91Ó°ÊÓ

The given differential equations are:

$\begin{array}{ll}{x}^{\text{'}\text{'}}+y^{\text{'}}=2;& x\left(0\right)=3\begin{array}{cc},& x^{\text{'}}\left(0\right)=0\end{array}\\ 4x+y^{\text{'}}=6;& y\left(1\right)=4.\end{array}$

Using$y\left(\text{0}\right)=c$we get the system

$\{\begin{array}{ll}{x}^{\text{'}\text{'}}+y^{\text{'}}=2,& x\left(0\right)=3,x^{\text{'}}\left(0\right)=0\\ 4x+y^{\text{'}}=6,& y\left(0\right)=c\end{array}$

Applying the Laplace transform we get

$\{\begin{array}{c}\mathcal{L}\{x^{\text{'}\text{'}}+y^{\text{'}}\}=\mathcal{L}\left\{2\right\}\\ \mathcal{L}\{4x+y^{\text{'}}\}=\mathcal{L}\left\{6\right\}\end{array}$

$\{\begin{array}{l}\mathcal{L}\left\{x^{\text{'}\text{'}}\right\}+\mathcal{L}\left\{y^{\text{'}}\right\}=\frac{2}{s}\\ 4\mathcal{L}\left\{x\right\}+\mathcal{L}\left\{y^{\text{'}}\right\}=\frac{6}{s}\end{array}$

$\{\begin{array}{l}{s}^{2}X\left(s\right)−sx\left(0\right)−x^{\text{'}}\left(0\right)+sY\left(s\right)−y\left(0\right)=\frac{2}{s}\\ 4X\left(s\right)+sY\left(s\right)−y\left(0\right)=\frac{6}{s}\end{array}$

Multiplying the second equation with -1 and adding to the first gives

$\begin{array}{c}{s}^{2}X\left(s\right)−4X\left(s\right)−3s=−\frac{4}{s}\\ (s^2−4)X\left(s\right)=3s−\frac{4}{s}\\ (s^2−4)X\left(s\right)=\frac{3}{s}\\ X\left(s\right)=\frac{3s^2−4}{s(s^2−4)}\end{array}$

Decompose the resulting equation as:

$X\left(s\right)=\frac{1}{s}+\frac{1}{s+2}+\frac{1}{s−2}$

Take inverse Laplace transform on both sides of $X\left(s\right)=\frac{1}{s}+\frac{1}{s+2}+\frac{1}{s−2}$, as:

$x\left(t\right)=1+e^{−2t}+e^{2t}$

Substituting $\frac{1}{s}+\frac{1}{s+2}+\frac{1}{s−2}$ for$X\left(s\right)$into the second equation in the system (1) we get

$\begin{array}{c}4(\frac{1}{s}+\frac{1}{s+2}+\frac{1}{s−2})+sY\left(s\right)−c=\frac{6}{s}\\ sY\left(s\right)−c=\frac{6}{s}−\frac{4}{s}−\frac{4}{s+2}−\frac{4}{s−2}\\ Y\left(s\right)=\frac{2}{s^2}−\frac{4}{s(s+2)}−\frac{4}{s(s−2)}+\frac{c}{s}\\ =\frac{2}{s^2}−\frac{2}{s}+\frac{2}{s+2}+\frac{2}{s}−\frac{2}{s−2}+\frac{c}{s}\\ =\frac{2}{s^2}+\frac{2}{s+2}−\frac{2}{s−2}+\frac{c}{s}\end{array}$

Applying the inverse Laplace, we get

$y\left(t\right)=2t+2e^{−2t}−2e^{2t}+c$

Using that $y\left(1\right)=4$we obtain

$\begin{array}{c}4=2+2e^{−2}−2e^2+c\\ c=2e^2−2e^{−2}+2\end{array}$

Therefore

$y\left(t\right)=2t+2e^{−2t}−2e^{2t}+2e^2−2e^{−2}+2$