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:



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

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

Take Laplace Transform:



`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:





Solve for `Y`:




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:



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:


and since


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





`=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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