Linear Constant Coefficient Differential Equations: Zero-Input and Zero-State responses
The two approaches should return the same solution. They are just two different ways to get to the same place.

In the ZSR/ZIR method, you are solving two different IVPs - they have the same expression on the left-side of the equality, but different initial conditions and different right-hand sides (one has a $0$, the another one has some function of $t$). In the more mathematical non-engineer like approach (homogeneous/particular), you solve just one IVP with both things: the right-hand side of the equation not set to $0$ and the original initial conditions. Thus one can intuitively see the second approach as some kind of superposition of the two IVPs proposed in the first one.

Just in case you don't believe this yet, let me just write down an example (without doing it in detail as that's not the point of this post) to show that both methods work. Let's solve the following differential equation:

$$\left\{
 \begin{array}{ll}
  y''(t)+4y'(t)+3y(t)=e^{-2t} \\
  y(0)=1 \ , \ y'(0)=1
 \end{array}
\right.$$

The mathematical approach ($y_p+y_h$)

I propose:

$$y_p(t)=-e^{-2t}$$

Check that it satisfies the differential equation and the initial conditions:

$$y_p''(t)+4y_p'(t)+3y_p(t)=-4e^{-2t}+8e^{-2t}-3e^{-2t}=e^{-2t} \ \implies\mathrm{OK}$$
$$y_p(0)=-e^0=1\ \implies\mathrm{OK}$$
$$y_p'(0)=-2e^0=1\ \implies\mathrm{OK}$$

Therefore we conclude that $y_p(t)$ is a valid particular solution. Then we should solve the homogeneous equation. I'm not going to do it step by step, but you can check thay you get something of the form $$y_h(t)=Ae^{-t}+Be^{-3t}$$

If we impose the initial conditions to $y_p(t)+y_h(t)$, the constants turn out to be $A=5/2$ and $B=-1/2$. So the solution can be written as:

$$y(t)=y_p(t)+y_h(t)=\frac52 e^{-t}-e^{-2t}-\frac12 e^{-3t}$$

The engineer approach ($y_{zs}+y_{zi}$)

As stated at the beginning, this approach requires us to solve two different IVPs. The first one is:

$$\left\{
 \begin{array}{ll}
  y_{zi}''(t)+4y_{zi}'(t)+3y_{zi}(t)=0 \\
  y_{zi}(0)=1 \ , \ y_{zi}'(0)=1
 \end{array}
\right.$$

The solution (the generic one, with the constants $A$ and $B$) to this problem is the same as in the previous case, but with different constants (now we only have the homogeneous solution term). In this case, $A=2$ and $B=-1$. Therefore:

$$y_{zi}(t)=2e^{-t}-e^{-3t}$$

Now we have the other IVP:

$$\left\{
 \begin{array}{ll}
  y_{zs}''(t)+4y_{zs}'(t)+3y_{zs}(t)=e^{-2t} \\
  y_{zs}(0)=0 \ , \ y_{zs}'(0)=0
 \end{array}
\right.$$

Note that his is exactly the same IVP as the one stated at the beginning, but with different initial conditions. So we can use the homogeneous/particular approach to solve this IVP as well, and find the constants $A$ and $B$ that suit the initial conditions in this case. We find that

$$y_{zs}(t)=\frac12 e^{-t}-2e^{-2t}+\frac12 e^{-3t}$$

If we sum the two solutions:

$$y(t)=y_{zs}(t)+y_{zi}(t)=\frac52 e^{-t}-e^{-2t}-\frac12 e^{-3t}$$

We got the same result.

Conclusions

As can be seen from the example, both methods arrive to the same solution because they are just two different ways of decomposing the problem into two easier ones, but they are equivalent. 

The approach using ZSR and ZIR is more practical in real life. Take for example a circuit. We usually know the initial conditions with no input. When we turn the input on (i.e. we make a sudden change in the circuit), voltages and currents could change abruptly. The ZIR uses these intitial conditions with no input, and the ZSR uses the input when we know all initial conditions are zero (for example, we discharge all capacitors before turning the circuit on). Performing the mathematical approach would not be simple, as we would have to know the initial conditions just when the input appeared, and that can be rather hard to measure or observe. With ZSR/ZIR, we can measure the two things independently and then add them up to get the total solution.

Regarding the steady state question, note that when you solve the differential equation you get the signal $y(t)$. Namely, you can evaluate the output at any instant $t$ you want. Thus, you don't have just the transient response or the steady state one: you have both. You can observe whatever you want as you have the signal as a function of time. The steady state will be reached at $t\to\infty$ (in real life, after some time, when all components in the circuit are acting as the input had been turned on for a long time).