How to Analytically Solve this PDE?

Fix an arbitrary value for $y$, say $y_0$. Then let $f(x) =u(x,y_0)$. You can immediately see that $f''+f=0$. This is very common ODE and most people just know that $\sin$ and $\cos$ are solutions, but if you're unhappy with that you can get there with $\exp(\omega x)$ solutions and some algebra.

Now we know that $f(x) = A \sin(x) + B \cos(x)$ for an arbitrary $y_0$. The only thing that can change when we change $y_0$ are the values of constants $A,B$.

This finally leads to conclusion $u(x,y) = A(y)\sin(x) + B(y)\cos(x)$ and the $A,B$ now functions of $y$ are determined by boundary conditions.

We want to solve $\frac{\partial^2}{\partial x^2} u(x,y)+u(x,y)=0.$ As you said, we can solve this like an ODE. Let us consider a guess $u(x,y)=c(y)e^{rx}.$ We make $c$ depend on $y$ because $c(y)$ is still constant in $x$, which is the variable that we're differentiating in. Plugging in such a guess, we get the characteristic equation $$r^2+1=0,$$ which has imaginary roots at $\pm i$, which yields the solution $$u(x,y)=c(y)\cos x+b(y)\sin x.$$ So, the mistakes in your interpretation were that your guess should depend on $y$ and that your guess was a perfectly valid one, but you must consider complex roots here.

There are a few facts that are always treated as "obvious" in ODEs (because, presumably, you've been thinking about them non-stop since Calc I):

• The derivative of a constant (with respect to any independent variable) is zero.
• The derivative of a polynomial (with respect to its variable) reduces the degree by one.
• The derivative of $f(x) = \mathrm{e}^{kx}$ with respect to $x$ is a constant multiple of $f$. (Precisely, $\dfrac{\mathrm{d}}{\mathrm{d}x} \mathrm{e}^{kx} = k \mathrm{e}^{kx}$.)
• The second derivatives of $g(x) = \sin kx$ and $h(x) = \cos kx$ with respect to $x$ are constant multiples of $g$ and $h$, respectively. (Precisely, $\dfrac{\mathrm{d}^2}{\mathrm{d}x^2} \sin kx = -k^2 \sin x$, and similarly for cosine.) Additionally, odd order derivatives of $g$ and $h$ swap them.

So, when you see "the second derivative of $y$ is the negative of $y$", you should be thinking "sine and cosine" pretty much immediately.

The bullet point about sine and cosine can be rolled into the one about exponentials, so you could have gotten there with your characteristic equation method, but you need to recall what exponentiation does to complex numbers. In particular, you recall $\mathrm{e}^{\mathrm{i}x} = \cos x + \mathrm{i} \sin x$. So what would happen with your characteristic equation is, from $$ y'' + y = 0 $$ you have the characteristic equation $$ x^2 + 1 = 0 \text{.} $$ Then the characteristic roots are $\pm \mathrm{i}$, so the solutions (with arbitrary constants $c_1$ and $c_2$) are $c_1 \mathrm{e}^{\mathrm{i} x}$ and $c_2 \mathrm{e}^{-\mathrm{i} x}$. These are sines and cosines in disguise: \begin{align*} \mathrm{e}^{\mathrm{i} x} &= \cos x + \mathrm{i} \sin x \\ \mathrm{e}^{-\mathrm{i} x} &= \cos x - \mathrm{i} \sin x \text{.} \end{align*}