91Ó°ÊÓ

The sets of equation are $y\text{'}+z\text{'}-3z=0y_0=y{\text{'}}_0=0$and$y"+z\text{'}=0z_0=\frac{4}{3}$

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 sides

$$\begin{gathered} y\text{'}+z\text{'}-3z=0 \\ L\left\{y\text{'}+z\text{'}-3z\right\}=L\left\{0\right\} \end{gathered}$$

Solve further

$L\left\{y\text{'}\right\}+L\left\{z\text{'}\right\}-3L\left\{z\right\}=0$ ……. (1)

Similarly

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

Solve further

$L\left\{y\text{'}\right\}+L\left\{z\text{'}\right\}=0$ ……. (2)

$L35$Laplace transform

$$\begin{gathered} L\left\{y"\right\}=p^{2}L\left\{y\right\}-p{y}_0-y{\text{'}}_0 \\ L\left\{y\text{'}\right\}=pL\left\{y\right\}-y_0 \\ L\left\{y\text{'}\right\}+L\left\{z\text{'}\right\}-3L\left\{z\right\}=0 \\ pL\left\{y\right\}-y_0+pL\left\{z\right\}-z_0-3L\left\{z\right\}=0 \end{gathered}$$

Solve further

role="math" localid="1659349382487" $$\begin{gathered} pY-0+pZ-\frac{4}{3}-3z=0 \\ pY+\left(p-3\right)Z=\frac{4}{3} \end{gathered}$$ ……. (3)

role="math" localid="1659349479014" $$\begin{gathered} L\left\{y"\right\}+L\left\{z\text{'}\right\}=0 \\ {p}^{2}Y-p{y}_0+y{\text{'}}_0+pZ-z_0=0 \\ {p}^{2}Y-p\left(0\right)-0+pZ-\frac{4}{3}=0 \end{gathered}$$

$p^{2}Y+pZ=\frac{4}{3}$ ..…(4)

Multiply equation (3) by.

$$\begin{gathered} pY+\left(p-3\right)Z=\frac{4}{3} \\ {p}^{2}Y+\left(p-3\right)pZ=\frac{4}{3}p \end{gathered}$$

Subtract equation (4) from (3)

$$\begin{gathered} p^{2}Y+\left(p-3\right)pZ-\left(p^{2}Y+pZ\right)=\frac{4}{3}p-\frac{4}{3} \\ \left(p-3-1\right)pZ=\frac{4}{3}p-\frac{4}{3} \\ \left(p-4\right)pZ=\frac{4}{3}p-\frac{4}{3} \\ z=\frac{4}{3}\frac{\left(p-1\right)}{p\left(p-4\right)} \end{gathered}$$

Solve further

$$\begin{gathered} z=\frac{4}{3}\frac{\left(p-1-3+3\right)}{p\left(p-4\right)} \\ z=\frac{4}{3}\frac{1}{p}+\frac{4-p+p}{p\left(p-4\right)} \\ z=\frac{4}{3}\frac{1}{p}+\frac{p+p}{p\left(p-4\right)}+\frac{p}{p\left(p-4\right)} \\ z=\frac{4}{3}\frac{1}{p}-\frac{1}{p}+\frac{1}{\left(p-4\right)} \end{gathered}$$

Solve further

$$\begin{gathered} z=\left(\frac{4}{3}-1\right)\frac{1}{p}+\frac{1}{\left(p-4\right)} \\ z=\frac{1}{3}.\frac{1}{p}+\frac{1}{\left(p-4\right)} \end{gathered}$$

Take the inverse Laplace transform

$$\begin{gathered} Z=\frac{1}{3}.\frac{1}{p}+\frac{1}{\left(p-4\right)} \\ z=\frac{1}{3}.L^{-1}\left(\frac{1}{p}\right)+L^{-1}\left(\frac{1}{p-4}\right) \end{gathered}$$

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

$$\begin{gathered} z=\frac{1}{3}.L^{-1}\left(\frac{1}{p}\right)+L^{-1}\left(\frac{1}{\left(p+4\right)}\right) \\ z=\frac{1}{3}\left(1\right)+e^{4t} \\ z=\frac{1}{3}+e^{4t} \end{gathered}$$

Put$z=\frac{4}{3}\frac{\left(p-1\right)}{p\left(p-4\right)}$in equation (4)

$$\begin{gathered} p^{2}Y+pZ=\frac{4}{3} \\ {p}^{2}Y+p\left(\frac{4}{3}\frac{\left(p-1\right)}{p\left(p-4\right)}\right)=\frac{4}{3} \\ {p}^{2}Y+p\left(\frac{4}{3}\frac{\left(p-1\right)}{\left(p-4\right)}\right)=\frac{4}{3} \\ {p}^{2}Y=\frac{4}{3}-\frac{4}{3}\frac{\left(p-1\right)}{\left(p-4\right)} \end{gathered}$$

Solve further

$$\begin{gathered} p^{2}Y=\frac{4}{3}\left(1-\frac{\left(p-1\right)}{\left(p-4\right)}\right) \\ {p}^{2}Y=\frac{4}{3}\left(\frac{p-4-p+1}{\left(p-4\right)}\right) \\ {p}^{2}Y=\frac{4}{3}\left(\frac{p-3}{\left(p-4\right)}\right) \\ {p}^{2}Y=\frac{4}{3}\left(\frac{p-3-1+1}{\left(p-4\right)}\right) \end{gathered}$$

Further solve for Y

$$\begin{gathered} Y=\frac{4}{3}\left(\frac{p-4}{\left(p-4\right)p^2}\right)+\frac{4}{3}\left(\frac{1}{\left(p-4\right)p^2}\right) \\ Y=\frac{4}{3}\left(\frac{1}{p^2}\right)+\frac{4}{3}\left(\frac{1}{\left(p-4\right)p^2}\right) \end{gathered}$$

Use partial fraction$\frac{1}{3\left(p-4\right)p^2}$

$\frac{4}{3(p-4)p^2}=\frac{A}{p}+\frac{B}{p^2}+\frac{C}{\left(p-4\right)}$

Compare the coefficients

$$\begin{gathered} A=\frac{1}{4} \\ B=-1 \\ C=\frac{-1}{4} \end{gathered}$$

Solve for Y

$$\begin{gathered} Y=\frac{4}{3}\left(\frac{1}{p^2}\right)+\frac{4}{3}\left(\frac{1}{\left(p-4\right)p^2}\right) \\ Y=\frac{4}{3}\left(\frac{1}{p^2}\right)+\frac{4}{3}\left(\frac{1}{4p}-\frac{1}{p^2}-\frac{1}{4\left(p-4\right)}\right) \\ Y=\left(\frac{4-1}{3}\right)\left(\frac{1}{p^2}\right)+\frac{1}{4}\left(\frac{1}{p}-\frac{1}{\left(p-4\right)}\right) \end{gathered}$$

Solve further

$$\begin{gathered} Y=\left(\frac{3}{3}\right)\left(\frac{1}{p^2}\right)+\frac{1}{4}\left(\frac{1}{p}-\frac{1}{\left(p-4\right)}\right) \\ Y=\left(\frac{3}{3}\right)\left(\frac{1}{p^2}\right)+\frac{1}{4}.\frac{1}{p}-\frac{1}{4}\frac{1}{\left(p-4\right)} \end{gathered}$$

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

$$\begin{gathered} L^{-1}\left(\frac{1}{p+a}\right)=e^{-at} \\ Y=\left(\frac{1}{p^2}\right)+\frac{1}{4}.\frac{1}{p}-\frac{1}{4}.\frac{1}{p-4} \\ Y=L^{-1}\left(\frac{1}{p^2}\right)+\frac{1}{4}{L}^{-1}\left(\frac{1}{p}\right)-\frac{1}{4}{L}^{-1}\left(\frac{1}{p-4}\right) \\ Y=t+\frac{1}{4}\left(1\right)-\frac{1}{4}{e}^{-4t} \end{gathered}$$

Thus, the value of given pair of linear equation is $y=t+\frac{1}{4}\left(1-e^{-4t}\right)$and $z=\frac{1}{3}+e^{4t}$.