91Ó°ÊÓ

The sets of equation are $z+2y=0$

$$\begin{gathered} y_0=z_0=0 \\ y\text{'}-2z=2 \end{gathered}$$

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

$$\begin{gathered} z\text{'}+2y=0 \\ y\text{'}-2z=2 \end{gathered}$$

The initial conditions are$y_0=0$, and$z_0=0$.

Take Laplace transform on both equations

$$\begin{gathered} z\text{'}+2y=0 \\ L\left\{z\text{'}\right\}+2L\left\{y\right\}=0 \end{gathered}$$

……. (1)

Similarly,

$$\begin{gathered} y\text{'}-2z=2 \\ L\left\{y\right\}-2L\left\{z\right\}=L\left\{2\right\} \end{gathered}$$

……….. (2)

$L35$Laplace transform

localid="1659409372775" $$\begin{gathered} L\left\{y\text{'}\right\}=pL\left\{y\right\}-y_0 \\ L\left\{z\text{'}\right\}=2L\left\{y\right\}=0 \\ pL\left\{z\right\}-z_0+2L\left\{y\right\}=0 \end{gathered}$$ ……….. (3)

Solve further

$$\begin{gathered} pZ-0+2Y=0 \\ pZ+2Y=0 \\ L\left\{y\text{'}\right\}-2L\left\{z\right\}=L\left(2\right) \\ pL\left\{y\right\}-y_0-2L\left\{z\right\}=2.\frac{1}{p} \\ pY-0-2Z=\frac{2}{p} \\ pY-2Z=\frac{2}{p} \end{gathered}$$……… (4)

Multiply equation (3) by (2)

$$\begin{gathered} pY+2Z=0 \\ 2pY+4Y=0 \end{gathered}$$

Multiply equation (4) by p

$$\begin{gathered} pY-2Z=\frac{2}{p} \\ {p}^{2}Y-2pZ=\frac{2P}{p} \\ {p}^{2}Y-2pZ=2 \end{gathered}$$

Add equation (4) and (3)

$$\begin{gathered} 2pZ+4Y+p^{2}Y-2pZ=0+2 \\ 4Y+p^{2}Y=2 \\ \left(4+p^2\right)Y=2 \\ Y=\frac{2}{\left(4+p^2\right)} \end{gathered}$$

Solve further

$Y=\frac{2}{\left(4+p^2\right)}$

Use inverse Laplace transform

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

Substitute$\frac{2}{\left(2^2+p^2\right)}$forin equation (3)

$$\begin{gathered} pZ+2Y=0 \\ pZ+2\left(\frac{2}{{\left(-2+\right)}^2}\right)=0 \end{gathered}$$

…….. (5)

Solve further

$pZ=\frac{-4}{\left(p^2+2^2\right)}$

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

$$\begin{gathered} \frac{-4}{p\left(p^2+4\right)}=\frac{A}{p}+\frac{Bp+c}{p\left(p^2+4\right)} \\ -4=A\left(p^2+4\right)+\left(Bp+c\right)p \\ -4=A{p}^2+4A+B{p}^2+Cp \end{gathered}$$

Compare coefficients

$$\begin{gathered} 4A=-4 \\ A=-1 \\ C=0 \\ A+B=0 \end{gathered}$$

Solve for B

$$\begin{gathered} -1+B=0 \\ B=1 \end{gathered}$$

Further solve

$$\begin{gathered} \frac{-4}{p\left(p^2+4\right)}=\frac{A}{p}+\frac{Bp+C}{\left(p^2+4\right)} \\ \frac{-4}{p\left(p^2+4\right)}=\frac{-1}{p}+\frac{1\left(p\right)+0}{\left(p^2+4\right)} \\ \frac{-4}{p\left(p^2+4\right)}=\frac{-1}{p}+\frac{p}{\left(p^2+4\right)} \end{gathered}$$

Use above relation in equation (5)

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

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

Thus, the value of given pair of linear equation is $y=\sin 2t$and$z=\cos 2t-1$