That is a simple case of division by 0 which normally creates a undefined number...
BUT you can use Calculus and a little Quantum theory to solve that equation aswell, but still proving it comes out 0=0...
here's one from some Quantum Physics
When we re-write a Hamiltonian using a Fock space and creation and annihilation operators, as in the previous example, the symbol N, which stands for the total number of particles, drops out. This means that the Hamiltonian is applicable to systems with any number of particles. Of course, in many common situations N is a physically important and perfectly well-defined quantity. For instance, if we are describing a gas of atoms sealed in a box, the number of atoms had better remain a constant at all times. This is certainly true for the above Hamiltonian. Viewing the Hamiltonian as the generator of time evolution, we see that whenever an annihilation operator ak destroys a particle during an infinitesimal time step, the creation operator a_k^\dagger to the left of it instantly puts it back. Therefore, if we start with a state of N non-interacting particles then we will always have N particles at a later time.
On the other hand, it is often useful to consider quantum states where the particle number is ill-defined, i.e. linear superpositions of vectors from the Fock space that possess different values of N. For instance, it may happen that our bosonic particles can be created or destroyed by interactions with a field of fermions. Denoting the fermionic creation and annihilation operators by c_k^\dagger and ck, we could add a "potential energy" term to our Hamiltonian such as:
