Vibrating Movement, Applications of Second Order Differential Equations An object weighing 16 pounds is suspended from a vertical spring with a fixed end and constant $k=4 
 lb/ft$. The object is moving in a medium that offers a resistance in pounds of $3$ times the instantaneous speed in feet per second. Being in the equilibrium position, it is propelled with an upward velocity of $2 ft/s$, in addition, an external force acts on it, given in pounds by $f(t)$, whose graphical representation is shown in figure 1. Determine the position and speed of the object as a function of time $t$. Consider the acceleration of gravity as $32 ft / s^2$.

I did the following
For $f(t)$ the external force
$f(t) =0$ if $0 \leq t < 1$
$f(t) =4$ if $1 \leq t$
So
$\dfrac{16}{32} x'' +3x'+4x=4 $
$\implies x(t) =C_1e^{-2t} +C_2e^{-4t} +1$
The question is,  what do I do with the conditions x(0) =0 and x(0)= 2 ft/s?
Considering that they only apply for the $0 \leq t <1$
 A: Laplace Transforms are of course the standard way to solve this. If you want to avoid them for whatever reason there's nothing stopping you from treating the solution as piecewise (which is exactly what the Heaviside function in the Laplace Transform formulation does).
Initially we have the ODE
$$ \dfrac{1}{2} x_1'' +3x_1'+4x_1=0\: \: \: \: \: \text{ when } \: \: 0\leq t <1$$
which solves as
$$ x_1(t) =C_1e^{-2t} +C_2e^{-4t}\text{ .} $$
Given the conditions $$\begin{align} x_1(0)&=0\\x_1'(0)&=2 \end{align}$$
we simply plug them into $x_1(t)$ and $x_1'(t)$ to obtain
$$\begin{align} C_1+C_2&=0\\-2C_1-4C_2 - 2&=0 \end{align}$$
which gives us that
$$\begin{align} C_1&=1\\C_2&=-1 \\ \\ x_1(t) &=e^{-2t} -e^{-4t} \: \: \: \: \: \text{ when } \: \: 0\leq t <1\text{ .}\end{align}$$
If we demand continuity of $x$ and $x'$ then we compute the new boundary conditions as
$$\begin{align} x_2(1) = x_1(1) &= e^{-2} - e^{-4} \\\\ x'_2(1) = x_1'(1) &= -2e^{-2} + 4e^{-4} \end{align}$$
and solve the "new" ODE
$$ \dfrac{1}{2} x_2'' +3x_2'+4x_2=4\: \: \: \: \: \text{ when } \: \: 1\leq t $$
using the above boundary conditions. Now, the solution for $x_2$ is
$$x(t) =C_3e^{-2t} +C_4e^{-4t} +1$$
and the boundary conditions give us
$$ \begin{align}  &C_3e^{-2} +C_4e^{-4} +1 = e^{-2} - e^{-4} \\\\ &-2C_3e^{-2} -4C_4e^{-4} = -2e^{-2} + 4e^{-4} \\\\\\ \implies&\\ &C_3 = 1-2e^2 \\&C_4 = -1+e^{4}  \text{ .} \end{align}$$
Plugging things in to $x_2$, we have
$$ x_2(t) =(1-2e^2)e^{-2t} +(-1+e^{4})e^{-4t} + 1 \: \: \: \: \: \text{ when } \: \: 0\leq t <1\text{ .} $$
Putting the solutions together, we have
$$x(t) =
\begin{cases}
e^{-2t} -e^{-4t},  & \text{if $0\leq t<1$} \\
(1-2e^2)e^{-2t} +(-1+e^{4})e^{-4t} + 1, & \text{if $1\leq t$}
\end{cases}$$
A: One usually solves an initial value differential equation with step functions involved using Laplace transforms.
$$x" + 6x' + 8x = 8\cdot u(t-1)$$
$$\mathscr{L}\{x\} = X$$
$$\mathscr{L}\{x'\} = sX-x(0)$$
$$\mathscr{L}\{x"\} = s^2X - sx(0) - x'(0)$$
$$\mathscr{L}\{u(t-1)\} = \frac{e^{-1s}}{s}$$
giving
$$s^2X-0-2 + 6(sX-0) + 8X = \frac{e^{-s}}{s}$$
which we solve algebraically for $X$ and then look up the inverse Laplace Transform to get the answer.
