Use the Laplace Transform solve the IVP begin{cases}x'+y=t 4x+y'=0end{cases} x(0)=1 y(0)=2

Step 1
Theory :
The Laplace transform of a second order linear differential equation : $y"+\alpha y^{\prime }+\beta y=f(t)$ subject to initial conditions y(0)=A and y'(0)=B is given by,
$Y(s)=\frac{(s+\alpha )A+B}{(s^2+\alpha s+\beta )}+\frac{(F(s))}{(s^2+\alpha s+\beta )}$
Taking inverse laplace transform on both sides we get, $y(t)=L^{-1}[\frac{(s+\alpha )A+B}{(s^2+\alpha s+\beta )},t]+L^{-1}[\frac{F(s)}{(s^2+\alpha s+\beta )},t]$
Step 2
Given two simultaneous differential equations with ICs,
$x^{\prime }+y=t\text{ and }4x+y^{\prime }=0\text{ with }x(0)=1\text{ and }y(0)=2$
Taking Laplace transforms on both sides of each equation and using notation $Y=L[y,s]$ we obtain,
$sX-x(0)+Y=L[t,s]=\frac{1}{s^2}$
$sY-y(0)+X=L[0,s]=0$
Using initial conditions, $sX-1+Y=\frac{1}{s^2}$
$sY-2+X=0$
$\Rightarrow sX+Y=\frac{1}{s^2}\dots (1)$
$X+sY=2\dots (2)$
Solving equations (1) and (2) for X and Y we obtain,
$X=\frac{1-2s}{s(s^2-1)}\dots (3)$
$Y=\frac{2s}{s^2-1}-\frac{1}{s^2(s^2-1)}\dots (4)$
Using partial fractions method in (3) and (4) we further get the values of X and Y as,
$X=-\frac{1}{s}+\frac{s}{(s^2-1)}-\frac{2}{(s^2-1)}$
$Y=\frac{5}{4}\cdot \frac{1}{s}+\frac{s}{(s^2-1)}$
Using inverse laplace transform we get,
$x=-1+\cos ht-\sin ht$
$y=\frac{5}{4}+\cos ht$
The answer is $x=-1+\cos ht-\sin ht$
$y=\frac{5}{4}+\cos ht$