Finding the Laplace transform (using the Table of Laplace Transforms) of each differential equation yields:

First equation: `(dx)/(dt)+x+4y=10`

Take Laplace Transform:

`sX-x(0)+X+4Y=10/s`

`sX-4+X+4Y=10/s`

`(s+1)X+4Y=10/s+4\ \ \ ...(1)`

Second equation: `x-(dy)/(dt)-y=0`

Take Laplace Transform:

`X-(sY-y(0))-Y=0`

`X-sY+3-Y=0`

`X-(s+1)Y=-3\ \ \ ...(2)`

We now have 2 equations in 3 unknowns (X, Y and s). We need to solve for X and Y and leave s on the right hand side of each expression. We choose to solve for Y first.

Equation `(2) × (s + 1)`:

`(s+1)X-(s+1)^2Y` `=-3(s+1)\ \ \ ...(3)`

`(1) − (3)` gives:

`(4+(s+1)^2)Y`

`=10/s+4+3s+3`

`=10/s+3s+7`

`=(3s^2+7s+10)/s`

Solve for `Y`:

`Y=(3s^2+7s+10)/(s(4+(s+1)^2))`

`=(3s^2+7s+10)/(s(s^2+2s+5))`

`=A/s+(Bs+c)/(s^2+2s+5)`

The last step above is using partial fractions.

`3s^2+7s+10` `=A(s^2+2s+5)+s(Bs+C)`

Comparing terms gives us:

s2 terms: A + B = 3

s terms: 2A + C = 7

Constant terms: 5A = 10

So `A=2, B=1, C=3`

We can now write Y as the sum of partial fractions:

`Y=2/s+(s+3)/(s^2+2s+5)`

`=2/s+(s+1)/((s+1)^2+2^2)+2/((s+1)^2+2^2)`

Finding the inverse Laplace of our expression for Y gives:

`y=2+e^(-t)cos\ 2t+e^(-t)sin\ 2t`

We now need to find the expression for x. Equation (2) gives us:

`X=-3+(s+1)Y`

and since

`Y=2/s+(s+3)/(s^2+2s+5)`

`=(3s^2+7s+10)/(s(s^2+2s+5))`, we have:

`X=-3+(s+1)((3s^2+7s+10)/(s(s^2+2s+5)))`

`=(-3s(s^2+2s+5)+(s+1)(3s^2+7s+10))/(s(s^2+2s+5))`

`=(4s^2+2s+10)/(s(s^2+2s+5))`

`=(2(s^2+2s+5)+2s^2-2s)/(s(s^2+2s+5))`

`=2/s+(2s)/((s^2+2s+5))-` `2/((s^2+2s+5))`

`=2/s+2(s+1)/((s+1)^2+2^2)-` `2 2/((s+1)^2+2^2)`

So `x=2+2e^(-t)cos\ 2t-` `2e^(-t)sin\ 2t`

Therefore, in summary:

`x=2(1+e^(-t)cos 2t-e^(-t)sin 2t)`

`y=2+e^(-t)cos 2t+e^(-t)sin 2t`

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