Use the Laplace transform to solve the given system of differential equations.

Let X(s) and Y(s) denote the Laplace transforms of x(t) and y(t), respectively. Then taking the transform of both sides of both equations gives

LT[dx/dt + 5x + dy/dt] = LT[1]

==>   s X(s) - x (0) + 5 X(s) + s Y(s) - y (0) = 1/s

==>   s X(s) + 5 X(s) + s Y(s) = 1/s

==>   (s + 5) X(s) + s Y(s) = 1/s

LT[dx/dt - x + dy/dt - y] = LT[exp(t )]

==>   s X(s) - x (0) - X(s) + s Y(s) - y (0) - Y(s) = 1/(s - 1)

==>   s X(s) - X(s) + s Y(s) - Y(s) = 1/(s - 1)

==>   (s - 1) X(s) + (s - 1) Y(s) = 1/(s - 1)

Solve for X(s) and Y(s). Using elimination, you would get

X(s) = (1 - 2s) / (5s (s - 1)²)

Y(s) = (7s - 1) / (5s (s - 1)²)

Now take the inverse transforms of each. Start by getting the partial fraction decompositions:

(1 - 2s) / (5s (s - 1)²) = 1/5 (a/s + b/(s - 1) + c/(s - 1)²)

-2s + 1 = a (s - 1)² + bs (s - 1) + cs

-2s + 1 = (a + b) s ² + (-2a - b + c) s + a

==>   a + b = 0, -2a - b + c = -10, a = 5

==>   a = 1, b = -1, c = -1

==>   X(s) = 1/5 (1/s - 1/(s - 1) - 1/(s - 1)²)

Similarly, you would find

Y(s) = -1/5 (1/s - 1/(s - 1) - 6/(s - 1)²)

Now for the inverse transforms:

LT⁻¹ [1/s] = 1

LT⁻¹ [1/(s - 1)] = exp(t )

LT⁻¹ [1/(s - 1)²] = t exp(t )

Putting everything together, we have

LT⁻¹ [X(s)] = x(t) = 1/5 - 1/5 exp(t ) - 1/5 t exp(t )

and

LT⁻¹ [Y(s)] = y(t) = -1/5 + 1/5 exp(t ) + 6/5 t exp(t )