Question

In Problems 1-12 Use the Laplace transform to solve the given system of differential equations

Short answer

$\begin{array}{l}\mathbf{x}\left(t\right)=\frac{4}{5}{e}^{2t}-\frac{4}{5}\cos \left(t\right)-\frac{8}{5}\sin \left(t\right)\\ \mathbf{y}\left(t\right)=1-2t-t^2+\frac{1}{5}{e}^{2t}-\frac{6}{5}\cos \left(t\right)+\frac{8}{5}\sin \left(t\right)\end{array}$

Step-by-step solution

Definition of Laplace Transform

Let f be a function defined for$t\ge 0$. Then the integral

$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^x{e}^{-xt}f\left(t\right)dt$

is said to be the Laplace transform of f , provided that the integral converges.

Use Laplace Transform

$\begin{array}{l}s\mathbf{X}\left(s\right)-x(0{)}^{\text{'}0}-4\mathbf{X}\left(s\right)+s^3\mathbf{Y}\left(s\right)-s^{2}y\left(0\right)-s{y}^{\text{'}}\left(0\right)-y^{\text{'}\text{'}}\left(0\right)=\frac{6}{1+s^2}\\ \Rightarrow (s-4)\mathbf{X}\left(s\right)+s^3\mathbf{Y}\left(s\right)=\frac{6}{1+s^2}---\left(1\right)\\ (s+2)\mathbf{X}\left(s\right)-2s^3\mathbf{Y}\left(s\right)=0--\left(2\right)\end{array}$

multiply in(1) by 2 and add it to equation (2)

$$\begin{gathered} (3s-6)\mathbf{X}\left(s\right)=\frac{12}{1+s^2}\Rightarrow \mathbf{X}\left(s\right)=\frac{4}{(s-2)\left(1+s^2\right)} \\ u\sin gpartialfractions\frac{4}{(s-2)\left(1+s^2\right)}=\frac{A}{s-2}+\frac{Bs+C}{1+s^2} \\ Bycoverupmethod\mathrm{A}=\frac{4}{5},B=\frac{-4}{5}andC=\frac{-8}{5} \\ hence,\mathbf{X}\left(s\right)=\frac{4/5}{s-2}-\frac{4s/5}{s^2+1}-\frac{8/5}{s^2+1} \end{gathered}$$

Substitute with x in the equation into equation (2).

$\begin{array}{l}\mathbf{Y}\left(s\right)=\frac{2s+4}{s^3(s-2)\left(1+s^2\right)}\\ \Rightarrow \mathbf{Y}\left(s\right)=\frac{1}{s}-\frac{2}{s^2}-\frac{2}{s^3}+\frac{1/5}{s-2}-\frac{6s/5}{s^2+1}+\frac{8/5}{s^2+1}---------\left(5\right)\end{array}$

Apply inverse Laplace transform to the equation

$\begin{array}{l}\mathbf{x}\left(t\right)=\frac{4}{5}{e}^{2t}-\frac{4}{5}\cos \left(t\right)-\frac{8}{5}\sin \left(t\right)\\ \mathbf{y}\left(t\right)=1-2t-t^2+\frac{1}{5}{e}^{2t}-\frac{6}{5}\cos \left(t\right)+\frac{8}{5}\sin \left(t\right)\end{array}$