Well there are many questions in your question and I'll try to address them one by one.
As you may know,
$L* di/dt = V (t)$
one important thing to note about inductors is that unless there is an impulsive excitation(that is, the input is bounded), it's current can not make instantaneous jumps.
Therefore, when you apply the voltage, it will still have a 0 current, it will have a high slope, but 0 current nonetheless at $0^+$ where $0^+$ denotes the instant right after 0.
Ok, so using $L* di/dt = V (t)$ and basic first order circuit analysis coupled with a bit of differential equations, you get the current equation:
$i(t) = V/R *(1 - e^ {-tR/L})$
R is the total thevenin resistance in the circuit. This includes the internal resistance of the inductor. This is the Zero state response of the inductor and it does not take into account any form of initial condition, however it is not hard to implement that into the formula. As you can see, the current will only reach its final value as t -> infinity.
If you assume the inductor circuit has zero resistance, it would be a short circuit between the terminals of the voltage source. If you assume there is an output resistance of voltage source, you can separate it from the voltage source and use in the calculation as R.
Change in the magnetic flux is basically the self induced EMF. The differential equation basically points out to the fact that most things in the inductor occur in an exponential decay kind-of fashion. I hope this answers at least some of your questions. Let me know in the comments if anything is not clear.
Edit: Just to be clear * is multiplication in my answer, not convolution or something like that.
When we have a DC voltage source with a switch in series with RL and the switch is closed at t=0 then it is said that current is zero initially, but the voltage across inductor is same as that of applied voltage( according to kirchhoff voltage law) so there should be current( according to v=L(di/dt) )but it contradicts the initial statement so how do I understand this?
You are right that right when we close the switch the voltage across the inductor is equal to the applied voltage. However, you are misinterpreting what a potential difference of magnitude $v=L\cdot\text di/\text dt$ means. This equation doesn't say if there is a potential difference across the inductor then there is current through the inductor. What it says is that a potential difference across the inductor is associated with a change in current through the inductor. Therefore, since the voltage across the inductor is non-zero at $t=0$, we know the current is changing at $t=0$.
...but addition of resistor makes the current increase exponential , how to understand this intuitively (I understand from the equations but not theoretically how it is happening)?
The current increases like
$$i=i_0\left(1-e^{-t/\tau}\right)$$
So it is increasing, and there is an exponential function, but usually "increasing exponentially" means it keeps growing and growing more rapidly without bound. This is not what is happening here.
As the current in the circuit increases the voltage across the resistor increases. Therefore, the voltage across the inductor decreases. Based on our previous discussion, this means that the change in current must be decreasing. Hence this "voltage trade-off" happens at a slower and slower rate. This causes the current to approach a steady value where the increase over time decays exponentially.
I understand that changing current causes the induced EMF which opposes the changing current, but what I don't understand is - won't it cause the current to be constant...
Keep in mind that "oppose" does not mean "block".
Everything else...
It seems like your confusion stems from what we discussed initially. You are mixing up the current and its derivative. The voltage across the inductor tells you nothing about the current in general. It tells you how the current is changing.
Also, you say that you understand things from the equations, but I would argue that if you don't understand how the equations model reality then you haven't truly understood the equations. It would help for you to look at how the equations are derived. Make sure you understand the physical significance and motivation for each step, each equation, etc. This is an important step in the learning process, so I will leave that job to you.
I hope this answer is a good scaffold to hold up the deeper understanding you will develop here.
Best Answer
Lenz's law states that the induced current is in such a direction as to oppose the change producing it.
Back emf and a complete "conducting" circuit will result in an induced current.
The back emf can never exactly equal the applied voltage as then the current would be zero and not changing which would mean that there cannot be an back emf.
So you can think of it as follows.
As soon as a current starts to flow an emf is induced which produces an induced current which tries to oppose that change of current in the circuit produced by the applied voltage.
That induced current slows down the rate at which the current in the circuit increases.
So when deriving equations relating current to time in such a circuit it is convenient to say that at time = 0, when the switch is closed, the current is zero because the applied voltage and the back emf are equal in magnitude.
What you are usually not concerned with is a time scale equal to that whilst the switch is being closed.
Here is an attempt to illustrate the complexity of what happens even as the switch is being closed.
The switch is a capacitor and when open has charges stored on it.
As the switch is being closed the capacitance of the switch increases and so a very small current starts to flow around the circuit as the capacitor charges up.
The change of current is opposed by the induced current produced by the back emf.
As the distance between the contact of the switch continues to get less, the capacitance of the switch continues to increase and a changing current continues to flow.
When the contacts finally close there is already a very small but changing current flowing around the circuit which is being opposed by the induced current produced by the back emf with the back emf slightly less than the applied voltage.
What actually happens during the switch closing process is complicated by the fact that now you have an inductor, resistor and capacitor in the circuit and also that lumped circuit element analysis might be inappropriate to use over such a small time scale?.