91Ó°ÊÓ

The sets of equation are $\stackrel{Ë™}{\mathrm{y}},\stackrel{Ë™}{\mathrm{z}}-2y=1y_0=z_0=1$and$z-\stackrel{Ë™}{y}=t$

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

$\stackrel{˙}{y}+\stackrel{˙}{z}-2z=1$ ……. (1)

$z-\stackrel{˙}{y}=t$ ……. (2)

The initial conditions are$y_0=z_0=1$.

Take Laplace transform on both side of equation (1)

$\stackrel{Ë™}{y}+\stackrel{Ë™}{z}-2z=1$

$L\left\{\stackrel{˙}{y}\right\}+L\left\{\stackrel{˙}{z}\right\}-2L\left\{z\right\}=L\left(1\right)$ ……. (3)

Take Laplace transform on both side of equation (2)

$z-\stackrel{Ë™}{y}=t$

$L\left\{z\right\}-L\left\{\stackrel{˙}{y}\right\}=L\left(t\right)$ ……. (4)

Use the formula

$L\left(\stackrel{Ë™}{y}\right)=pL\left(y\right)-y$

Use the above formula in equation (3)

$$\begin{gathered} L\left(\stackrel{Ë™}{y}\right)+L\left(\stackrel{Ë™}{z}\right)-2L\left(z\right)=L\left(1\right) \\ pL\left(y\right)-y\left(0\right)+pL\left(z\right)-z\left(0\right)-2L\left(y\right)=L\left(1\right) \\ pL\left(y\right)-1+pL\left(z\right)-1-2L\left(y\right)=L\left(1\right) \end{gathered}$$

Use the properties of Laplace transformation

$\begin{array}{l}pY-1+pZ-1-2Y=\frac{1}{p}\\ pY+pZ-2-2Y=\frac{1}{p}\end{array}$

Further solve

$(p-2)Y+pZ=\frac{1}{p}+2$ ……. (5)

Use the formula $L\left(\stackrel{Ë™}{y}\right)=pL\left(y\right)-y\left(0\right)$in equation (3)

$$\begin{gathered} L\left(z\right)+L\left(\stackrel{Ë™}{y}\right)=L\left(t\right) \\ L\left(z\right)-pL\left(y\right)+y\left(0\right)=L\left(t\right) \end{gathered}$$

Use the properties of Laplace transformation.

$Z-pY+1=\frac{1}{p^2}$ ……. (6)

Multiplying equation (6) by

$PZ-p^{2}Y=\frac{1}{p^2}-p$ ……. (7)

Subtract equation (7) from (5)

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

The quadratic equation is

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

Use partial fraction

$$\begin{gathered} \frac{2+p}{\left(p-1\right)\left(p+2\right)}=\frac{A}{\left(p-1\right)}+\frac{B}{\left(p+2\right)} \\ =\frac{A\left(p+2\right)+B\left(p-1\right)}{\left(p-1\right)\left(p+2\right)} \\ =\frac{Ap+2A+Bp-B}{\left(p-1\right)\left(p+2\right)} \end{gathered}$$

Compare the numerator

$\begin{array}{l}A+B=1\\ 2A-B=2\end{array}$

Rewrite above equation as

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

Substitute the values$A=1-B$in equation$2A-B=2$.

$$\begin{gathered} 2A-B=2 \\ \begin{array}{l}2(1-B)-B=2\\ 2-2B-B=2\end{array} \end{gathered}$$

Further solve

$\begin{array}{l}B=0\\ A=1\end{array}$

The equation will be

$\begin{array}{l}Y=\frac{1}{\left(p-1\right)}+\frac{0}{(p+2)}\\ Y=\frac{1}{\left(p-1\right)}\end{array}$

Use inverse Laplace transform on both sides

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

Substitute$Y=\frac{2+p}{\left(p-1\right)\left(p+2\right)}$in equation (6).

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

Solve further value of Z

$Z=\frac{1}{p^2}+\frac{1}{(p-1)}$

Take inverse Laplace on both sides.

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

Thus, the value of given pair of linear equation is$y=e^{t}andz=t+e^t$