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 )

answer:

3 bags of apples and 2 bags of oranges.

step-by-step explanation:

well what you have to do is divide 84 by 6 so theanswer

step-by-step explanation:

84 divided by 6 = 14

and that lightbulb changing party must be freaking lit!