \arccos(\frac{\xi_1}{A})$
In the formula for B, there is no solution when $A=0$, also $A=0 \Rightarrow y_2(t)=0$
We can choose $\xi_1=1,\ \xi_2=i$ to let $A=0$ and we now see that:
$y_2(t)=0\cdot \sin(\omega t + B) = 0 \neq \sin(\omega t)+i\cos(\omega t)$.
Since $\xi_1, \xi_2$ cannot assume values that result in $A=0$, $\xi_1, \xi_2$ do not span $\mathbb{C^2}$. We conclude that the general solution of your lecture notes does not cover all the solutions your own general solution does.
