How to calculate step response for $y''(t) - y(t) = x'(t) - x(t)$ in time domain?
It is indeed possible to convert the system to an equivalent first order system, but let's first directly solve the given problem. The system's step response satisfies

$$y''(t)-y(t)=\delta(t)-u(t)\tag{1}$$

This implies that we can compute the step response of the given system by considering another system

$$y''(t)-y(t)=x(t)\tag{2}$$

and compute its impulse response ($x(t)=\delta(t)$) and its step response ($x(t)=u(t)$), and obtain the step response $(1)$ as the difference of the impulse response and the step response of the new system $(2)$.

For the step response we need to solve

$$y''(t)-y(t)=u(t)\tag{3}$$

Assuming the system is initially at rest, we get initial conditions $y(0)=y'(0)=0$. The homogeneous solution of $(3)$ is $y_h(t)=Ae^{t}+Be^{-t}$, $t&gt;0$, whereas a particular solution is $y_p(t)=-1$, $t&gt;0$. The complete solution is then

$$y(t)=y_h(t)+y_p(t)=Ae^{t}+Be^{-t}-1,\quad t&gt;0\tag{4}$$

The constants $A$ and $B$ are determined by the initial conditions:

$$\left.\begin{align}y(0)&amp;=A+B-1=0\\y'(0)&amp;=A-B=0\end{align}\right\}\Rightarrow A=B=\frac12\tag{5}$$

Consequently, the step response of system $(2)$ is

$$y(t)=\frac12\left(e^t+e^{-t}\right)-1,\quad t&gt;0\tag{6}$$

Its impulse response satisfies

$$y''(t)-y(t)=\delta(t)\tag{7}$$

and it can be obtained by differentiating the step response $(6)$:

$$y(t)=\frac12\left(e^t-e^{-t}\right),\quad t&gt;0\tag{8}$$

Finally, according to $(1)$ and $(2)$, the step response of the original system is the difference between the impulse response and the step response given in $(8)$ and $(6)$:

$$y(t)=\frac12\left(e^t-e^{-t}\right)-\frac12\left(e^t+e^{-t}\right)+1=1-e^{-t},\quad t&gt;0\tag{9}$$


Another approach would be to see that the original system can also be represented by a first-order system. Note that

$$y''(t)-y(t)=(y(t)+y'(t))'-(y(t)+y'(t))\tag{10}$$

So if $y''(t)-y(t)=x'(t)-x(t)$, it follows from $(10)$ that

$$y(t)+y'(t)=x(t)\tag{11}$$

With $x(t)=u(t)$ you obtain from $(11)$ the same step response as in $(9)$:

$$\begin{align}y_h(t)&amp;=Ce^{-t}, &amp;t&gt;0\\
y_p(t)&amp;=1, &amp;t&gt;0
\end{align}\\y(t)=y_h(t)+y_p(t)=Ce^{-t}+1,\quad t&gt;0$$

From $y(0)=0$ we get $C=-1$, and the step response becomes

$$y(t)=1-e^{-t},\quad t&gt;0$$

just like in $(9)$.


As a final check it is useful to see what happens when you solve the problem using the Laplace transform. The original difference equation transforms to

$$Y(s)(s^2-1)=X(s)(s-1)$$

which gives for the transfer function

$$H(s)=\frac{s-1}{s^2-1}=\frac{1}{s+1}$$

This shows that you have a pole-zero cancellation, and the system is actually a first-order system, just as described by Eq. $(11)$. For $x(t)=u(t)$ you get $X(s)=1/s$, and, consequently, for the Laplace transform of the step response

$$Y(s)=\frac{1}{s}\frac{1}{s+1}=\frac{1}{s}-\frac{1}{s+1}$$

which corresponds to the time domain function

$$y(t)=u(t)-e^{-t}u(t)=(1-e^{-t})u(t)$$

which is the same as $(9)$.