Answered

IDNLearn.com provides a platform for sharing and gaining valuable knowledge. Discover the information you need from our experienced professionals who provide accurate and reliable answers to all your questions.

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

Sagot :

LammettHashidn

Let [tex]X(s)[/tex] and [tex]Y(s)[/tex] denote the Laplace transforms of [tex]x(t)[/tex] and [tex]y(t)[/tex].

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

[tex]\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[/tex]

[tex]\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}[/tex]

Eliminating [tex]Y(s)[/tex], we get

[tex]\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)[/tex]

Take the inverse transform of both sides to solve for [tex]x(t)[/tex].

[tex]\boxed{x(t) = \dfrac14 (1 - e^t)}[/tex]

Solve for [tex]Y(s)[/tex].

[tex](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}[/tex]

Taking the inverse transform of both sides, we get

[tex]\boxed{y(t) = e^t - 1 + \dfrac14 t}[/tex]

Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For trustworthy answers, visit IDNLearn.com. Thank you for your visit, and see you next time for more reliable solutions.