how to solve the ODE DON'T use laplace transform $u(t)$ is step function.
$$
\frac{d^2}{dt^2}y(t) +2\frac{d}{dt}y(t) + y(t) = \frac{du(t)}{dt} + u(t)
$$
I don't know how to deal with $u(t)$.
 A: Hint: Start by letting $$z(t)=\frac{d}{dt}y(t)+y(t).$$ How does this substitution let us rewrite the left-hand side? Now, gather the differential terms on one side and the non-differential terms on the other. Everything should fall readily into place.
A: Write the equation as
$$
                   (D+1)^{2}y = \frac{du}{dt}+u,
$$
where $D$ is the Heaviside differentiation operator, which seems appropriate because $u$ is the Heaviside step function (Heaviside first formulated the derivative of the step function as an impluse (delta) function, before Dirac.)
To finish, notice that
$$
          e^{t}(D+1)f=e^{t}f'+e^{t}f=(e^{t}f)'=D(e^{t}f).
$$
Therefore, multiplying both sides of the original equation by $e^{t}$ gives
$$
          e^{t}(D+1)^{2}y=e^{t}\frac{du}{dt}+e^{t}u, \\
              D^{2}(e^{t}y) = e^{t}\frac{du}{dt}+e^{t}u. \\
$$
Choose $a < 0$ and integrate over $[a,t]$. Following the comment I left you: If $[a,t)$ includes $0$, then the integral of $e^{t}\frac{du}{dt}$ is $e^{0}=1$ and it is $0$ otherwise. So, you obtain
$$
           D(e^{t}y)=D(e^{t}y)|_{t=a}+\int_{a}^{t}e^{s}\frac{du}{ds}ds+\int_{a}^{t}e^{s}u(s)\,ds. \\
     = e^{a}y'(a)+e^{a}y(a)+\left\{\begin{array}{lc}1+(e^{t}-1) & t > 0\\
                                   0 & t < 0\end{array}\right. \\
     = e^{a}y'(a)+e^{a}y(a)+u(t)e^{t}.
$$
Then integrate again over $[a,t]$:
$$
     e^{t}y(t)=e^{a}y(a)+\{e^{a}y'(a)+e^{a}y(a)\}(t-a)+u(t)(e^{t}-1).
$$
The final answer is
$$
       y(t) = e^{a}y(a)e^{-t}+\{e^{a}y'(a)+e^{a}y(a)\}(t-a)e^{-t}+u(t)(1-e^{-t})
$$
In other words, there are constants $C$ and $D$ such that
$$
         y(t) = \left\{\begin{array}{lc}
                    Ce^{-t}+Dte^{-t}+(1-e^{-t}) & t > 0 \\
                    Ce^{-t}+Dte^{-t} & t < 0
                      \end{array}\right.
$$
