It looks like the ODE is
[tex]y'+5y=\begin{cases}0&\text{for }0\le t<3\\1&\text{for }3\le t<5\\0&\text{for }5\le t<\infty\end{cases}[/tex]
with the initial condition of [tex]y(0)=4[/tex].
Rewrite the right side in terms of the unit step function,
[tex]u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t<c\end{cases}[/tex]
In this case, we have
[tex]\begin{cases}0&\text{for }0\le t<3\\1&\text{for }3\le t<5\\0&\text{for }5\le t<\infty\end{cases}=u(t-3)-u(t-5)[/tex]
The Laplace transform of the step function is easy to compute:
[tex]\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s[/tex]
So, taking the Laplace transform of both sides of the ODE, we get
[tex]sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s[/tex]
Solve for [tex]Y(s)[/tex]:
[tex](s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}[/tex]
We can split the first term into partial fractions:
[tex]\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs[/tex]
If [tex]s=0[/tex], then [tex]1=5a\implies a=\frac15[/tex].
If [tex]s=-5[/tex], then [tex]1=-5b\implies b=-\frac15[/tex].
[tex]\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}[/tex]
[tex]\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}[/tex]
Take the inverse transform of both sides, recalling that
[tex]Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)[/tex]
where [tex]F(s)[/tex] is the Laplace transform of the function [tex]f(t)[/tex]. We have
[tex]F(s)=\dfrac1s\implies f(t)=1[/tex]
[tex]F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}[/tex]
We then end up with
[tex]y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}[/tex]
