Solve the following sets of equations by the Laplace transform method

.$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{z}\mathbf{\text{'}}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}_{\mathbf{0}}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}{\mathbf{\text{'}}}_{\mathbf{0}}\mathbf{=}\mathbf{1} \\ \mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{z}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\mathbf{-}{\mathit{e}}^{\mathbf{t}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{z}}_{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{z}{\mathbf{\text{'}}}_{\mathbf{0}}\mathbf{=}\mathbf{1} \end{gathered}$$

The sets of equation are$$\begin{gathered} y\text{'}\text{'}+z\text{'}\text{'}-z\text{'}=0y_0=0,y{\text{'}}_0=1\mathrm{and} \\ \mathrm{y}\text{'}\text{'}+\mathrm{z}\text{'}\text{'}-2\mathrm{z}\text{'}=1-{\mathrm{e}}^{\mathrm{t}}{\mathrm{z}}_0=1,\mathrm{z}{\text{'}}_0=1 \end{gathered}$$

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.

The given pair of linear equation is

$$\begin{gathered} y\text{'}\text{'}+z\text{'}\text{'}-z\text{'}=0 \\ y\text{'}+z\text{'}-2z=1-e^t \end{gathered}$$

The initial conditions are$y_0=0,1;z_0\text{'}=1$and$z_0\text{'}=1$.

Take Laplace transform on both equations.

$$\begin{gathered} y\text{'}\text{'}+z\text{'}\text{'}-z\text{'}=0 \\ L\left\{y\text{'}\text{'}\right\}+L\left\{z\text{'}\text{'}\right\}-L\left\{z\text{'}\right\}=0 \end{gathered}$$

…….. (1)

Similarly,

$$\begin{gathered} y\text{'}+z\text{'}-2z-1-e^t \\ L\left\{y\text{'}\right\}+L\left\{z\text{'}\right\}-2L\left\{z\text{'}\right\}=L\left\{1\right\}-L\left\{e^t\right\} \end{gathered}$$

……. (2)

Laplace transform

$$\begin{gathered} L\left\{y\text{'}\text{'}\right\}=p^{2}L\left\{y\right\}-p{y}_0-y{\text{'}}_0 \\ L\left\{y\text{'}\right\}=pL\left\{y\right\}-y{\text{'}}_0 \\ L\left\{y\text{'}\text{'}\right\}+L\left\{z\text{'}\right\}-L\left\{z\text{'}\right\}=0 \end{gathered}$$

Solve further

$$\begin{gathered} p^{2}L\left\{y\right\}-p{y}_0-y{\text{'}}_0+p^{2}L\left\{z\right\}-p{z}_0-z{\text{'}}_0-\left(pL\left\{z\right\}-z_0\right)=0 \\ {p}^{2}Y-p\left(0\right)-1+p^{2}Z-p-pZ=0 \\ {p}^{2}Y+\left(p^2-p\right)Z=\left(p+1\right) \\ \mathrm{Similarly} \\ L\left\{y\text{'}\right\}+L\left\{z\text{'}\right\}-2L\left\{z\right\}=L\left\{1\right\}=L\left\{1\right\}-L\left\{e^t\right\} \\ pL\left\{y\right\}-y{\text{'}}_0+pL\left\{z\right\}-z0-2Z=\frac{1}{p}-\frac{1}{p-1} \end{gathered}$$

……. (3)

Further solve

$pY+\left(p-2\right)Z=\frac{1}{p}-\frac{1}{p-1}+1$ ……. (4)

Multiply equation (4) by

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

Subtract the equation (4) from (3)

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

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

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

Substitute $\frac{1}{\left(p-1\right)}$in equation (3)

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

Solve further

$$\begin{gathered} p^{2}Y=1 \\ Y=\frac{1}{p^2} \end{gathered}$$

Use inverse Laplace transform$L^{-1}\left(\frac{1}{p^{k+1}}\right)=t^k$.

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

Thus, the value of given pair of linear equation is y=t and $z=e^t$.