Finding solution of ODE using Laplace transform Solve the following ODE with initial condition $y(0)=0$.
$y'+2y=f(x)$ where $f(x)=0$ if $x>1$ and $f(x)=1$ if $0\leq x\leq 1$.
I applied Laplace transform on both sides which gets
$sY(s)+2Y(s)=\frac{1-e^{-s}}{s}$
$\Rightarrow Y(s)=\frac{1-e^{-s}}{s(s+2)}$
$=\frac{1}{s(s+2)}-\frac{e^{-s}}{s(s+2)}$
Now I know inverse Laplace transform of $\frac{1}{s(s+2)}$ but how to find inverse Laplace transform of second term?
Or can we solve above ODE using different method?
 A: The Laplace transform is a little bit of an overkill for tis equation. Variation of parameters does the trick:
the general solution of your equation is:
$$
y(x) = c\cdot e^{-2 x} + e^{-2 x} \int_0^x e^{2 s} f(s) ds
$$
Using your intial data and $f$ (it is not clearly defined) we end up with
\begin{align}
y(x) &= e^{-2 x} \int_0^{\min(x,1)} e^{2 s} ds\\
&= e^{-2 x} \frac{e^{2\min(x,1)}-1}{2}\\
&= \frac{e^{2\min(0,1-x)}-e^{-2 x}}{2}
\end{align}

That being said of course you equation can be solved with Laplace transform. You could handle the second term with a partial fraction decomposition and than use a table of standard transformations.
A: Assuming you still want to solve the problem using the method of Laplace transform. From your work, the Laplace transform $Y(s)$ of $y(t)$ satisfies
$$ Y(s) = \frac{1 - e^{-s}}{s(s + 2)} = \frac{1}{s(s + 2)} - \frac{e^{-s}}{s(s + 2)}. $$
You can check that
$$ \frac{1}{s(s + 2)} = \frac{1}{2s} - \frac{1}{2(s + 2)}. $$
To transform the second term, you would need to use the so-called Translation on the $t$-axis property: If the inverse Laplace transform of $F(s)$ is $f(t)$, then
$$ \mathcal{L}^{-1}\left\{F(s)e^{-cs}\right\} = f(t - c)H(t - c), $$
where $H(t - c)$ is the shifted Heaviside function. You can check that $y(t)$ is given by

 $$ y(t) = \frac{1}{2} - \frac{1}{2}e^{-2t} - H(t -1)\left[\frac{1}{2} - \frac{1}{2}e^{-2(t - 1)}\right]. $$

