Question

Solve the following sets of equations by the Laplace transform method

$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{z}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathit{z}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{y}}_{\mathbf{0}}\mathbf{=}\mathit{y}{\mathbf{\text{'}}}_{\mathbf{0}}\mathbf{=}\mathbf{0} \\ \mathit{y}\mathbf{"}\mathbf{+}\mathit{z}\mathbf{\text{'}}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{z}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}} \end{gathered}$$

Short answer

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

Step-by-step solution

Given information from question

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

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

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)

Use L35 Laplace transform

$L35$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

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

Apply inverse Laplace transform

Take the inverse Laplace transform

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

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

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

Solve further by partial fraction

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 $z=\frac{1}{3}+e^{4t}$.