Question

Solve the following sets of equations by the Laplace transform method

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

Short answer

The value of given pair of linear equation is y=t and $z=e^t$.

Step-by-step solution

Given information from question

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

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

$$\begin{gathered} y\text{'}\text{'}+z\text{'}\text{'}-z\text{'}=0 \\ y\text{'}+z\text{'}-2z=1-e^t \end{gathered}$$

The initial conditions are$y_0=0,1;z_0\text{'}=1$and$z_0\text{'}=1$.

Take Laplace transform on both equations.

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

…….. (1)

Similarly,

$$\begin{gathered} y\text{'}+z\text{'}-2z-1-e^t \\ L\left\{y\text{'}\right\}+L\left\{z\text{'}\right\}-2L\left\{z\text{'}\right\}=L\left\{1\right\}-L\left\{e^t\right\} \end{gathered}$$

……. (2)

Use L35 Laplace transform

Laplace transform

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

Solve further

$$\begin{gathered} p^{2}L\left\{y\right\}-p{y}_0-y{\text{'}}_0+p^{2}L\left\{z\right\}-p{z}_0-z{\text{'}}_0-\left(pL\left\{z\right\}-z_0\right)=0 \\ {p}^{2}Y-p\left(0\right)-1+p^{2}Z-p-pZ=0 \\ {p}^{2}Y+\left(p^2-p\right)Z=\left(p+1\right) \\ \mathrm{Similarly} \\ L\left\{y\text{'}\right\}+L\left\{z\text{'}\right\}-2L\left\{z\right\}=L\left\{1\right\}=L\left\{1\right\}-L\left\{e^t\right\} \\ pL\left\{y\right\}-y{\text{'}}_0+pL\left\{z\right\}-z0-2Z=\frac{1}{p}-\frac{1}{p-1} \end{gathered}$$

……. (3)

Further solve

$pY+\left(p-2\right)Z=\frac{1}{p}-\frac{1}{p-1}+1$ ……. (4)

Multiply equation (4) by

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

Subtract the equation (4) from (3)

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

Apply inverse Laplace transform

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

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

Substitute $\frac{1}{\left(p-1\right)}$in equation (3)

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

Solve further

$$\begin{gathered} p^{2}Y=1 \\ Y=\frac{1}{p^2} \end{gathered}$$

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

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

Thus, the value of given pair of linear equation is y=t and $z=e^t$.