Question

Use the laplace transform to solve the given system of differential equations. dx dt + 3x + dy dt = 1 dx dt − x + dy dt − y = et x(0) = 0, y(0) = 0

Answer 1

Let $X(s)$ and $Y(s)$ denote the Laplace transforms of $x(t)$ and $y(t)$.

Taking the Laplace transform of both sides of both equations, we have

$\dfrac{dx}{dt} + 3x + \dfrac{dy}{dt} = 1 \implies \left(sX(s) - x(0)\right) + 3X(s) + \left(sY(s) - y(0)\right) = \dfrac1s \\\\ \implies (s+3) X(s) + s Y(s) = \dfrac1s$

$\dfrac{dx}{dt} - x + \dfrac{dy}{dt} = e^t \implies \left(sX(s) - x(0)\right) - X(s) + \left(sY(s) - y(0)\right) = \dfrac1{s-1} \\\\ \implies (s-1) X(s) + s Y(s) = \dfrac1{s-1}$

Eliminating $Y(s)$, we get

$\left((s+3) X(s) + s Y(s)\right) - \left((s-1) X(s) + s Y(s)\right) = \dfrac1s - \dfrac1{s-1} \\\\ \implies X(s) = \dfrac14 \left(\dfrac1s - \dfrac1{s-1}\right)$

Take the inverse transform of both sides to solve for $x(t)$.

$\boxed{x(t) = \dfrac14 (1 - e^t)}$

Solve for $Y(s)$.

$(s - 1) X(s) + s Y(s) = \dfrac1{s-1} \implies -\dfrac1{4s} + s Y(s) = \dfrac1{s-1} \\\\ \implies s Y(s) = \dfrac1{s-1} + \dfrac1{4s} \\\\ \implies Y(s) = \dfrac1{s(s-1)} + \dfrac1{4s^2} \\\\ \implies Y(s) = \dfrac1{s-1} - \dfrac1s + \dfrac1{4s^2}$

Taking the inverse transform of both sides, we get

$\boxed{y(t) = e^t - 1 + \dfrac14 t}$