Use Laplace transforms to solve the initial value problems.yy′=δ(t)−δ(t−3),y(0)=1,y′(0)=0

$L\{\delta (t-a)\}=e^{-as}$
$L^{-1}\left\{\frac{1}{s-a}\right\}=e^{at}$
$L^{-1}\left\{\frac{e^{-as}}{s}\right\}=u(t-a)$
Given
${\displaystyle y{y}^{\prime }=\delta \left(t\right)-\delta (t-3),y\left(0\right)=1,y^{\prime }\left(0\right)=0}$
Taking laplace transformation both sides:
$L\left\{y"\right\}+L\left\{y^{\prime }\right\}=L\{\delta (t)\}-L\{\delta (t-3)\}$
${\displaystyle [s^{2}Y\left(s\right)-sy\left(0\right)-y^{\prime }\left(0\right)]+[sy\left(s\right)-y\left(0\right)]=1-e^{-3s}}$
${\displaystyle (s^{2}y\left(s\right)-s)+(sy\left(s\right)-1)=1-e^{-3s}}$
${\displaystyle y\left(s\right)[s^2+s]-s-1=1-e^{-3s}}$
${\displaystyle y\left(s\right)[s^2+s]=2-e^{-3s}+s}$
${\displaystyle y\left(s\right)=\frac{2+s-e^{-3s}}{s(s+1)}}$
${\displaystyle y\left(s\right)=\frac{2+s}{s(s+1)}-\frac{e^{-3s}}{s(s+1)}}$
${\displaystyle y\left(s\right)=\frac{2}{s}-\frac{1}{s+1}-\frac{e^{-3s}}{s}+\frac{e^{-3s}}{s+1}}$
Now taking inverse laplace transformation both sides:
$L^{-1}\{y(s)\}=2L^{-1}\left\{\frac{1}{s}\right\}-L^{-1}\left\{\frac{1}{s+1}\right\}-L^{-1}\left\{\frac{e^{-3s}}{s}\right\}+L^{-1}\left\{\frac{e^{-3s}}{s+1}\right\}$
${\displaystyle y\left(t\right)=2\times 1-e^{-t}-u(t-3)+e^{3-t}u(t-3)}$