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

`(dx)/(dt) + 3x + (dy)/(dt) = 1 \implies \left(sX(s) - x(0)\right) + 3X(s) + \left(sY(s) - y(0)\right) = \frac1s \\\\ \implies (s+3) X(s) + s Y(s) = \frac1s`

`(dx)/(dt) - x + (dy)/(dt) = e^t \implies \left(sX(s) - x(0)\right) - X(s) + \left(sY(s) - y(0)\right) = \frac1{s-1} \\\\ \implies (s-1) X(s) + s Y(s) = \frac1{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) = \frac1s - \frac1{s-1} \\\\ \implies X(s) = \frac14 \left(\frac1s - \frac1{s-1}\right)`

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

`\boxed{x(t) = \frac14 (1 - e^t)}`

Solve for
`Y(s)`.

`(s - 1) X(s) + s Y(s) = \frac1{s-1} \implies -\frac1{4s} + s Y(s) = \frac1{s-1} \\\\ \implies s Y(s) = \frac1{s-1} + \frac1{4s} \\\\ \implies Y(s) = \frac1{s(s-1)} + \frac1{4s^2} \\\\ \implies Y(s) = \frac1{s-1} - \frac1s + \frac1{4s^2}`

Taking the inverse transform of both sides, we get

`\boxed{y(t) = e^t - 1 + \frac14 t}`