Question

Solve the following sets of equations by the Laplace transform method

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

Short answer

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

Step-by-step solution

Given information from question

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

Laplace transform

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.

Use Laplace transform on both sides of given 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)

Use Laplace transform

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)

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

Apply inverse Laplace transform

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