Solve the following sets of equations by the Laplace transform method

$$\begin{gathered} \dot{\mathbf{y}}\mathbf{+}\mathit{z}\mathbf{=}\mathbf{2}\mathit{c}\mathit{o}\mathit{s}\mathit{t}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{y}}_{\mathbf{0}}\mathbf{=}\mathbf{-}\mathbf{1} \\ \dot{\mathbf{z}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{1}\mathbf{}\mathbf{}\mathbf{}{\mathit{z}}_{\mathbf{0}}\mathbf{=}\mathbf{1} \end{gathered}$$

The sets of equation are$\dot{y}+z=2cost{y}_0=-1$ and$\dot{\mathrm{z}}-y=1z_0=1$

Application of Laplace transform:

It's used to simplify complicated differential equations by using polynomials. It's used to convert derivatives into numerous domain variables, then utilise the Inverse Laplace transform to convert the polynomials back to the differential equation.

Take Laplace transform on both side of equation

$y\text{'}+z=2\cos t$

$L\mathrm{\{}\mathrm{y}\text{'}\}+L\{z\}=2L\{\cos t\}$ ……. (1)

Similarly,

$y\text{'}+z=2\cos t$

$L\mathrm{\{}\mathrm{z}\text{'}\}-L\{y\}=L\{1\}$ ……. (2)

Use Laplace transform

$\begin{array}{l}L\left\{y\text{'}\right\}=pL\left\{y\right\}-y_0\\ L\left\{y\text{'}\right\}+L\left(z\right)=2L\left\{\cos t\right\}\\ pY-y_0+Z=\frac{2p}{p^2+1}\end{array}$

Solve further

$pY+Z=\frac{2p}{p^2+1}-1$ ……. (3)

$\begin{array}{l}L\left\{z\text{'}\right\}-L\left\{y\right\}=L\left\{1\right\}\\ pZ-z0-Y=\frac{1}{p}\\ pZ-1-Y=\frac{1}{p}\end{array}$

Solve further

$pZ-Y=\frac{1}{p}+1$ …….. (4)

Multiply equation (4) by p

$\begin{array}{l}pZ-Y=\frac{1}{p}+1\\ p^{2}Z-pY=\frac{p}{p}+p\\ p^{2}Z-pY=1+p\end{array}$

Subtract equation (3) from (4)

$$\begin{gathered} p^{2}Z-pY+pZ+Z=1+p+\frac{2p}{p^2+1}-1 \\ pZ+Z=p+\frac{2p}{p^2+1} \\ \left(p^2+1\right)Z=p+\frac{2p}{p^2+1} \\ Z=\frac{2p}{\left(p^2+1\right)}+p+\frac{2p}{{\left(p^2+1\right)}^2} \end{gathered}$$

Take inverse Laplace transform $L^{-1}\left(\cos at\right)=\frac{p}{(p^2+a^2)}$, and$L^{-1}\left(t\sin at\right)=\frac{2ap}{{(p^2+a^2)}^2}$

$$\begin{gathered} Z=\frac{p}{\left(p^2+1\right)}+\frac{2p}{{(p^2+1)}^2} \\ z=L^{-1}\left(\frac{p}{\left(p^2+1\right)}\right)+L^{-1}\left(\frac{2p}{{\left(p^2+1\right)}^2}\right) \\ z=\cos t+t\sin t \end{gathered}$$

Put$Z=\frac{p}{\left(p^2+1\right)}+\frac{2p}{(p^2+1^2}$ in equation (4)

$$\begin{gathered} pZ-Y=\frac{1}{p}+1 \\ p\left(\frac{p}{\left(p^2+1\right)}+\frac{2p}{{\left(p^2+1\right)}^2}\right)-y=\frac{1}{p}+1 \\ \frac{p}{\left(p^2+1\right)}+\frac{2p}{{\left(p^2+1\right)}^2}-y=\frac{1}{p}+1 \end{gathered}$$

Solve further

$$\begin{gathered} Y=\frac{p^2}{\left(p^2+1\right)}+\frac{2p^2}{{\left(p^2+1\right)}^2}-\frac{1}{p}-1 \\ Y=\frac{p^2\left(p^2+1\right)+2p^2-{\left(p^2+1\right)}^2}{{\left(p^2+1\right)}^2}-\left(\frac{1}{p}\right) \\ Y=\frac{p^2-1}{\left(p^2+1\right)}-\frac{1}{p} \end{gathered}$$

Use inverse Laplace transform$L^{-1}\left(\frac{1}{p}\right)=1$ and$L^{-1}\left\{t\cos at\right\}=\frac{p^2-a^2}{{\left(p^2+a^2\right)}^2}$

$$\begin{gathered} L^{-1}\left(\frac{1}{p+a}\right)=e^{-at} \\ Y=\frac{p^2-1}{\left(p^2+1\right)}-\frac{1}{p} \\ y=L^{-1}\left(\frac{p^2-1}{\left(p^2+1\right)}\right)-L^{-1}\left(\frac{1}{p}\right) \\ y=t\cos t-1 \end{gathered}$$

Thus, the value of given pair of linear equation is$y=t\cos t-1$ and $z=\cos t+t\sin t$.