The complete solution for the time dependent equation with an infinite potential step is found by the method of images. Given any initial wavefunction
$$ \psi_0(x) $$
for x<0, you write down the antisymmetric extension of the wavefunction
$$ \psi_0(x) = \psi_0(x) - \psi_0(-x) $$
And you solve the free Schrodinger equation. So any solution of the free Schrodinger equation gives a solution for the infinite potential step. This is not completely trivial to make, because the solutions do not vanish in any region. But, for example, the spreading delta-function
$$ \psi(x,t) = {1\over \sqrt{2\pi it}} e^{-(x-x_0)^2\over it } $$
Turns into the spreading, reflecting, delta function
$$ \psi(x,t) = {1\over \sqrt{2\pi it}} e^{-(x-x_0)^2\over it } - {1\over \sqrt{2\pi it}} e^{-(x+x_0)^2\over it } $$
You can do the same thing with the spreading Gaussian wavepacket, just subtract the solution translated to +x from the solution translated to -x. In this case, normalizing the wavefunction is hard when the wavefunction start out close to the reflection wall.
Time independent infinite potential wall
The solution to the time independent problem of the infinite potential wall are all wavefunctions of the form
$$ \sin(kx) $$
for all k>0. Superposing these solutions gives all antisymmetric functions on the real line.
To find this solution, note that the time independent problem (eigenvalue problem) for the Schrodinger equation is solved by sinusoidal waves of the form $e^{ikx}$, and you need to superposes these so that they are zero at the origin, to obey the reflection condition. This requires that you add two k-waves up with opposite signs of k and opposite sign coefficients.
The opposite sign of k just means that the wave bounces off the wall (so that k changes sign), while the opposite sign of the coefficient means that the phase is opposite upon reflection, so that the wave at the wall cancels.
General solution
The time dependent problem for a time independent potnetial is just the sum of the solutions to the time independent problem with coefficients that vary in time sinusoidally.
If the eigenfunctions $\psi_n$ are known, and their energies $E_n$ are known, and the potential doesn't change in time, then the,
$$ \psi(t) = \sum_n C_n e^{-iE_n t} \psi_n(x) $$
is the general solution of the time dependent problem. This is so well known that generally people don't bother saying they solved the time-dependent problem once they have solved the eigenvalue problem.
The general solution of the time-dependent Schrodinger equation for time dependent potentials doesn't reduce to an eigenvalue problem, so it is a different sort of thing. this is generally what people understand when you say solving the time-dependent equation, and this reflects the other answers you are getting. I don't think this was the intent of your question, you just wanted to know how to solve the time dependent equation for a time independent potential, in particular, for an infinite reflecting potential wall. This is just the bouncing solution described above.
Note: I have tried to make my answer a little more general, with detail, so that it will be useful for more people.
The question is what boundary conditions do we apply to our wavefunction either side of a Dirac delta function?
In your example we have the potential $$V(x)=\begin{cases}\infty &\text{ if } x < 0\\ \alpha~\delta(x-a) &\text{ if } x \geq 0 \end{cases}$$ We are interested in the boundary conditions either side of $x=a$. What information do we have? Well, due to the probabilistic interpretation of the wavefunction we require continuity of the wavefunction. That is, our first boundary condition is
$$(1)~~~~~~~~~~~~\boxed{\psi_{-}(a) = \psi_{+}(a)}$$
where the $\pm$ subscripts represent the right and left sides of $x=a$ respectively. What other conditions can we set? Well, usually we would ask that the first derivative is also matched either side of $x=a$ (you should be asking yourself why do we do this?), but in this case this is not the right condition to impose. Let's see why.
Where does the boundary condition on $\frac{\partial \psi}{\partial x}$ come from?
Our wavefunction is a solution to the 1-D time-independent Schrödinger Equation:
$$H~\psi(x) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)+V(x)\psi(x) = E~\psi(x)$$
Looking at this equation, we see that we can get a boundary condition on $\frac{\partial \psi}{\partial x}$ at any point $a$ by integrating it w.r.t $x$ over the region $[a-\epsilon,a+\epsilon]$, taking $\epsilon\rightarrow 0$:
$$\lim_{\epsilon\rightarrow 0}\left[-\int^{a+\epsilon}_{a-\epsilon}dx\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)+\int^{a+\epsilon}_{a-\epsilon}dx~V(x)\psi(x)\right] = E~\lim_{\epsilon\rightarrow 0}~\int^{a+\epsilon}_{a-\epsilon}dx~\psi(x)\\
\implies -\frac{\hbar^2}{2m}\left[\frac{\partial \psi_{+}(a)}{\partial x}-\frac{\partial \psi_{-}(a)}{\partial x}\right] + \lim_{\epsilon\rightarrow 0}\int^{a+\epsilon}_{a-\epsilon}dx~V(x)\psi(x) = 0$$
where we have used the continuity of $\psi$ to evaluate the RHS as zero. Note that for any $V(x)$ the second term doesn't necessarily vanish. But, if $V(x)$ is continuous, this term will vanish for the same reason the RHS did, and in those cases we yield the usual boundary condition $$\boxed{\frac{\partial \psi_{+}(a)}{\partial x}=\frac{\partial \psi_{-}(a)}{\partial x}}~~~~~~~(\mbox{when $V(a)$ is continuous at $x=a$})$$
We note that the for example given here, $V(x)$ is definitely not continuous at $x=a$, where it diverges to infinity. So this second term doesn't vanish, in fact
$$\lim_{\epsilon\rightarrow 0}\int^{a+\epsilon}_{a-\epsilon}dx~V(x)\psi(x) = \alpha\lim_{\epsilon\rightarrow 0}\int^{a+\epsilon}_{a-\epsilon}dx~\delta(x-a)\psi(x) = \alpha~\lim_{\epsilon\rightarrow 0}~\psi(a) = \alpha~\psi(a)$$
so rearranging our results, the second boundary condition for the problem is
$$(2)~~~~~~~~~~~~\boxed{\frac{\partial \psi_{+}(a)}{\partial x}-\frac{\partial \psi_{-}(a)}{\partial x} = \frac{2m\alpha}{\hbar^2}\psi(a)}$$
Explicitly using the form of your wavefunctions (using $B=-A$ to eliminate $B$)
$$\psi_{+}(x) = Fe^{ikx},~~~~~~\psi_{-}(x) = A\left(e^{ikx}-e^{-ikx}\right)$$
this boundary condition becomes
$$\boxed{ikFe^{ika}-ik\left(Ae^{ika}+e^{-ika}\right) = \frac{2m\alpha}{\hbar^2}\psi(a)}$$
as required.
Best Answer
Solution using a basis set expansion
Let $H_0|n\rangle=E_n|n\rangle$ and lets make the following ansatz for our time dependent state, $$ |\psi(t)\rangle = \sum_n c_n(t) \exp\left(-\frac{i}{\hbar }E_nt\right)|n\rangle . $$ Inserting this into the time dependent Schrödinger equation $i\hbar \frac{d}{dt}|\psi(t)\rangle = H(t)|\psi(t)\rangle$ with $H(t)=H_0 +\delta(t)V$ leads after cancelation of some terms to
$$ i\hbar \sum_n \dot c_n(t) \exp\left(-\frac{i}{\hbar }E_nt\right)|n\rangle = \sum_l c_l(t) \exp\left(-\frac{i}{\hbar }E_lt\right)V\delta (t)|l\rangle $$ Projection unto state $\langle k|$ yields with the definition $V_{kl}= \langle k|V|l\rangle$ and using the orthonormality of states $\langle k|n\rangle = \delta_{kn} $ $$ i\hbar \dot c_k(t) \exp\left(-\frac{i}{\hbar }E_kt\right)= \sum_l V_{kl} c_l(t) \exp\left(-\frac{i}{\hbar }E_lt\right)\delta(t) $$ Now we multiply with $$-\frac{i}{\hbar}\exp\left(\frac{i}{\hbar }E_kt\right) $$ and define $\tilde V_{kl}(t)=V_{kl}\exp\left(-\frac{i}{\hbar }(E_l-E_k)t\right)$ which leads to $$ \dot c_k = -\frac{i}{\hbar } \sum_l \tilde V_{kl}(t) c_l(t)\delta(t) $$ This can be written as vector equation $$ \mathbf{ \dot c} = -\frac{i}{\hbar } \mathbf{\tilde V(t)c(t)}\delta(t) $$ We can now integrate the equation, $$\begin{aligned} \int_0^t dt' \mathbf{ \dot c} &= -\frac{i}{\hbar } \int^t_0 dt' \mathbf{\tilde V(t')c(t')}\delta(t')\\ \mathbf{c(t)} - \mathbf{c(0)} &= -\frac{i}{\hbar } \mathbf{\tilde V(0)c(0)}\\ \mathbf{c(t)} &= \left(\mathbf 1 -\frac{i}{\hbar } \mathbf{\tilde V(0)}\right )\mathbf{c(0)} \end{aligned}$$ This equations holds for $t \geq 0$. If we assume that our system propagates freely before $t=0$, we can introduce the Heaviside theta function to obtain $$ \mathbf{c}(t) = \left(\mathbf 1 -\frac{i}{\hbar } \mathbf{\tilde V(0)}\theta(t)\right )\mathbf{c(0)} $$
Solution without basis set expansion
We could also do the same without expansion in an explicit basis(which is just a bad habit of mine due to my Comp. Chem. background ...)
For that define the state in the interaction picture as $$\begin{aligned} |\psi(t)\rangle = \exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle \\ |\psi(0)\rangle = |\psi_I(0)\rangle \end{aligned}$$
Now insert this into the TDSE, $$\begin{aligned} i\hbar \frac{d}{dt}|\psi(t)\rangle &= (H_0 + V\delta(t) )|\psi(t)\rangle \\ i\hbar \frac{d}{dt} \left(\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle \right) &= (H_0 + V\delta(t) )\left(\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle \right) \\ i\hbar \left( -\frac{i}{\hbar}H_0 \exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle +\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\frac{d}{dt}\psi_I(t)\rangle \right) &= (H_0 + V\delta(t) )\left(\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle \right)\\ H_0\left(\exp\left({-\frac{i}{\hbar}}H_0 t\right)|\psi_I(t)\rangle \right)+i\hbar\left(\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\frac{d}{dt}\psi_I(t)\rangle \right) &=(H_0 + V\delta(t) )\left(\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle \right)\\ \text{Cancel }H_0 \text{ term on both sides}\\ i\hbar\left(\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\frac{d}{dt}\psi_I(t)\rangle \right) &= V\delta(t)\left(\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle \right)\\ \text{Multiply with }\exp\left(\frac{i}{\hbar}H_0 t\right)\text{ from the left}\\ i\hbar |\frac{d}{dt}\psi_I(t)\rangle &= \exp\left(\frac{i}{\hbar}H_0 t\right)V\delta(t)\exp\left({-\frac{i}{\hbar}}H_0 t\right) |\psi_I(t)\rangle \\ \int^t_0 dt' |\frac{d}{dt'}\psi_I(t')\rangle & =-\frac{i}{\hbar} \int^t_0 dt' \exp\left(\frac{i}{\hbar}H_0 t'\right)V\delta(t')\exp\left({-\frac{i}{\hbar}}H_0 t'\right) |\psi_I(t')\rangle \\ |\psi_I(t)\rangle - |\psi_I(0)\rangle &= -\frac{i}{\hbar} \exp\left(\frac{i}{\hbar}H_0 \cdot 0\right)V\exp\left({-\frac{i}{\hbar}}H_0 \cdot 0\right) |\psi_I(0)\rangle \\ |\psi_I(t)\rangle &= |\psi_I(0)\rangle -\frac{i}{\hbar}V|\psi_I(0)\rangle\\ |\psi_I(t)\rangle &= (\mathbf{1} -\frac{i}{\hbar}V)|\psi_I(0)\rangle\\ \exp\left({\frac{i}{\hbar}}H_0 t\right) |\psi(t)\rangle &= (\mathbf{1} -\frac{i}{\hbar}V)|\psi(0)\rangle\\ |\psi(t)\rangle &=\exp\left(-{\frac{i}{\hbar}}H_0 t\right)(\mathbf{1} -\frac{i}{\hbar}V)|\psi(0)\rangle\\ \end{aligned}$$
This is equivalent to my first calculation. The line $$ \mathbf{c}(t) = \left(\mathbf 1 -\frac{i}{\hbar } \mathbf{\tilde V(0)}\right )\mathbf{c(0)} $$ is equivalent to $$ |\psi_I(t)\rangle = (\mathbf{1} -\frac{i}{\hbar}V)|\psi_I(0)\rangle\\ $$