If you don't want the simplifications, you have to fall back to the general model of a diode:
$$
I = I_o \left( e^{\frac{eV}{nkT}}-1 \right)
$$
This equation relates the diode current to the diode voltage (it's V-I characteristic)
- Io - is the diode reverse saturation current
- k - Boltzmann's constant = 1.38e-23 Joules per Kelvin
- T - Analysis temperature (Kelvin)
- e - Magnitude of electric charge
- n - Ideality factor (for silicon diodes, n=2 for small currents and approaches n=1 for large currents; in theory should always = 1)
You can now solve your circuit via the system of equations that it produces. Although you now have continuous V-I functions to describe your elements, a closed-form solution is not always guaranteed to exist.
It is often necessary to use an iterative solution technique such as Newton-Raphson to approximate/approach the answer. This is what SPICE solvers do in the general case... and why they ask you for initial conditions (which can dramatically speed up the solution time).
I think the easiest method to solve such problems is to assume that the diodes are off (both, and then one of the two), compute the voltages across the diodes and see if there's a contradiction with your assumption. Let's call the top left diode \$D_1\$ and the diode in the middle \$D_2\$.
Case 1: \$D_1\$ off, \$D_2\$ off: Since \$D_1\$ is off there is no current through the top 5k resistor, and since \$D_2\$ is off, there is also no current through the bottom left 10k resistor. So \$V=0\$ and the voltage at the anode of \$D_1\$ is 15 Volts. Contradiction! (\$D_1\$ should be on).
Case 2: \$D_1\$ off, \$D_2\$ on: again no current through top 5k resistor. Voltage \$V\$ is
$$V=\frac{15V\cdot(5k||10k)}{10k+(5k||10k)}=3.75V$$
Contradiction! (Because the voltage across \$D_1\$ would be \$15V-3.75V=11.25V\$ and it should be on.)
Case 3: \$D_1\$ on, \$D_2\$ off: Voltage \$V\$ is
$$V=\frac{15V\cdot 10k}{5k+10k}=10V$$
The voltage at the anode of \$D_2\$ is \$15V\cdot 5k/15k=5V\$. This agrees with our assumption, because with these voltages \$D_2\$ must be off. So your solution is
$$I=0A,\quad V=10V$$
Best Answer
They are both just two different ways of saying that you don't need to worry about the variation of voltage across the diode as a function of the current through that diode. Whether or not you assume \$V_D=0\:\textrm{V}\$ or \$V_D=700\:\textrm{mV}\$, doesn't really matter. (One might be said to be "more ideal" than the other, but that's moot.)
So in one case the author is saying that the diode is not ideal in the sense that \$V_D=700\:\textrm{mV}\$ but implying that it is still ideal in the sense that \$V_D\$ doesn't vary with current (implied.) The other author is taking note of the fact that \$V_D\$ doesn't vary with current (now made explicit), so it is "ideal" in that sense, while taking note that they want you to use \$V_D=700\:\textrm{mV}\$, anyway.