Solve the following sets of equations by the Laplace transform method

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

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

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.

The given pair of linear equation is

y' + 2z = 1

2y -z' =2t

The initial conditions are $y_0=0$ and$z_0=1$

Take Laplace transform on both side of both equations

role="math" localid="1659252471750" $$\begin{gathered} y\text{'}+2z=1 \\ L\left\{y\text{'}\right\}+2L\left\{z\right\}=L\left\{1\right\} \end{gathered}$$

……. (1)

Similarly,

$$\begin{gathered} 2y-z=1 \\ 2L\left\{y\right\}+L\left\{z\right\}=L\left\{2t\right\} \end{gathered}$$ ……. (2)

L35 Laplace transform

$$\begin{gathered} L\left\{y\text{'}\right\}=pL\left\{y\right\}-y_0 \\ L\left\{y\text{'}\right\}+2L\left\{z\right\}=L\left(1\right) \\ pL\left\{y\right\}-y_0+2L\left\{z\right\}=\frac{1}{p} \end{gathered}$$

Solve further

$pY+2Z=\frac{1}{p}$ ……. (3)

Solve further

$$\begin{gathered} 2L\left\{y\right\}-L\left\{z\text{'}\right\}=L\left(2t\right) \\ 2L\left\{y\right\}-\left(pL\left\{z\right\}-z_0\right)=2L\left(t\right) \\ 2Y-pZ=\frac{2}{p^2}-1 \end{gathered}$$ …….. (4)

Multiply equation (3) by (2)

$$\begin{gathered} pY+2Z=\frac{1}{p} \\ 2pY+4Z=\frac{2}{p} \end{gathered}$$

……. (5)

Multiply equation (4) by p

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

Solve further

$2pY=p^{2}Z=\frac{2}{p^2}-p$ ……. (6)

Subtract equation (5) from (6)

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

Take the inverse Laplace transform

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

$$\begin{gathered} L^{-1}\left(\cos at\right)=\frac{p}{\left(p^2+a^2\right)} \\ Z=\frac{p}{\left(p^2+4\right)} \\ z=L^{-1}\left(\frac{p}{p^2+4}\right) \\ z=L^{-1}\left(\frac{p}{p^2+2^2}\right) \\ \mathrm{z}=\cos 2\mathrm{t} \\ \mathrm{Put}\mathrm{Z}=\frac{\mathrm{p}}{{\mathrm{p}}^2+4}\mathrm{in}\mathrm{equation}\left(3\right) \\ \mathrm{pY}+2\mathrm{Z}=\frac{1}{\mathrm{p}} \end{gathered}$$

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

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

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

Solve further for the value of y

y=t-sin 2t

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