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?

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.

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